CHNG I: HM S LNG GIC V PHNG TRNH LNG GIC Ngy son: 23/0/2011 Lp 11CB: 25/01/2011 Ngy dy:Lp 10B://2011 Lp 10D://2011 Tit 1 1. HM S LNG GIC I.Mc tiu: Qua bi hc HV cn nm: 1. V kin thc: - Hiu c khi nim hm s lng gic (ca bin s thc). 2. V k nng: - Xc nh c: Tp xc nh, tp gi tr, tnh cht chn, l, tnh tun hon, chu k , khong ng bin, nghc bin ca cc hm s- V c th cc hm s3. V t duy: - Hiu th no l hm s lng gic . - Xy dng t duy lgc, linh hot . 4. V thi :- Cn thn trong tnh ton v trnh by .- Qua bi hc HV bit c ton hc c ng dng trong thc tin II. Chun b: 1, Gio vin: - Gio n, SGK, STK, phn mu. - Bng ph - Phiu tr li cu hi 2, Hc vin: Hc bi, lm bi tp, c trc bi, n tp kin thc lin quan, my tnh b ti. III.Nhng iu cn lu : - Coi trng vic n l thuyt hc vin h thng c cc kin thc c bn- Ch trng vic rn luyn k nng gii ton. IV.Tin trnh bi hc v cc hot ng: 1. Kim tra bi c: Hot ng 1: Kim tra bi c (10) Hot ng ca GVHot ng ca HV - n tp kin thc c gi tr lg ca cung gc c bit- H1 (sgk) ? a) Y/c HV s dng my tnh (lu my ch rad l) b) S dng ng trn lg biu din cung AM theo bi - Ln bng tr li.- Tt c cc HV cn li tr li vo v nhp. - Nhn xt. 2. Bi mi: Hot ng 2 : Hm s sin v csin (10) Hot ng ca GVHot ng ca HVNi dung - t mi s thc x tng ng im M trn ng trn lg m s cung AM bng x . Nhn xt s im M . Xc nh gi tr sinx, cosx tng ng -Sdngngtrnlgthit lp . -CduynhtimMctung l sinx, honh im M l cosx,I. Cc nh ngha: 1. Hm s sin v csin: a) Hm s sin: (sgk) sin :R R sin ; cos ; tan ; cot y x y x y x y x = = = =sin ; cos ; tan ; cot y x y x y x y x = = = =- Sa cha, un nn cch biu t ca HV. - nh ngha hm s sin nh sgk- Tp xc nh, tp gi tr ca hm s y = sinx - Nhn xt, ghi nhn - Suy ngh tr li- Nhn xt- Ghi nhn kin thc Tp xc nh l R Tp gi tr l | | 1 ; 1 Hot ng 3 : Hm s csin (5) Hot ng ca GVHot ng ca HVNi dung - Xy dng nhhm s sin? - Pht biu nh ngha hm s csin- Tp xc nh, tp gi tr ca hm s y = cos x - Cng c kn Hvy = sin x,y = cos x - Xem sgk, tr li- Nhn xt - Ghi nhn kin thc

b) Hm s csin: (sgk) cos :R R x y x cos = Tp xc nh l R Tp gi tr l| | 1 ; 1 Hot ng 4 : Hm s tang v ctang (7) Hot ng ca GVHot ng ca HVNi dung - nh ngha nh sgk- Tp xc nh?- HV tr li - Nhn xt - Ghi nhn kin thc 2. Hm s tang v ctang: a) Hm s tang: (sgk) K hiu: y = tan x Tp xc nh lD = R\ { Z k k e + ,2tt} Hot ng 5 : Hm s ctang (8) Hot ng ca GVHot ng ca HVNi dung - nh ngha nh sgk-Tp xc nh?- Hot ng 2 sgk ?- Th no l hm s chn, l ? - Chnh sa- Tr li - Nhn xt - Ghi nhn kin thc sin(-x) = - sinx cos(-x) = cosx b) Hm s ctang: (sgk)

K hiu : y = cot x Tp xc nh l D = R \ { Z k k e , t } Nhn xt : sgk V/ Hng dn bi v nh : (5) Cng c : Cu 1: Ni dung c bn c hc? Cu 2: Tp xc nh, tp gi tr cc hm sy = sin x, y = cos x, y = tan x, y = cot x?Dn d: - Xem bi v BT gii.- Lm BT 1, 2/SGK/17 - Xem trc s bin thin v th ca hm s lng gic sin x y x =sin( cos 0)cosxy xx= =cos(sin 0)sinxy xx= = Ngy son: 23/02/2011 Lp 11CB: 25/01/2011 Ngy dy:Lp 10B://2011 Lp 10D://2011 Tit 2 1. HM S LNG GIC (tip theo) I.Mc tiu :Qua bi hc HV cn nm: 1. V kin thc: - Hiu c khi nim hm s lng gic (ca bin s thc). 2. V k nng: - Xc nh c: Tp xc nh, tp gi tr, tnh cht chn, l, tnh tun hon, chu k , khong ng bin, nghc bin ca cc hm s- V c th cc hm s3. V t duy: - Hiu th no l hm s lng gic . - Xy dng t duy lgc, linh hot . 4. V thi : - Cn thn trong tnh ton v trnh by .- Qua bi hc HV bit c ton hc c ng dng trong thc tin II. Chun b: 1. Gio vin: - Gio n, SGK ,STK , phn mu. - Bng ph,- Phiu tr li cu hi 2. Hc vin:- Hc bi, lm bi tp, c trc bi, my tnh b ti. - dng hc tp: Thc k, compa,... III. Nhng iu cn lu : - Coi trng vic n l thuyt hc vin h thng c cc kin thc c bn . - Ch trng vic rn luyn k nng gii tonIV. Tin trnh bi hc v cc hot ng: 1. Kim tra bi c: Hot ng 1: Kim tra bi c (10) HGVHHV -Tp xc nh, tp gi tr, tnh chn, l ca hm s LG? -Treo bng ph kt qu - HV tr li- Tt c cc HV cn li tr li vo v nhp - Nhn xt2. Bi mi: Hot ng 2 :Tnh tun hon ca hm s lng gic(10) Hot ng ca GVHot ng ca HVNi dung - Hot ng 3 sgk ?- Xem sgk, tr liII.Tnh tun hon ca hm s sin ; cos ; tan ; cot y x y x y x y x = = = =sin ; cos ; tan ; cot y x y x y x y x = = = = - Chnh sa hon thin - Nhn xt - Ghi nhn kin thcHm s y = sin x, y = cos x tun hon vi chu k t=2tHm s y = tan x, y = cot x tun hon vi chu k t =t lng gic : (sgk) Hot ng 3 : S bin thin v th ca hm s lng gic (20) Hot ng ca GVHot ng ca HVNi dung -Xt trn on [0,1] nh sgk? - Nu sbt v th ca hm s y = sin x trn cc on [-2t ; -t ]; [2t ; 3t ]; R? - Chnh sa hon thin - Suy ngh tr li- Nhn xt- Ghi nhn kin thc III. S bin thin v th ca hm s lng gic: 1. Hm s y = sinx: BBT V/ Hng dn bi v nh: (5) Cng c: Cu 1: Ni dung c bn c hc? Cu 2: BT6/SGK/18 ?Dn d : - Xem bi v VD gii X, BT 3, 4, 5, 7, 8 (SGK/T17,18) - Xem trc bi lm bi Ngy son: 24/02/2011 Lp 11CB: 25/01/2011 Ngy dy:Lp 10B://2011 Lp 10D://2011Tit 3 1 HM S LNG GIC (tip theo) I. Mc tiu:Qua bi hc HV cn nm: 1. V kin thc: - Hiu c khi nim hm s lng gic (ca bin s thc). 2. V k nng: - Xc nh c: Tp xc nh, tp gi tr, tnh cht chn, l, tnh tun hon, chu k , khong ng bin, nghc bin ca cc hm s- V c th cc hm s3. V t duy: - Hiu th no l hm s lng gic. - Xy dng t duy lgc, linh hot. 4. V thi :- Cn thn trong tnh ton v trnh by . - Qua bi hc HV bit c ton hc c ng dng trong thc tin II. Chun b: 1, Gio vin:- Gio n, SGK ,STK, phn mu. - Bng ph - Phiu tr li cu hi x0ty = sinx0012tsin ; cos ; tan ; cot y x y x y x y x = = = =sin ; cos ; tan ; cot y x y x y x y x = = = =2, Hc vin:- Hc bi, lm bi tp, c trc bi, my tnh b ti. - dng hc tp: Thc k, compa,... III. Nhng iu cn lu - Coi trng vic n l thuyt hc vin h thng c cc kin thc c bn- Ch trng vic rn luyn k nng gii tonIV. Tin trnh bi hc v cc hot ng: 1. Kim tra bi c: Hot ng 1 : Kim tra bi c (10) Hot ng ca GVHot ng ca HV -Tp xc nh, tp gi tr, tnh chn, l v tnh tun hon ca hm s lg? -Treo bng ph kt qu - HV tr li- Tt c cc HV cn li tr li vo v nhp -Nhn xt2. Bi mi: Hot ng 2 : Hm s y = cosx (10) Hot ng ca GVHot ng ca HVNi dung - Xt trn on [0;t ] nh? - Nu sbt v th ca hm sy = sin x trn cc on[-t ; 0 ]; [t ; 2t ]; R ? x e R ta c sin (x + 2t) = cos x tnh tin th y = sin x theo vct) 0 ;2(t = uc th hm s y = cos x - Suy ngh tr li - Nhn xt - Ghi nhn kin thc 2. Hm s y = cosx: BBT Hot ng 3 : Hm s y = tanx (10) Hot ng ca GVHot ng ca HVNi dung - Xt trn na khong ? - S dng tnh cht hm s l c th trn khong - Suy ra th hm s trn D - Chnh sa hon thin - Suy ngh tr li - Nhn xt - Ghi nhn kin thc 3. Hm s y = tanx: BBT x0 2t

y = tan x + 0 Hot ng 4 : Hm s y = cotx (10) Hot ng ca GVHot ng ca HVNi dung - Xt trn na khong? - S dng tnh cht hm s l - Suy ngh tr li - Nhn xt4. Hm s y = cotx: tng tBBT x0 2t

x0ty = cosx11 02t0;2t||

.;2 2t t||

.0;2t||

.c th trn khong - Suy ra th hm s trn D - Chnh sa hon thin - Ghi nhn kin thc + y = cot x 0 V/ Hng dn bi v nh: (5) Cng c: Cu 1: Ni dung c bn c hc? Cu 2: Tp xc nh, tp gi tr cc hm s y = tan x, y = sin x, y = cos x, y = cot x?Dn d: - Xem bi v VD gii, lm BT 3, 4, 5, 7, 8(SGK/T17,18) - Xem trc bi lm biNgy son: 07/02/2011 Lp 11CB: 09/02/2011 Ngy dy:Lp 10B://2011 Lp 10D://2011 Tit 5LUYN TP HM S LNG GIC. I. Mc tiu: Qua bi hc HV cn nm vng : 1. V kin thc : - Tp xc nh ca hm s lng gic - V th ca hm s - Chu k ca hm s lng gic 2. V k nng : -Xc nh c: Tp xc nh, tp gi tr, tnh chn, l, tnh tun hon, chu k , khong ng bin, nghch bin ca cc hm s- V c th cc hm s3. V t duy : - Hiu th no l hm s lng gic. - Xy dng t duy lgc, linh hot. 4. V thi :- Cn thn trong tnh ton v trnh by . - Qua bi hc HV bit c ton hc c ng dng trong thc tin. II. Chun b: 1. Gio vin: - Gio n, SGK ,STK , phn mu. - Bng ph- Phiu tr li cu hi 2. Hc vin:- Hc bi, lm bi tp, c trc bi, my tnh b ti. - dng hc tp: Thc k, compa,... III. Nhng iu cn lu :-Coi trng vic n l thuyt hc vin h thng c cc kin thc c bn-Ch trng vic rn luyn k nng gii tonIV. Tin trnh bi hc v cc hot ng: 1. Kim tra bi c: Hot ng 1 : Kim tra bi c (10) ;2 2t t||

.sin ; cos ; tan ; cot y x y x y x y x = = = =sin ; cos ; tan ; cot y x y x y x y x = = = =Hot ng ca GVHot ng ca HV - n tp kin thc c gi tr lg ca cung gc c bit - HV trnh by ni dung - Nhn xt - Chnh sa hon thin nu c - Ghi nhn kt qu 2. Bi mi: Hot ng 2: BT1/SGK/17 (15) Hot ng ca GVHot ng ca HVNi dung - BT1/sgk/17 ? - Cn c th y = tanxtrn on [-23,tt ] - HV trnh by bi lm - Tt c cc HV cn li tr li vo v nhp - Nhn xt - Chnh sa hon thin nu c - Ghi nhn kt qu 1) BT1/sgk/17 : a) b) c) d) Hot ng 3 : BT2/SGK/17 (15) Hot ng ca GVHot ng ca HVNi dung - BT2/sgk/17 ? - iu kin: sin x=0 - iu kin: 1 cosx > 0 hay Cos x=1. - iu kin:x -R k k e + = ,2 3tt t - iu kin: x +R k k e = ,6tt - Xem BT2/sgk/17. - HV trnh by bi lm.- Tt c cc HV cn li tr li vo v nhp. - Nhn xt. - Chnh sa hon thin nu c. - Ghi nhn kt qu. 2) BT2/sgk/17 : a) D = R \ { kt, keZ}. b) D = R \ { k2t , k eZ } c)D = R \ {65t + kt , k eZ}. d) D = R \ {-6t + kt , k eZ}. V/ Hng dn bi v nh: (5) Cng c : Ni dung c bn c hc? Dn d : Xem bi v VD gii. BT 3, 4, 5, 7, 8 (SGK/T17,18). Xem trc bi lm bi Ngy son: 09/02/2011 Lp 11CB: 11/02/2011 Ngy dy:Lp 10B://2011 Lp 10D://2011 Tit 6LUYN TP HM S LNG GIC (tip theo) { } ;0; x t t e3 5; ;4 4 4xt t t e ` )3; 0; ;2 2 2xt t tt t| | | | | |e |||\ . \ . \ .; 0 ;2 2xt tt| | | |e ||\ . \ .I/ Mc tiu:Qua bi hc HV cn nm vng: 1) V kin thc: -Tp xc nh ca hm s lng gic. -V th ca hm s. -Chu k ca hm s lng gic. 2) V k nng : -Xc nh c: Tp xc nh, tp gi tr, tnh chn, l, tnh tun hon, chu k , khong ng bin, nghch bin ca cc hm s-V c th cc hm s3) V t duy: -Hiu th no l hm s lng gic.-Xy dng t duy lgc, linh hot.4) V thi : - Cn thn trong tnh ton v trnh by . - Qua bi hc HV bit c ton hc c ng dng trong thc tin. II/ Chun b: 1. Gio vin: -Gio n, SGK, STK, phn mu. -Bng ph- Phiu tr li cu hi. 2. Hc vin:- Hc bi, lm bi tp, c trc bi, my tnh b ti. - dng hc tp: Thc k, compa,... III/ Nhng iu cn lu :-Coi trng vic n l thuyt hc vin h thng c cc kin thc c bn-Ch trng vic rn luyn k nng gii ton.IV/ Tin trnh bi hc v cc hot ng: 1. Kim tra bi c: Xen ln trong bi dy 2. Bi mi: Hot ng 1: BT3/SGK/17 (10) Hot ng ca GVHot ng ca HVNi dung - BT3/sgk/17 ? M sin x < 0 R k k k x e + + e ), 2 2 , ( t t t t- Ly i xng qua Ox phn th Hs y = sin x trn cc khong ny. - Xem BT3/sgk/17 - HV trnh by bi lm- Tt c cc HV cn li tr li vo v nhp - Nhn xt - Chnh sa hon thin nu c - Ghi nhn kt qu 3) BT3/sgk/17 : th ca hm s y =| sin x| Hot ng 2 : BT4/SGK/17 (10) Hot ng ca GVHot ng ca HVNi dung - BT4/sgk/17 ? - Hm s y = sin 2x l tun hon chu kt ta xt trn on [0, 2t ]- Ly i xng qua O c - Xem BT4/sgk/17. - HV trnh by bi lm.- Tt c cc HV cn li tr li vo v nhp. - Nhn xt. - Chnh sa hon thin nu c. - Ghi nhn kt qu. 4) BT4/sgk/17 : Sin 2(x + kt )= sin (2x + 2kt )= sinx 2x,k eZ. sin ; cos ; tan ; cot y x y x y x y x = = = =sin ; cos ; tan ; cot y x y x y x y x = = = =sin, sin 0sinsin,sin 0x xxx x> = t Hot ng 3 : BT5/SGK/18 (10) Hot ng ca GVHot ng ca HVNi dung - BT5/sgk/18 ? - Ct th hm sy = cos x bi ng thng y = 21 c giao im x = 3t + k2t , x = - 3t + k2t(keZ). - Xem BT5/sgk/18 - HV trnh by bi lm- Tt c cc HV cn li tr li vo v nhp - Nhn xt - Chnh sa hon thin nu c - Ghi nhn kt qu 5) BT5/sgk/18 :

Hot ng 4: BT 6, 7/SGK/18 (12) Hot ng ca GVHot ng ca HVNi dung - BT6/sgk/18 ? - sin x > 0 ng phn th nm trn trc Ox. - BT7/sgk/18 ? - cos x < 0 ng phn th nm di trc Ox - BT8/sgk/18 ? a) T k: - Xem BT6,7/sgk/18 - HV trnh by bi lm.- Tt c cc HV cn li tr li vo v nhp. - Nhn xt. - Chnh sa hon thin nu c. - Ghi nhn kt qu.b) 6) BT6/sgk/18 : ( k2t , t+ k2t ), k e Z. 7) BT7/sgk/18 : ( 2t +k2t , 23t + k2t ),k e Z. 8) BT8/sgk/18 : a) GTLN y = 3 khi v chi khicosx = 1 x = k2t , k e Z. b) GTLN y = 5 khi v ch khi sinx = -1 x = - 2t +k2t,k e Z. V/ Hng dn bi v nh: (3) Cng c :Ni dung c bn c hc? Dn d : Xem bi v BT gii, xem trc bi phng trnh lng gic c bn Ngy son: 09/02/2011 Lp 11CB: 11/02/2011 Ngy dy:Lp 10B://2011 Lp 10D://2011 Tit 7 2. PHNG TRNH LNG GIC C BN. 0 cos 1 2 cos 2 x x s s s2 cos 1 3hay3 x y + s ssin 1 sin 1 x x > s3 2sin 5 hay 5 x y s s I/ Mc tiu :Qua bi hc HV cn nm: 1) Kin thc: -Bit cc phng trnh lng gic c bn: sin x = a, cos x= a, tan x = a, cot x = a v cng thc tnh nghim . 2) K nng: -Gii c cc phng trnh lng gic c bn . -Bit s dng my tnh b ti tm nghim gn ng ca phng trnh lng gic c bn. 3) T duy:-Xy dng t duy lgic, sng to . -Hiu c cng thc tnh nghim . 4) Thi : - Cn thn trong tnh ton v trnh by. - Qua bi hc HV bit c ton hc c ng dng trong thc tin. II/ Chun b: 1. Gio vin: -Gio n, SGK ,STK , phn mu. -Bng ph - Phiu tr li cu hi. 2. Hc vin:- Hc bi, lm bi tp, c trc bi, my tnh b ti. - dng hc tp: Thc k, compa,... III/ Nhng iu cn lu : -Coi trng vic n l thuyt hc vin h thng c cc kin thc c bn-Ch trng vic rn luyn k nng gii ton. IV/ Tin trnh bi hc v cc hot ng: 1. Kim tra bi c: Hot ng 1 : Kim tra bi c (15) Hot ng ca GVHot ng ca HVNi dung - Tm gi tr ca x sin x = 21?- Cch biu din cung AM trn ng trn lng gic? - H1 sgk? - Ptlg c bn. - Ln bng tr li- Tt c cc HV cn li tr li vo v nhp - Nhn xt. 2. Bi mi: Hot ng 2 : Phng trnh sinx = a (25) Hot ng ca GVHot ng ca HVNi dung - H2 sgk?- Phng trnh sinx = a nhn xt a? |a| > 1 nghim pt ntn? |a|s 1 nghim pt ntn? -1s sin xs 1 ? - Minh ho trn trn lg - Kt lun nghim - Xem H2 sgk. - Trnh by bi gii . - Nhn xt.- Chnh sa hon thin. - Ghi nhn kin thc. 1. Phng trnh sinx = a (sgk) Ch : (sgk) Trng hp c bit : (sgk) x k2x k2sinx = sin= o+ t o = to+ t-Nu =s s a ototsin2 2 th x = arcsinx + k2t , k e Z; x =t- arrcsinx + k2t , k e Z. - VD1 sgk ? N1,2 a) N3,4 b) - H3 sgk? - Trnh by bi gii, nhn xt.- Chnh sa, ghi nhn kin thc. V/Hng dn hc v lm bi nh: (5) Dn d: Xem bi v VD gii, lm BT1->BT4/SGK/28, Ngy son: 14/02/2011Lp 11CB: 16/02/2011 Ngy dy:Lp 10B://2011 Lp 10D://2011 Tit 92. PHNG TRNH LNG GIC C BN (tip theo) I/ Mc tiu bi dy:Qua bi hc HV cn nm vng: 1. V kin thc: - Hiu c khi nim hm s lng gic (ca bin s thc). 2. V k nng: - Xc nh c: Tp xc nh, tp gi tr, tnh cht chn, l, tnh tun hon, chu k , khong ng bin, nghc bin ca cc hm s- V c th cc hm s3) T duy:-Xy dng t duy lgic, sng to . -Hiu c cng thc tnh nghim . 4) Thi : Cn thn trong tnh ton v trnh by .Qua bi hc HV bit c ton hc c ng dng trong thc tin. II/ Chun b: 1. Gio vin: -Gio n, SGK ,STK , phn mu. -Bng ph,- Phiu tr li cu hi. 2. Hc vin:- Hc bi, lm bi tp, c trc bi, n tp kin thc lin quan, my tnh b ti. - dng hc tp: Thc k, compa,... III/ Nhng iu cn lu : -Coi trng vic n l thuyt hc vin h thng c cc kin thc c bn-Ch trng vic rn luyn k nng gii ton.IV/ Tin trnh bi hc v cc hot ng:1. Kim tra bi c: Hot ng 1: Kim tra bi c (15) arcsina o =asincosOM'Msin ; cos ; tan ; cot y x y x y x y x = = = =sin ; cos ; tan ; cot y x y x y x y x = = = =Hot ng ca GVHot ng ca HVNi dung - Gii phng trnh: sin x = 21 - Chnh sa hon thin. - Ln bng tr li . - Tt c cc HV cn li tr li vo v nhp. - Nhn xt . - Ghi nhn kin thc. 2. Bi mi: Hot ng 2 : Phng trnh cosx = a (15) Hot ng ca GVHot ng ca HVNi dung - Yu cu HV c SGK T 21. - GV nhn mnh v a ra cng thc nghim. Phng trnh cosx = a c cc nghim lx =Z k k e + , 2t o . - Nh vy so vi phng trnh sinx = a th phng trnh cosx =a c nghim l i ca nhau. - Xem VD2 sgk.- H 4 sgk? N1,2 a) N3,4 b) - Xem sgk - Ghi nhn kin thc - Trnh by bi gii- Nhn xt- Chnh sa- Ghi nhn kin thc1. Phng trnh cosx = a: (sgk) Phng trnh cosx = coso , viol s cho trc, c cc nghim l : x =Z k k e + , 2t o . Ch : (sgk) Trng hp c bit : (sgk) V/Hng dn hc v lm bi v nh: (15) Cng c: Cu 1: Ni dung c bn c hc? CT nghim? Cu 2: Gii ptlg:Dn d: Xem bi v VD gii. Lm BT1->BT4/SGK/28. Ngy son: 15/02/2011 Lp 11CB: 17/02/2011 Ngy dy:Lp 10B://2011 Lp 10D://2011 Tit 102. PHNG TRNH LNG GIC C BN(tip theo) I/ Mc tiu:Qua bi hc HV cn nm vng: 1. V kin thc: - Hiu c khi nim hm s lng gic (ca bin s thc). 2. V k nng: - Xc nh c: Tp xc nh, tp gi tr, tnh cht chn, l, tnh tun hon, chu k , khong ng bin, nghc bin ca cc hm s- V c th cc hm s3) T duy:-Xy dng t duy lgic, sng to . asincosOM'M1 3 1 3sin ;sin ; ; cos2 2 2 2x x cox x = = = =sin ; cos ; tan ; cot y x y x y x y x = = = =sin ; cos ; tan ; cot y x y x y x y x = = = =-Hiu c cng thc tnh nghim . 4) Thi : Cn thn trong tnh ton v trnh by .Qua bi hc HV bit c ton hc c ng dng trong thc tin. II/ Chun b: 1. Gio vin: -Gio n, SGK, STK,phn mu. -Bng ph, Phiu tr li cu hi. 2. Hc vin:- Hc bi, lm bi tp, c trc bi, my tnh b ti. - dng hc tp: Thc k, compa,... III/ Nhng iu cn lu -Coi trng vic n l thuyt hc vin h thng c cc kin thc c bn -Ch trng vic rn luyn k nng gii ton.IV/ Tin trnh bi hc v cc hot ng: 1. Kim tra bi c: Hot ng 1: Kim tra bi c (10) Hot ng ca GVHot ng ca HV - Gii phng trnh:

- Chnh sa hon thin. - Ln bng tr li.- Tt c cc HV cn li tr li vo v nhp. - Nhn xt . - Ghi nhn kin thc.2. Bi mi: Hot ng 2 : Phng trnh tanx = a(13) Hot ng ca GVHot ng ca HVNi dung - iu kin tanx c ngha? - Trnh by nh sgk. - Minh ho trn th. - Giao im ca ng thng y = a v th hm sy = tanx? - Kt lun nghim.-Nu 2 2t t< < x th x = arctan a + kt , k e Z. - VD3 sgk ?- H5 sgk? N1,2 a) N3,4 b) - Xem H2 sgk. - Trnh by bi gii.- Nhnn xt . - Chnh sa hon thin. - Ghi nhn kin thc. - Trnh by bi gii, nhn xt.- Chnh sa, ghi nhn kin thc.1. Phng trnh tanx = a: (sgk) iu kin:D = R \ {2t+kt ,k eZ}. x = arctan a + kt , k e Z. Ch : (sgk)Phng trnh tan x = tan o , vi ol s cho trc, c cc nghim lx =o+ kt , k e Z. Hot ng 3: Phng trnh cotx = a (12) Hot ng ca GVHot ng ca HVNi dung - iu kin cotx c ngha? - Trnh by nh sgk. - Minh ho trn th. - Giao im ca ng thngy = a v th hm s- Xem H2 sgk. - Trnh by bi gii . -.Nhn xt- Chnh sa hon thin. 1. Phng trnh cotx = a: (sgk) iu kin: D = R \ { kt , k e Z } x = arccotan a + kt , k e Z. 1cos2x=y = cot x ? - Kt lun nghim.- Nu gi tr x ca phng trnh cot x = a tho mn0 < x BT4/SGK/28. -------------------------------------------------------------------------------------------------------------------- Ngy son: 15/02/2011 Lp 11CB: 17/02/2011 Ngy dy:Lp 10B://2011 Lp 10D://2011 Tit 11LUYN TP PHNG TRNH LNG GIC C BN. I/ Mc tiu: Qua bi hc HV cn nm vng: 1. V kin thc: - Hiu c khi nim hm s lng gic (ca bin s thc). 2. V k nng: - Xc nh c: Tp xc nh, tp gi tr, tnh cht chn, l, tnh tun hon, chu k , khong ng bin, nghc bin ca cc hm s- V c th cc hm s3) T duy:-Xy dng t duy lgic, sng to. -Hiu c cng thc tnh nghim. 4) Thi :Cn thn trong tnh ton v trnh by. Qua bi hc HV bit c ton hc c ng dng trong thc tin. II/ Chun b: 1. Gio vin: -Gio n, SGK ,STK , phn mu. -Bng ph,- Phiu tr li cu hi. 2. Hc vin:- Hc bi, lm bi tp, c trc bi, n tp kin thc lin quan, my tnh b ti. - dng hc tp: Thc k, compa,... III/ Nhng iu cn lu :-Coi trng vic n l thuyt hc vin h thng c cc kin thc c bn-Ch trng vic rn luyn k nng gii ton.sin ; cos ; tan ; cot y x y x y x y x = = = =sin ; cos ; tan ; cot y x y x y x y x = = = =IV/ Tin trnh bi hc v cc hot ng: 1. Kim tra bi c: Xen ln trong bi dy 2. Bi mi: Hot ng 1: BT1/SGK/28 (10) Hot ng ca GVHot ng ca HVNi dung - n tp kin thc c gi tr lg ca cung gc c bit. - BT1/sgk/28 ? - Cn c cng thc nghim gii. - HV trnh by bi lm. - Tt c cc HV cn li tr li vo v nhp. - Nhn xt. - Chnh sa hon thin nu c. - Ghi nhn kt qu. 1) BT1/sgk/17 : a) sin (x + 2) = 1/3 V x + 2 = arcsin 1/3 nnsin (x+2) = 1/3 c nghim lx = arcsin 1/3 - 2+ k2t , k eZ. x =t - arcsin 1/3- 2 + k2t , ke Z. c) sin (3 32 tx) = 0 3 32 tx = kt , k e Z. x =2t + 3/2kt , k e Z. d) sin (2x + 23) 200 =V - 23= sin (-600) nnsin (2x + 200)= sin (-600) c nghim l ; x =- 400 + k1800; x = 1100 +k1800, ke Z Hot ng 2: BT2/SGK/28 (10) Hot ng ca GVHot ng ca HVNi dung - BT2/sgk/28 ? - Gii pt: sin3x = sinx - Chnh sa hon thin. - Xem BT2/sgk/28. - HV trnh by bi lm.- Tt c cc HV cn li tr li vo v nhp. - Nhn xt. - Chnh sa hon thin nu c. - Ghi nhn kt qu. Phng trnh c hai nghim l : x = kt , k e Z; x = 4t+ k2t, k e Z. Hot ng 3: BT3/SGK/28 (15) Hot ng ca GVHot ng ca HVNi dung - BT3/sgk/28 ? - Cn c cng thc nghim gii. - Xem BT3/sgk/28. - HV trnh by bi lm. - Tt c cc HV cn li tr li vo v nhp. - Nhn xt. - Chnh sa hon thin nu c. - Ghi nhn kt qu. 3) BT3/sgk/28 :a) cos ( x- 1) = 2/3V x 1 = arccos 2/3 nncos(x- 1) = 2/3 c nghim lx =arccos 2/3 +1 + k2t , k e Z. c) cos (21)4 23 = t x V = cos 32t nncos ( )4 23 tx = cos 32tc nghim l : Z k k xZ k k xe + =e + =,34185; ,341811t tt t. d) cos2x = 22 cos 1 x + v Cos2x = 22 cos 1 x + nncos22x =1/4 22 cos 1 x + = 1/4 cos2x = -1/2 cos2x = cos 32t x = 3t + k2t , k e Z. V/Hng dn hc v lm bi nh: (10')Cng c : Ni dung c bn c hc? Dn d : Xem li bi vBT gii --------------------------------------------------------------------------------------------------------------------- Ngy son: 16/02/2011 Lp 11CB: 18/02/2011 Ngy dy:Lp 10B://2011 Lp 10D://2011 Tit 13 LUYN TP PHNG TRNH LNG GIC C BN ( tip theo). I/ Mc tiu:Qua bi hc HV cn nm vng: 1. V kin thc: - Hiu c khi nim hm s lng gic (ca bin s thc). 2. V k nng: - Xc nh c: Tp xc nh, tp gi tr, tnh cht chn, l, tnh tun hon, chu k , khong ng bin, nghc bin ca cc hm ssin ; cos ; tan ; cot y x y x y x y x = = = =- V c th cc hm s3) T duy:-Xy dng t duy lgic, sng to. -Hiu c cng thc tnh nghim. 4) Thi :Cn thn trong tnh ton v trnh by. Qua bi hc HV bit c ton hc c ng dng trong thc tin. II/Chun b: 1. Gio vin: -Gio n, SGK,STK, phn mu. -Bng ph, Phiu tr li cu hi. 2. Hc vin:- Hc bi, lm bi tp, c trc bi, my tnh b ti. - dng hc tp: Thc k, compa,... III/ Nhng iu cn lu : - Coi trng vic n l thuyt hc vin h thng c cc kin thc c bn. - Ch trng vic rn luyn k nng gii ton.IV/ Tin trnh bi hc v cc hot ng: 1. Kim tra bi c: Xen ln trong bi dy 2. Bi mi: Hot ng1 (12'): BT4/SGK/29 Hot ng ca GVHot ng ca HVNi dung - BT4/sgk/29 ? - Tm iu kin ri gii? - iu kin: sin 2x= 1 Z k k x e + = ,4tt - Gii pt: cos 2x = 0 - KL nghim? LoiZ k k x e + = ,4tt do iu kin - Xem BT4/sgk/29. - HV trnh by bi lm.- Tt c cc HV cn li tr li vo v nhp. - Nhn xt. - Chnh sa hon thin nu c. - Ghi nhn kt qu. 4) BT4/sgk/29 : Cos 2x = 0) ( ,4) (44) (222222Z k k xZ kk xk xZ kk xk xe + = e

+ =+ =e

+ =+ =tttttttttt Hot ng 2 (18'): BT5/SGK/29 Hot ng ca GVHot ng ca HVNi dung - BT5/sgk/29 ? - Cn c cng thc nghim gii. - iu kin c) v d) ? - Xem BT5/sgk/29. - HV trnh by bi lm. - Tt c cc HV cn li tr li vo v nhp. - Nhn xt. 5) BT5/sgk/29 : a) tan ( x 150) = 33 v 33= tan 300nntan ( x 150) = tan 300 c nghim l : x 150 = 300 + k1800, k e Z x = 450 + k1800, k e Z b) cot (3x 1) = -3v - 3 = cot (- 300) nn cot (3x 1) = cot sin ; cos ; tan ; cot y x y x y x y x = = = =- Chnh sa hon thin nu c. - Ghi nhn kt qu.c) k : cos x=0. d) k : sin x=0. (-300) c nghim l : 3x 1 = - 300 + k1800, k e Z x = - 100 + 1/3 + k600, k e Z. c ) cos 2x tan x = 0

==0 tan0 2 cosxxe + =

=e + =Z k k xk xZ k k x,2,2 4tt tt t Vy x =Z k k x k e = + , ,.2 4tt t d)sin 3x cot x = 0 e =

e + =e =

==Z k k xZ k k xZ k k xxx,,2,30 cot0 3 sintttt VyZ k k xZ m m k k xe + =e = =,2; , 3 ,3ttt Hot ng 3 (10'): BT6/SGK/29 Hot ng ca GVHot ng ca HVNi dung - BT6/sgk/29 ? - Tm iu kin? - Gii pt:tan (4t- x) = tan 2x th Z m m k k x e = + = , 1 3 ,3 12t t - Xem BT6,7/sgk/29. - HV trnh by bi lm.- Tt c tr li vo v nhp, ghi nhn kt qu. 6) BT6/sgk/29 : K : V/Hng dn hc v lm bi nh:(5') Cng c : Ni dung c bn c hc? Dn d: Xem li bi v BT gii .

-------------------------------------------------------------------------------------------------------------------- Ngy son: 17/02/2011 Lp 11CB: 19/02/2011 Ngy dy:Lp 10B://2011 Lp 10D://2011 cos 2 0, cos 04x xt | |= = |\ .Tit 14 LUYN TP PHNG TRNH LNG GIC C BN. (tip theo) I. Mc tiu : Qua bi hc HV cn nm: 1. V kin thc: - Hiu c khi nim hm s lng gic (ca bin s thc). 2. V k nng: - Xc nh c: Tp xc nh, tp gi tr, tnh cht chn, l, tnh tun hon, chu k , khong ng bin, nghc bin ca cc hm s- V c th cc hm s3. V t duy : - Xy dng t duy l gc, sng to. - Hiu c cng thc tnh nghim. 4. V thi : - Cn thn trong tnh ton v bi hc. - Bit ng dng ton hc trong thc tin. II. Chun b phng tin dy hc : 1. Hc vin: - HV lm bi tp trc khi n lp, n tp l thuyt. - dng hc tp phi y . 2. Gio vin: - Gio n, bng ph tng hp l thuyt cng thc nghim ca cc phng trnh lng gic. - Compa, thc, phn mu. III. Tin trnh dy hc: 1. Kim tra bi c : Kt hp trong tit luyn tp. 2. Bi mi:Hot ng ca GVHot ng ca HVNi dung H 1 : (18) Bi 7 : (SGK T29) i vi dng bi ny mun gii cc em phi nh n cc cng thc cung ph nhau. Cc ng thc lng gic v mi lin h gia hm tan v hm cot. Yu cu hv nhc li. Vi s chun b bi nh cng s gip ca Gv, gi 2 HV ln lm bi, cc em cn li t lm vo v hoc lm li bng cch khc. Cng thc cc cung ph nhau l: sin (2t- x ) = cosx;cos (2t - x) = sinx tan (2t - x) = cotx;cot (2t - x) = tan x.tan x cot x = 1. - Xem BT7/sgk/29. - HV trnh by bi lm. Bi 7 ( sgk t 29) a) sin 3x cos 5x = 0 sin 3x = cos 5x sin 3x = sin (2t - 5x) c nghim l : 3x = 2t - 5x + k2 Z k e , tv3x =t - 2t + 5x + k2 Z k e , tVy x =Z k k e + ,4 16t t

v x = - Z k k e ,4tt b) iu kincos 3x0 cos , 0 = = x tan 3x tan x = 1tan 3x = x tan1 tan 3x = cot xsin ; cos ; tan ; cot y x y x y x y x = = = =sin ; cos ; tan ; cot y x y x y x y x = = = = Gv nhn xt v nh gi bi lm. - Tt c tr li vo v nhp, ghi nhn kt qu. tan 3x = tan (2t- x) x =Z k k e + ,4 8t t. H 2 :(25) Bi tp : Gii cc phng trnh sau : a) (1 + 2cos x)(3 cos x) = 0 b) sin (x 600) = 1/2 c)(3tan x +3 )(2sin x 1) = 0 d) sin 2x cot x = 0 Chia lp thnh 4 nhm v cho cc em trao i nhm a ra phng n ng v c ngi i din ln trnh by Gv hng dn thm cho cc em. Nhn mnh vi mi dng bi c cch gii nh th no. nh gi bi lm cho cc em. Cc nhm trao i v c ngi i din ln trnh by. - Xem BT. - HV trnh by bi lm. - Tt c tr li vo v nhp, ghi nhn kt qu. Bi tp : Gi cc phngtrnh sau. a) 1 + 2cos x = 0 hoc 3 cos x = 0 cos x = - 1/2hoc cos x = 3 Ta c: Cos x = - 1/2 x =Z k k e + , 232tt

Cos x = 3 v nghim. b) sin ( x 600) = 1/2x = 900 +k1800, k e Z. vx = 2100 +k1800, k e Z. c)(3 tan x +3 )(2sin x 1) = 0 iu kin : xZ k k e + = ,2tt ta c : (3 tan x +3 )(2sin x 1) = 0 3tan3tan 3 032sin 1 0 1sin262 ( )6526xxxxx kx k k Zx ktttttt

=

+ =

= =

= +

= + e

= +

Vy tp nghim ca phng trnh cho l : ) ( ; 265; 26;6Z k k xZ k k x k xe + =e + = + =tttttt d) sin 2x cot x = 0iu kin : xZ k k e = , tta c : ) (20 cot0 2 sinZ kk xk xxxe

+ ==

==ttt Kt hp vi iu kin th phng trnh cho c nghim l : x =Z k k e + ,2tt. V/Hng dn hc v lm bi nh:(5') Cng c : Ni dung c bn c hc? Dn d: - V nh n tp l thuyt v n tp cc dng bi tp cha. - c bi mi. Ngy son: 17/02/2011 Lp 11CB: 19/02/2011 Ngy dy:Lp 10B://2011 Lp 10D://2011 Tit 15 3. MT S PHNG TRNH THNG GP. I/ Mc tiu: Qua bi hc HV cn nm: 1) Kin thc: -Bit c dng v cch gii phng trnh: bc nht, bc hai i vi mt hm s lng gic; phng trnh asin x + bcosx = c. 2) K nng: -Gii c phng trnh thuc cc dng nu trn. 3) T duy:Nm c dng v cch gii cc phng trnh n gin. 4) Thi : Cn thn trong tnh ton v trnh by. Qua bi hc HV bit c ton hc c ng dng trong thc tin. II/ Chun b: 1. Gio vin: -Gio n, SGK, STK, phn mu. -Bng ph, - Phiu tr li cu hi. 2, Hc vin:-Hc bi, lm bi tp, c trc bi, n tp kin thc lin quan, my tnh b ti. - dng hc tp: Thc k, compa,... III/ Nhng iu cn lu :-Coi trng vic n l thuyt hc vin h thng c cc kin thc c bn. -Ch trng vic rn luyn k nng gii ton.IV/ Tin trnh bi hc v cc hot ng:1. Kim tra bi c: Xem ln trong bi dy 2. Bi mi: Hot ng 1 (10'): nh ngha phng trnh bc nht i vi mt hm s lng gic. Hot ng ca GVHot ng ca HVNi dung - N pt bc nht? n pt bc nht i vi mt hm s lng gic ? - Cho vd ? - H1 sgk? - Chnh sa hon thin.- N, nhn xt, ghi nhn.- Nu v d - H 1 sgk. - Trnh by bi gii.- Nhn xt.- Chnh sa hon thin. I. Phng trnh bc nht i vi mt hm s lng gic:1) nh ngha : (sgk) VD : (sgk) 2sin 2 03 tan 1 0xx =+ =- Ghi nhn kin thc. Hot ng 2 (15') : Cch gii Hot ng ca GVHot ng ca HVNi dung - Cch gii? VD2 sgk ?VD : Gii cc phng trnh sau : a) 3cos x + 6 = 0 b)2 cot x 1 = 0 c) 2sin x -1 = 0 Hng dnHV lm v a ra phng php gii v gi 3 HV ln bng lm. - Nghe, suy ngh. - Tr li.- Ghi nhn kin thc. - c VD2 sgk. - Trnh by bi gii. - Nhn xt.- Chnh sa hon thin. - Ghi nhn kin thc.2) Cch gii: (sgk) VD2 : (SGK) VD: Gii cc phng trnh sau :a) 3cos x + 6 = 03cosx = -6 cos x = -2 V 2 < -1 nn phng trnh v nghim. b) iu kin sin x=0 2 cot x 1 = 02 cot x = 1 cot x = 21 V 21 = cot (450) nn cot x = cot 450 c nghim lx = 450 + k1800,k e Z ( tho mn iu kin) c) 2sin x 1 = 02sin x = 1 sin x = 21 V 21 = sin 300 nn sin x = sin 300 c nghim l : x = 300 + k 3600 vx = 1500 + k3600, k e Z Hot ng 3 (15') : Phng trnh a v bc nht i vi mt hm s lng gic. Hot ng ca GVHot ng ca HVNi dung - VD3 sgk ? V d: Gii cc phng trnh sau : a) 4sinx 2 sin 2x = 0 b) 4sin x cos x cos 2x = -1 Hng dnHV lm v a ra phng php gii v gi 2 HV ln bng lm. - Xem sgk, tr li. - Nhn xt. - Ghi nhn kin thc. - Suy ngh, trnh by bi gii. 3)Phngtrnhavbc nht i vi mt hm s lng gic: (sgk) VD3 : SGK V d: Gii cc phng trnh sau : a) 4 sin x 2 sin 2x = 0 4 sin x 2. 2 sin xcos x = 0 4 sin x ( 1 cos x) = 0 ===== =tt21 cos0 sin0 cos 10 sin 4k xk xxxxx

k e Z b) 4 sin x cos x cos 2x = -1 2 sin 2x cos 2x = -1 sin 4x = -1 4x = -2t+ k2 Z k e , tx = - Z k k e + ,2 8t t V/Hng dn hc v lm bi nh (5'): Cng c : Cu 1: Ni dung c bn c hc? Cu 2: Gii phng trnh: 2cos x + 1 = 0 Dn d : Xem bi v VD gii, n cc cng thc lng gic LmBT1/SGK/36, xem trc bi . Ngy son: 20/02/2011 Lp 11CB: 22/02/2011 Ngy dy:Lp 10B://2011 Lp 10D://2011 Tit 17 3 MT S PHNG TRNH THNG GP (tip theo). I/ Mc tiu: Qua bi hc HV cn nm: 1) Kin thc: - Bit c dng v cch gii phng trnh: bc nht, bc hai i vi mt hm s lng gic, phng trnh asinx + bcosx = c. 2) K nng: - Gii c phng trnh cc dng trn. 3) T duy:Nm c dng v cch gii cc phng trnh n gin. 4) Thi : Cn thn trong tnh ton v trnh by . - Qua bi hc HV bit c ton hc c ng dng trong thc tin. II/ Thit b dy hc: 1. Gio vin: - Gio n, SGK , STK, phn mu. - Bng ph,- Phiu tr li cu hi. 2. Hc vin:- Hc bi, lm bi tp, c trc bi, n tp kin thc lin quan, my tnh b ti. - dng hc tp: Thc k, compa,... III/ Nhng iu cn lu : -Coi trng vic n l thuyt hc vin h thng c cc kin thc c bn. -Ch trng vic rn luyn k nng gii ton.IV/ Tin trnh bi dy: 1. Kim tra bi c: Xen ln trong bi dy 2. Bi mi: Hot ng 1 (8'): nh ngha phng trnh bc hai i vi mt hm s lng gic Hot ng ca GVHot ng ca HVNi dung - N pt bc hai? n pt bc hai i vi mt hm s lng gic ? - Cho vd ? - H2 sgk? - Chnh sa hon thin. -N,nhnxt,ghinhnkt qu.- Nu v d. - H 2 sgk. - Trnh by bi gii.- Nhn xt.- Chnh sa hon thin. - Ghi nhn kin thc.II. Phng trnh bc hai i vi mt hm s lng gic:1) nh ngha : (sgk) VD : (sgk) Hot ng 2 (8'): Cch gii Hot ng ca GVHot ng ca HVNi dung - Cch gii? -VD5 sgk ? - Nghe, suy ngh. - Tr li.- Ghi nhn kin thc. - c VD5 sgk. - Trnh by bi gii.- Nhn xt. - Chnh sa hon thin. - Ghi nhn kin thc.2) Cch gii: (sgk) Hot ng3 (12'): Phng trnh a v bc hai i vi mt hm s lng gic Hot ng ca GVHot ng ca HVNi dung - H3 sgk?- Cc cng thc lg? - VD6 sgk ? - VD7 sgk ? - VD8 sgk ?- Xem sgk, tr li. - Nhn xt. - Ghi nhn kt qu.- Trnh by bi gii.- Nhn xt.- Chnh sa hon thin. - Ghi nhn kin thc.3) Phng trnh a v bc haiivimthms lng gic: (sgk) 222sin 3sin 2 03cot 5cot 7 0x xx x+ = =Hot ng 4 (7'): Cng thc bin i asinx + bcosx Hot ng ca GVHot ng ca HVNi dung - Bin i : vi -Gii thch s xut hin - S dng cng thc cng bin i - Cng thc cng. - Nhn xt. - c sch nm qui trnh bin i. - Chnh sa hon thin. - Ghi nhn kin thc, III. Phng trnh bc nht i vi sinx v cosx:1) Cng thc bin i : (sgk) Hot ng 5 (8'): Phng trnh dng: asinx + bcosx = c Hot ng ca GVHot ng ca HVNi dung - Xt phng trnh: - C th a v ptlgcb? - VD9 sgk ?- Ta c: - Nghe, suy ngh. - Tr li.- Ghi nhn kin thc. - c VD9 sgk, - Trnh by bi gii,- Nhn xt,- Chnh sa hon thin. - Ghi nhn kin thc. 2) Phng trnh dngasin x + b cos x = c : (sgk) Cch gii : (SGK) VD : (sgk) V/Hng d hc v lm bi nh: (2')Cng c: Cu 1: Ni dung c bn c hc? Cu 2: Gii phng trnh lng gic sin x + 2 cos x = 1 ?Dn d: Xem bi v VD gii, lm BT1->BT6/SGK/37

Ngy son: 20/02/2011 Lp 11CB: 22/02/2011 Ngy dy:Lp 10B://2011 Lp 10D://2011 Tit 18LUYN TP 3 ( )2 2sin cossina x b xa b x o+= + +2 2cosaa bo =+2 2sinba bo =+2 2a b +( )2 2sin cos0a x b x ca b+ =+ =sin 3 cos 2sin3x x xt | |+ = + |\ .sin 3 cos 12sin 13x xxt+ =| | + = |\ . I/ Mc tiu:Qua bi hc HV cn nm vng: 1) Kin thc: - Cch gii phng trnh: bc nht, bc hai i vi mt hm s lng gic, phng trnh asinx + bcosx = c. 2) K nng: - Gii c phng trnh cc dng trn . - S dng my tnh b ti gii pt n gin. 3) T duy: Nm c dng v cch gii cc phng trnh n gin. 4) Thi : Cn thn trong tnh ton v trnh by . Qua bi hc HV bit c ton hc c ng dng trong thc tin. II/ Chun b: 1. Gio vin: - Gio n, SGK ,STK , phn mu. - Bng ph, - Phiu tr li cu hi. 2. Hc vin:- Hc bi, lm bi tp, c trc bi, n tp kin thc lin quan, my tnh b ti. - dng hc tp: Thc k, compa,... III/ Nhng iu cn lu : - Coi trng vic n l thuyt hc vin h thng c cc kin thc c bn. - Ch trng vic rn luyn k nng gii ton.IV/ Tin trnh bi hc v cc hot ng: 1. Kim tra bi c: Xen ln trong bi dy 2. Bi mi: Hot ng 1 (12'): BT1/SGK/36 Hot ng ca GVHot ng ca HVNi dung - BT1/sgk/36 ? - a v phng trnh lng gic c bn gii. - HV trnh by bi lm. - Tt c cc HV cn li tr li vo v nhp. - Nhn xt. - Chnh sa hon thin nu c. - Ghi nhn kt qu. 1) BT1/sgk/36 : Sin2x sin x = 0 sin x ( sin x 1) = 0 ) (221 sin0 sinZ kk xk xxxe+ ====ttt KL: phng trnh c hai nghimx = k Z k e , tvx =2t +k2 Z k e , t Hot ng 2 (15'): BT2/SGK/36 Hot ng ca GVHot ng ca HVNi dung - BT2/sgk/36 ? - Gii pt: - Chnh sa hon thin nu - Xem BT2/sgk/36. - HV trnh by bi lm.- Tt c cc HV cn li tr li vo v nhp. - Nhn xt. - Ghi nhn kt qu. 2) BT2/sgk/28 : a) 2 cos2x 3 cos x + 1 = 0 ) (23221cos1 cosZ kk xk xxxe

+ ==

==tttb) 2 sin2x +2 sin 4x = 0 2)2cos 3cos 1 0)2sin 2 2 sin 4 0a x xb x x + =+ =c. 2sin 2x + 2 2 sin 2x cos 2x = 0 2sin 2x ( 1 +2 cos 2x ) = 0 ) (832212 cos0 2 sinZ kk xkxxxe

+ ==

==ttt Hot ng 3 (15'): BT3/SGK/37 Hot ng ca GVHot ng ca HVNi dung - BT3/sgk/37 ? - a v ptlgcb gii. a) a v thun cos b) a v thun sin - t n ph ntn? d) t t = tanx - Xem BT3/sgk/37. - HV trnh by bi lm. - Tt c tr li vo v nhp. - Nhn xt. - Chnh sa hon thin nu c. - Ghi nhn kt qu.3) BT3/sgk/37 :a) sin22x - 2 cos 2x + 2 = 0 (1) ta c : sin22x = 1 cos2 2x th vo (1) ta c:1 cos2 2x

-2 cos 2x + 2 = 0 cos2 2x + 3 cos 2x - 1 = 0 (1) t cos 2x = t , vi iu kin -1s ts 1 Pt (1) t2 + 3t 1 = 0 Gii phng trnh trn c nghimt1 = 212 3 + , t2 = 212 3 Kt hp vi iu kin th nghim t2 b loi, phng trnh c nghim lt = 212 3 + Vi t = 212 3 + ta c:cos 2x= 212 3 + x=2acrcos212 3 + +k4 Z k e , t . d)Ta c: tan x cot x = 1 tan x = 1\ cot x, (cot x= 0) th vo phng trnh cho ta c: 1\ cot x 2 cot x + 1 = 0 2 cot2x cot x 1 = 0 t cot x = t, Ta c 2 t2 t 1 = 0 c nghim l: t1 = 1, t2 = - 1\2 Vi t = 1 ta c: cot x = 1x =Z k k e + ,4tt Vi t = -1\2 ta c: cot x = - 1\2 x = acrcot (-1\2) + k Z k e , t . V/Hng d hc v lm bi nh: (3') Cng c :Cu 1: Ni dung c bn c hc? Cu 2: Cng thc lng gic?Dn d : Xem li bi gii, lm BT4->BT6/SGK/37 --------------------------------------------------------------------------------Ngy son: 21/02/2011 Lp 11CB: 23/02/2011 Ngy dy:Lp 10B://2011 Lp 10D://2011 Tit 19 LUYN TP 3 I/ Mc tiu: Qua bi hc HV cn nm vng: 1) Kin thc: - Cch gii phng trnh: bc nht, bc hai i vi mt hm s lng gic, phng trnh asinx + bcosx = c. 2) K nng: - Gii c phng trnh cc dng trn . - S dng my tnh b ti gii pt n gin . 3) T duy: Nm c dng v cch gii cc phng trnh n gin . 4) Thi : Cn thn trong tnh ton v trnh by. - Qua bi hc HV bit c ton hc c ng dng trong thc tin. II/ Chun b: 1. Gio vin: - Gio n, SGK, STK\, phn mu. - Bng ph, phiu tr li cu hi. 2. Hc vin:- Hc bi, lm bi tp, c trc bi, n tp kin thc lin quan, my tnh b ti. - dng hc tp: Thc k, compa,... III/ Nhng iu cn lu : - Coi trng vic n l thuyt hc vin h thng c cc kin thc c bn . - Ch trng vic rn luyn k nng gii ton. IV/ Tin trnh bi hc v cc hot ng: 1. Kim tra bi c: Xen ln trong bi dy 2. Bi mi: Hot ng 1 (20'): BT4/SGK/37 Hot ng ca GVHot ng ca HVNi dung - BT4/sgk/37 ? - Tm xem cosx = 0 nghim ng pt khng? - Chia hai v pt cho cos2x ? - Gii pt ntn? - KL nghim? - Xem BT4/sgk/37. - HV trnh by bi lm. - Tt c cc HV cn li tr li vo v nhp. - Nhn xt. 4) BT4/sgk/37 : a) cos x= 0 khng l nghim ca phng trnh. Chia c hai v phng trnh cho cos 2 x ta c phng trnh tng ng l : 2tan2x + tan x 3 = 0 tan 14( )33 tanarctan( )22x x kk Zxx kttt

= = +

e

=

= +

Tng t cc em v hon thnh phn c v d. - Chnh sa hon thin nu c. - Ghi nhn kt qu. b) cos x = 0 khng l nghim. Chia c hai v phng trnh cho cos2x ta c: 3tan2x 4tanx + 5 = x2cos2 Ta c:x2cos1 = 1 + tan2x Th vo phng trnh ta c: 3tan2x 4tanx + 5=2(1 + tan2x) tan2x 4tan x + 3 = 0 tan 1( )4tan 3arctan3x x kk Zxx kttt

= = +

e

== + Hot ng 2 (20'): BT5/SGK/37 Hot ng ca GVHot ng ca HVNi dung - BT5/sgk/37 ? - Bin i v ptlgcb gii ? Tng t v cc em tm cc bi tng t v hon thnh phn d ca bi 5/sgk/37 - Xem BT5/sgk/37. - HV trnh by bi lm. - Tt c cc HV cn li tr li vo v nhp. - Nhn xt. - Chnh sa hon thin. - Ghi nhn kt qu. 5) BT5/sgk/37 : a)cosx -3 sinx =2 . +) a = 1, b = 3 , c = 2 ,2 2b a +=2 +)Chia hai v phng trnh cho 2 ta c: 21sinx - 23cosx = 22 4sin 3 cos3sin 3 sin3cost t t= x x) (32361332367) (24 3324 334sin )33 sin(Z kk xk xZ kk xk xxe

+ =+ =e

+ = + = = tttttttt tt t b)3sin3x 4cos3x = 5 +) a = 3, b = 4, c = 5.2 2b a += 5 +) Chia c hai v phng trnh cho vi 5 ta c : 1 3 cos543 sin53= x xt 53= coso , sino = 54 th vo phng trnh ta c: coso sin3x - sino cos3x = 1 ) (326 3) ( 2232sin ) 3 sin(Z k k xZ k k xxe + + = e + = = tt ottoto c) 2sinx + 2cosx - 2= 0 2sinx + 2cosx =2 (*) Ta c : a = 2, b = 2 , c =22 2b a + = 2 2Chia c hai v phng trnh (*) cho 22ta c : ) (2122127) (26 426 46sin )4sin(6sin cos4cos sin4sin21cos21sin21Z kk xk xZ kk xk xxx xx xe

+ =+ =e

+ = ++ = += + = + = +tttttt tttttt tt t t V/Hng d hc v lm bi nh: (5) Cng c : Ni dung c bn c hc? Dn d : Xem li cc BT gii. Hon thin cc bi tp cn li. Ngy son: 22/02/2011 Lp 11CB: 24/02/2011 Ngy dy:Lp 10B://2011 Lp 10D://2011 Tit 21 LUYN TP 3 (tip theo) I/ Mc tiu: Qua bi hc HV cn nm vng: 1) Kin thc: -Bit c dng v cch gii phng trnh: bc nht, bc hai i vi mt hm s lng gic, phng trnh asinx + bcosx = c. 2) K nng: -Gii c phng trnh cc dng trn. 3) T duy:Nm c dng v cch gii cc phng trnh n gin. 4) Thi : Cn thn trong tnh ton v trnh by . Qua bi hc HV bit c ton hc c ng dng trong thc tin. II/ Chun b: 1. Gio vin: -Gio n, SGK , STK, phn mu. -Bng ph,- Phiu tr li cu hi. 2. Hc vin:- Hc bi, lm bi tp, c trc bi, n tp kin thc lin quan, my tnh b ti. - dng hc tp: Thc k, compa,... III/ Nhng iu cn lu : -Coi trng vic n l thuyt hc vin h thng c cc kin thc c bn. -Ch trng vic rn luyn k nng gii ton.IV/ Tin trnh bi dy: 1. Kim tra bi c: Hot ng 1 (8): Kim tra bi c. Hot ng ca GVHot ng ca HV ? Ta hc bao nhiu cng thc cng lng gic ?Suy ngh, tr li. Ghi nhn kin thc. 2. Bi mi: . Ngy son: 23/02/2011 Lp 11CB: 25/02/2011 Ngy dy:Lp 10B://2011 Lp 10D://2011 Tit 22 THC HNH GII PHNG TRNH LNG GIC BNG MY TNH B TI. I/ Mc tiu: Qua bi hc HV cn nm vng: 1) Kin thc: - Bit s dng my tnh vo gii cc phng trnh lng gic. 2) K nng: - Gii c cc phng trnh lng gic c bn bng my tnh b ti v gii c pt bc nht, bc hai i vi mt hm s lng gic, phng trnh asinx + bcosx = c. 3) T duy:- Hiu c phng trnh lng gic c bn, phng trnh bc nht v bc hai i vi mt hm s lng gic, phng trnh dng asinx + bcosx = c v cch gii. 4) Thi :- Cn thn trng trong tnh ton.- Qua bi hc HV bit c ton hc c ng dng trong thc tin. II/ Chun b: 1. Gio vin: - Gio n, SGK, STK, phn mu. - Bng ph, my tnh FX 500 MS 2. Hc vin:- Hc bi, lm bi tp, c trc bi, n tp kin thc lin quan, my tnh b ti. - dng hc tp: Thc k, compa,... III/ Nhng iu cn lu : Ch trng vic rn luyn k nng gii ton bng my tnh.IV/ Tin trnh bi hc v cc hot ng: 1. Kim tra bi c: Xen ln trong bi dy 2. Bi mi: Hot ng 1 (15'): Hot ng ca GVHot ng ca HVNi dung - Hng dn hc vin s dng my tnh c nhn (FX500 MS) vo gii phng trnh lng gic c bn.- Ghi nhn cch gii phng trnh lng gic c bn trn my tnh c naaan (FX500 MS). - Cng thc hnh v d SGK T 27-28. V d: dng my tnh c nhn (FX500 MS) gii phng trnh lng gic c bn sau: a) sin x =0,5 b)cos x = -1/3 c) tan x =3 Li gii: (SGK T 27-28 ) Hot ng 2 (25'): Hot ng ca GVHot ng ca HVNi dung - GV chia hc vin trong lp thnh cc nhm theo dy bn; - Giao bi tp theo nhm: dng my tnh c nhn d (FX500 MS) gii phng trnh lng gic c bn sau: - HV trao i nhm trnh by bi lm vo v nhp. - Nhn xt. - Chnh sa hon thin nu c. Nhm 1:a) sin x = 2/3 Nhm 2: b)cos x = - 3/2 Nhm 3:c) tan x =5 Nhm 4:d) cot x = 3 - Hng dn cc nhm thc hnh.- Cho cc nhm tho lun; Kt lun : - Ch : SGK T 28. - Ghi nhn kt qu. V/Hng d hc v lm bi nh:(5') Cng c: Ngoi my tnh c nhn (FX500 MS) ta c th s dng cc loi my tnh c nhn khc vo gii phng trnh lng gic; Dn d: Xem bi v BT gii. Xem trc lm bi tp n tp chng I Ngy son: 23/02/2011 Lp 11CB: 25/02/2011 Ngy dy:Lp 10B://2011 Lp 10D://2011 Tit 23 N TP CHNG I I/ Mc tiu:Qua bi hc HV cn nm vng: 1) Kin thc: - Hm s lng gic, tp xc nh, tnh chn l, tnh tun hon v chu k. th ca hm s lng gic. - Phng trnh lng gic c bn. - Phng trnh bc nht v bc hai i vi mt hm s lng gic. - Phng trnha v phng trnh bc hai i vi mt hm s lng gic. - Phng trnh dng asinx + bcosx = c.2) K nng: - Bit dng th cc hm s lng gic. - Bit s dng th xc nh cc im ti th nhn gi tr m, dng v cc gi tr c bit. - Gii c cc phng trnh lng gic c bn. - Gii c pt bc nht, bc hai i vi mt hm s lng gic, phng trnhasinx + bcosx = c. 3) T duy:- Hiu c hm s lng gic. Tp xc nh, tnh chn l, tnh tun hon v chu k. th ca hm s lng gic . - Hiu c phng trnh lng gic c bn, phng trnh bc nht v bc hai i vi mt hm s lng gic, phng trnh dngasinx + bcosx = c v cch gii. 4) Thi : Cn thn trong tnh ton v trnh by. - Qua bi hc HV bit c ton hc c ng dng trong thc tin. II/ Chun b: 1. Gio vin: - Gio n, SGK ,STK , phn mu. - Bng ph, phiu tr li cu hi. 2. Hc vin:- Hc bi, lm bi tp, c trc bi, n tp kin thc lin quan, my tnh b ti. - dng hc tp: Thc k, compa,... III/ Nhng iu cn lu : -Coi trng vic n l thuyt hc vin h thng c cc kin thc c bn. -Ch trng vic rn luyn k nng gii ton.IV/ Tin trnh bi hc v cc hot ng: 1. Kim tra bi c: 2. Bi mi: Hot ng 1 (15'): BT1/sgk/40 Hot ng ca GVHot ng ca HVNi dung - Th no l Hv chn? BT1a/sgk/40 ? - Th no l Hv l? BT1b/sgk/40 ? - Ln bng tr li.- Tt c cc HV cn li tr li vo v nhp. - Trnh by bi lm. - Nhn xt. BT1/40/sgk : a) Chn . V cos (-3x) = cos 3x xe R b) Khng l . V ti x = 0 Hot ng 2 (13'): BT2/40/sgk HGVHHVNI DUNG - BT2/40/sgk ? - Da vo th tr li. - Ln bng trnh by li gii. - HV cn li tr li vo v nhp. - Nhn xt. - Chnh sa hon thin.- Ghi nhn kin thc.BT2/40/sgk : a) x e {23;2t t } b) x e (-t , 0)(t ,2t ) Hot ng 3 (15'): BT3/41/sgk HGVHHVNI DUNG - BT3/41/sgk ? - Da vo tp gi tr ca hm s cosx v sinx lm - Ln bng trnh by li gii. - HV cn li tr li vo v nhp. - Nhn xt. - Chnh sa hon thin. - Ghi nhn kin thc. BT3/41/sgk :a) Ta c 1 + cos xs 2. Du ng thc xy ra khi v ch khi cos x = 1 x = k2t , k e Z. Vy GTLN ca hm s l Y = 3 khi x = k2t , k e Z. b) ta c sin ( x - 6t )s 1, du ng thc xy ra khi sin ( x - 6t ) = 1 hay x =32t+ k2t , k e Z. vy GTLN ca hm s ly = 1 t c khi v ch khix =32t+ k2t , k e Z. V/Hng d hc v lm bi nh (2'): Cng c :Ni dung c bn c hc? Dn d : Xem bi v BT gii. Xem vlm tip bi tp n tp chng I.. tan tan5 5x xt t | | | | + = + ||\ . \ . Ngy son: 20/08/2010 Lp 11CB: 23/08/2011 Ngy dy:Lp 10B://2011 Lp 10D://2011 Tit 25N TP CHNG I ( tip theo). I/ Mc tiu: Qua bi hc HV cn nm vng: 1) Kin thc: -Hm s lng gic, tp xc nh, tnh chn l, tnh tun hon v chu k. th ca hm s lng gic. -Phng trnh lng gic c bn. -Phng trnh bc nht v bc hai i vi mt hm s lng gic . -Phng trnha v phng trnh bc hai i vi mt hm s lng gic. -Phng trnh dng asinx + bcosx = c.2) K nng: -Bit dng th cc hm s lng gic. -Bit s dng th xc nh cc im ti th nhn gi tr m, dng v cc gi tr c bit. -Gii c cc phng trnh lng gic c bn. -Gii c pt bc nht, bc hai i vi mt hm s lng gic, phng trnh asinx + bcosx = c. 3) T duy:-Hiu c hm s lng gic. Tp xc nh, tnh chn l, tnh tun hon v chu k. th ca hm s lng gic . -Hiu c phng trnh lng gic c bn, phng trnh bc nht v bc hai i vi mt hm s lng gic, phng trnh dng asinx + bcosx = c v cch gii . 4) Thi : Cn thn trong tnh ton v trnh by. Qua bi hc HV bit c ton hc c ng dng trong thc tin. II/ Chun b: 1. Gio vin: -Gio n, SGK, STK, phn mu. -Bng ph, phiu tr li cu hi. 2. Hc vin:- Hc bi, lm bi tp, c trc bi, n tp kin thc lin quan, my tnh b ti. - dng hc tp: Thc k, compa,... III/ Nhng iu cn lu : -Coi trng vic n l thuyt hc vin h thng c cc kin thc c bn . -Ch trng vic rn luyn k nng gii ton.IV/ Tin trnh bi hc v cc hot ng: 1. Kim tra bi c: 2. Bi mi: Hot ng 1 (13'): BT4/41/sgk Hot ng ca GVHot ng ca HVNi dung - BT4/41/sgk? - a v ptlgcb gii. - Ln bng trnh by. - HV cn li tr li vo v nhp. - Nhn xt. - Chnh sa hon thin. - Ghi nhn kin thc. BT4/41/sgk :a) sin ( x + 1) = 32 ) (232arcsin 1232arcsin 1Z kk xk xe

+ = ++ = +t tt ) (2 132arcsin2 132arcsinZ kk xk xe

+ =+ =t tt d) tan ( 3 ) 1212 = + xt Z kkxZ k k xxe + = e + = + = + ,12 1445,31212)3tan( ) 1212tan(t ttt tt t Hot ng 2 (15'): BT5/41/sgk Hot ng ca GVHot ng ca HVNi dung - BT5/41/sgk? - a v ptlgcb gii. - Ln bng trnh by li gii. - HV cn li tr li vo v nhp. - Nhn xt. - Chnh sa hon thin. - Ghi nhn kin thc. BT5/41/sgk :a) 2cos2x 3cosx + 1 = 0 2 cos 1( )12 cos3 2x k xk Zx k xttt= = e = + = c) 2sinx + cosx = 1 51cos52sin52= + x xto cos52= , o sin52=Ta c: coso sinx + sino cosx =51 sin ( x +o ) = 51 ) (251arcsin251arcsin) (251arcsin251arcsinZ kk xk xZ kk xk xe

+ =+ =e

+ = ++ = +t o tt ot t ot o Hot ng 3 (15'): BTTN/41/sgk Hot ng ca GVHot ng ca HVNi dung - BTTN/41/sgk? - Tr li. - Nhn xt. - Chnh sa hon thin.- Ghi nhn kin thc.Bi tp trc nghim /41/sgk :6 A,7A,8C.9 , B,10C V/Hng d hc v lm bi nh: (2') Cng c:Ni dung c bn c hc trong chng I Dn d: Xem bi v BT gii; Chun b kim tra 45' . Ngy son: 20/08/2010 Lp 11CB: 23/08/2011 Ngy dy:Lp 10B://2011 Lp 10D://2011 TIT 26 : KIM TRA 45 CHNG I. I/ Mc tiu : -Kim tra k nng gii ton v kin thc c bn ca chng I.-Rn luyn t duy l gc khi gii ton. -H/v t nh gi kt qu hc tp, tch cc hn trong hc tp. II/ Chun b ca GV v HV : -GV : kim tra, p n -Tr : My tnh b ti, giy nhp.n tp chng I. III/ Ni dung kim tra : 1/ n nh lp. - Kim tra s s lp. 2/ bi : Cu 1 :Gii phng trnh: Cos2x -5 cosx + 4 =0 Cu 2 : Nu cch gii phng trnh sau bng my tnh b ti: CASIO 500MS p dng gii phng trnh sau :Sin (x - 3) =

3/ p n v biu im : Cu 1:( 5 im) Gii phng trnh: Cos2x -5 cosx + 4 = t : t = cosx ( k -1s ts 1)1 Ta c phng trnh: t2 -5t +4 =0 1 Suy ra : t =1hoct =4 1 T cch t ta c pt: cosx = 11 Vy: x=k2 Z k e , t 1 Cu 2 :(5 im) Gii phng trnh bng my tnh b ti: CASIO 500MS + 3 ModModMod2SHIFTSin (v2)=KQ: 0,7854 + Ta c: Nghim ca phng trnh: 2 x =3,7854 + k2 Z k e , thocx =t- 2,2146 + k2 Z k e , t ---------------------------------------------------------------------------------------------------------- CHNG II: T HP V XC XUT Ngy son:01/03/2011Lp 11CB:03/03/2011 Ngy dy:Lp 10B://2011 Lp 10D://2011 Tit 271. QUY TC M I/ Mc tiu: Qua bi hc HV cn nm: 1) Kin thc: Bit: - Qui tc cng v qui tc nhn.- Phn bit v vn dng cc tnh hung s dng qui tc cng, qui tc nhn.2) K nng: - Bit vn dng qui tc cng v qui tc nhn gii mt s bi ton. 3) T duy: Bit kt hp c hai qui tc a bi ton phc tp v bi ton n gin. 4) Thi : Cn thn trong tnh ton v trnh by.Tch cc tham gia vo bi hc. Qua bi hc HV bit ng dng ton hc vo thc tin. II. Chun b: 1. Gio vin: - Gio n, SGK, STK , phn mu. - Bng ph,- Phiu tr li cu hi 2. Hc vin:- Hc bi, c trc bi, my tnh b ti. - dng hc tp: Thc k, compa,... III/ Nhng iu cn lu : - Lm cho hc vin nm vng c cc kin thc c bn. - Ch trng vic rn luyn k nng gii ton.IV/ Tin trnh bi hc v cc hot ng: 1. Kim tra bi c: Xen ln trong bi dy 2. Bi mi: Hot ng 1 (10') : Quy tc cng Hot ng ca GVHot ng ca HVNi dung - Gii thiu cch ghi s phn t ca tp hp nh sgk. - Tm A\B cu b) ? - Xem VD1 sgk ? c bao nhiu cch chn qu cu en? trng? - Pht biu quy tc cng?- H1 sgk?- Tm s phn t v so snh tng s phn t ca A v B? - HV xem sgk. - Nhn xt.A = {a, b, c}. Tp hp A c 3 phn t . Vit: n(A) = 3 hay |A| = 3. - Xem sgk. - Pht biu. - Nhn xt. - Ghi nhn kt qu. 1 . Quy tc cng: (sgk) Nu(khng giao nhau) th Ch : (sgk) Hot ng 2 (10'): VD2 Hot ng ca GVHot ng ca HVNi dung - VD2 sgk ?- C th c hnh vung cnh bao nhiu t hcn cho? - S hnh vung cnh 1cm? 2cm?- Chnh sa.-cVD2sgk,nhnxt, ghi nhn kt qu. VD2: (sgk) Hot ng 3 (13'): Quy tc nhn Hot ng ca GVHot ng ca HVNi dung A B A B =C( ) ( ) ( ) n A B n A n B= +- Xem VD3 sgk ? c bao nhiu cch chn o? chn qun? - Chn o hoc qun khng p ng y /c bi ton cha? - Pht biu quy tc nhn? - H2 sgk?- C my cch i tA ti B? my cch i t B ti C? - Xem sgk.- Nghe, suy ngh, tr li. - Nhn xt. - Ghi nhn kin thc. - c H2 sgk/45.- Trnh by bi gii.- Nhn xt.- Chnh sa hon thin. - Ghi nhn kin thc.2. Quy tc nhn:(sgk) Ch : (sgk) Hot ng 4 (10'): VD4 Hot ng ca GVHot ng ca HVNi dung - VD4 sgk ?- Chn s hng n v my cch? s hng chc my cch?- c VD4 sgk, nhn xt, ghi nhn kt qu. VD4 : (sgk) V/Hng dn hc v lm bi nh (2') : - Cng c : Ni dung c bn c hc? - Dn d : Lm BT 1, 2, 3, 4 sgk- T46 ------------------------------------------------------------------------------------------------------------------ Ngy son: 03/03/2011 Lp 11CB: 08/03/2011 Ngy dy:Lp 10B://2011 Lp 10D://2011 TIT 29 LUYN TP 1 I/ Mc tiu: Qua bi hc HV cn nm vng: 1) Kin thc: - Bit phn bit v vn dng vo cc tnh hung, khi no s dng qui tc cng, qui tc nhn. 2) K nng: - Bit vn dng qui tc cng v qui tc nhn gii mt s bi ton. 3) T duy: Bit kt hp c hai qui tc a bi ton phc tp v bi ton n gin. 4) Thi : Cn thn trong tnh ton v trnh by. Tch cc tham gia vo bi hc. Qua bi hc HV bit ng dng ton hc vo thc tin. II. Chun b: 1. Gio vin: - Gio n, SGK, STK, phn mu. - Bng ph- Phiu tr li cu hi 2. Hc vin:- Hc bi, lm bi tp, c trc bi, my tnh b ti. - dng hc tp: Thc k, compa,... III/ Nhng iu cn lu - Lm cho hc vin nm vng c cc kin thc c bn . - Ch trng vic rn luyn k nng gii ton.IV/ Tin trnh bi hc v cc hot ng: 1. Kim tra bi c: Hot ng 1 (10'): Kim tra bi c Hot ng ca GVHot ng ca HVNi dung - Pht biu quy tc cng ?- V d?- Nhn xt, cho im- Pht biu quy tc nhn ?- V d?- Nhn xt, cho im * Pht biu quy tc cng?- Nu cc V d?- Nhn xt. *Pht biu quy tc cng?- Nu cc V d?- Nhn xt. 1 . Quy tc cng: (sgk) 2 . Quy tc nhn: (sgk) 2. Bi mi: Hot ng 2 (10'):BT1/sgk/46 Hot ng ca GVHot ng ca HVNi dung BT1/sgk/46 ?Hng dn: a) Cch chn s c mt ch s. b) Cch chn s c haich so: gm s hng chc v hng n v t4 s cho. c) Cch chn s c haich so: gm s hng chc v hng n vkhc nhu t4 s cho. - Gii bi tp.- Nhn xt, ghi nhn. p n a) 4 b) 4.4 = 16 c) 4.3 = 12 Hot ng 3 (10):BT2/sgk/46 Hot ng ca GVHot ng ca HVNi dung BT2/sgk/46 ?Hng dn: - Cch chn s c mt ch s t 6 s cho? - Cch chn s c haich s: gm s hng chc v hng n v t6 s cho? - S dng quy tc cng ta c KQ? - Gii bi tp - Nhn xt, ghi nhn p n: 6 + 62 = 42 (s) Hot ng4 (13'):BT3/sgk/46 Hot ng ca GVHot ng ca HVNi dung BT3/sgk/46 ?Hng dn: a) Cch chn i t A n D m qua B, C ch mt ln (ta s dng quy tc nhn) b) Cch chn i t A n D ri quay li A (ta s dng quy tc nhn)- Gii bi tp- Nhn xt, ghi nhn. p n a) 4.2.3 = 24 (cch) b) 4.2.3.3.2.4 = 242 = 576 (cch) V/Hng dn hc v lm bi nh: (2') - Cng c:Cu 1: Ni dung c bn c hc? Cu 2: BT4/sgk/46 ?Hng dn: 3.4 = 12 (cch) - Dn d: - Xem bi gii.- Xem trc bi v hot ng hon v, chnh hp, t hp. Ngy son: 03/03/2011 Lp 11CB: 08/03/2011 Ngy dy:Lp 10B://2011 Lp 10D://2011 Tit 30 2. HON V. CHNH HP. T HP. I/ Mc tiu: 1) Kin thc: Bit: - Khi nim hon v, s hon v, chnh hp, s chnh hp. 2) K nng: - Tnh c s cc hon v, chnh hp. 3) T duy: Hiu honv, s hon v, chnh hp, s chnh hp. 4) Thi : Cn thn trong tnh ton v trnh by . Qua bi hc HV bit c ton hc c ng dng trong thc tin. II. Chun b: 1. Gio vin: - Gio n, SGK, STK, phn mu. - Bng ph,- Phiu tr li cu hi 2. Hc vin:- Hc bi, c trc bi, my tnh b ti. - dng hc tp: Thc k, compa,... III/ Nhng iu cn lu - Lm cho hc vin nm vng c cc kin thc c bn . - Ch trng vic rn luyn k nng gii ton.IV/ Tin trnh bi hc v cc hot ng: 1. Kim tra bi c: Xen ln trong bi dy 2. Bi mi: Hot ng 1 (10'): Hon v Hot ng ca GVHot ng ca HVNi dung - VD1 sgk ?- Nu mt vi cch sp xp ? - Chnh sa hon thin.- H1 sgk? - c VD1 sgk. - Tr li. - Nhn xt, ghi nhn. 3! = 6 (cch) I/ Hon v: 1) nh ngha: (sgk) Nhn xt : (sgk) Hot ng 2 (10'): S cc hon v Hot ng ca GVHot ng ca HVNi dung - VD2 sgk ?- K cccch sp xp? - Cch lm khc? - S cch chn ngi v tr 1, 2, 3, 4 ? - CM sgk. - H2 sgk?- Xem sgk.- Nghe, suy ngh. - Tr li.- Ghi nhn kin thc. - Quy tc nhn:4.3.2.1 = 24 (cch) 10! (cch) 2) S cc hon v:(sgk) K hiu: Pn s hon v n phn t nh l: Pn = n(n 1) . . . 2.1 Ch : (sgk) Pn = n! Hot ng 3 (10'): Chnh hp Hot ng ca GVHot ng ca HVNi dung - VD3 sgk ?- K cccch sp xp? - S cch chn bn qut nh, bn lau bng, bn sp bn gh ? - H3 sgk? - Xem sgk, tr li - Nhn xt. - Ghi nhn kin thc-cVD4sgk,nhnxt,ghi nhn. - C A24 vct. II/ Chnh hp:1) nh ngha: (sgk) K hiu: Aknchnh hp chp k ca n phn t. Hot ng 4 (10'): S cc chnh hp Hot ng ca GVHot ng ca HVNi dung - Nh sgk- Quy tc nhn? - Cm sgk. - VD4 sgk ?- Xem sgk, tr li. - Nhn xt. - Quy tc nhn: 5.4.3 = 60 (cch) -cVD4sgk,nhnxt, ghi nhn. 2) S cc chnh hp:(sgk) nh l:

Ch : (sgk)a) Qui c 0! = 1, Ta c: b) Hon v n phn t V/Hng dn hc v lm bi nh: (5')- Cng c: Ni dung c bn c hc? - Dn d: Xem bi v VD gii. Lm BT2/SGK/54 . Ngy son:03/03/2011 Lp 11CB: 08/03/2011 Ngy dy:Lp 10B://2011 Lp 10D://2011 Tit 31 2. HON V. CHNH HP. T HP. ( Tip theo) I/ Mc tiu: Qua bi hc HV cn nm vng: 1) Kin thc:-Khi nim hon v , s hon v, chnh hp, s chnh hp, t hp, s t hp. 2) K nng:-Tnh c s cc hon v, chnh hp, t hp chp k ca n phn t. 3) T duy:-Hiu hon v, s hon v, chnh hp, s chnh hp, t hp, s t hp. 4) Thi : Cn thn trong tnh ton v trnh by.vi Qua bi hc HV bit c ton hc c ng dng trong thc tin. II. Chun b: 1. Gio vin: - Gio n, SGK, STK, phn mu. - Bng ph- Phiu tr li cu hi 2. Hc vin:- Hc bi, lm bi tp, c trc bi, my tnh b ti. - dng hc tp: Thc k, compa,... III/ Nhng iu cn lu :- Lm cho hc vin nm vng c cc kin thc c bn . - Ch trng vic rn luyn k nng gii ton.599.8.7.6.5 15120 A= =( ) ( ) 1 ... 1knA n n n k = +( )! 1!knnA k nn k= s snn nP A =IV/ Tin trnh bi hc v cc hot ng: 1. Kim tra bi c: Hot ng 1 (10'): Kim tra bi c Hot ng ca GVHot ng ca HVNi dung - Pht biu nh ngha chnh hp? - C bao nhiu chnh hp chp 2ca 5 phn t? - Ln bng tr li. - Tt c cc HV cn li tr li vo v nhp. - Nhn xt.* nh ngha: (sgk) *A25 = 20 2. Bi mi: Hot ng 2 (15'): T hp Hot ng ca GVHot ng ca HVNi dung - VD5 sgk ?- K cc tam gic? - nh ngha? - H4 sgk? - Xem sgk, tr li. - Nhn xt. - Ghi nhn kin thc. III/ T hp:1) nh ngha: (sgk) K hiu: Cnk t hp chp k ca n phn t ( ) Ch : (sgk) Hot ng3 (10'): S cc t hp Hot ng ca GVHot ng ca HVNi dung -Nh sgk-Cm sgk -VD6 sgk ?-H5 sgk?(trn) -Xem sgk, tr li -Nhn xt -cVD6sgk,nhnxt,ghi nhna)b) 2) S cc t hp:(sgk) nh l: Hot ng 4 (8'): Tnh cht HGVHHVNI DUNG - Tnh cht sgk? - VD7 sgk ? - Xem sgk. - Tr li. - Nhn xt. - Ghi nhn kin thc.3) Tnh cht: (sgk) a) TC1 : b) V/Hng dn hc v lm bi nh: (2')- Cng c: Cu 1: Ni dung c bn c hc? Cu 2: Cng thc tnh hon v, chnh hp, t hp? Lin h gia cc cng thc? - Dn d: Xem bi v VD gii. LmBT1 ->BT7/SGK/T.54,55 1 k n s s21616!1202!.14!C = =51010!2525!.5!C = =3 26 4. 20.6 120 C C = =( )!! !knnCk n k=( )0k n kn nC C k n= s s( )11 1 1k k kn n nC C C k n + = s < Ngy son: 09/03/2011Lp 11CB: 10/03/2011 Ngy dy:Lp 10B://2011 Lp 10D://2011 Tit 33 LUYN TP 2 I/ Mc tiu: Qua bi hc HV cn nm vng: 1) Kin thc: - Khi nim hon v, s hon v, chnh hp, s chnh hp, t hp, s t hp, cc cng thc tnh. 2) K nng: - Vn dng hon v, chnh hp, t hp vo gii bi ton thc t. - Dng my tnh tnh hon v, chnh hp, t hp. 3) T duy:Hiu hon v, s hon v, chnh hp, s chnh hp, t hp, s t hp. 4) Thi : Cn thn trong tnh ton v trnh by. Qua bi hc HV bit c ton hc c ng dng trong thc tin. II. Chun b: 1. Gio vin: - Gio n, SGK, STK, phn mu. - Bng ph- Phiu tr li cu hi. 2. Hc vin:- Hc bi, lm bi tp, c trc bi, my tnh b ti. - dng hc tp: Thc k, compa,... III/ Nhng iu cn lu :- Lm cho hc vin nm vng c cc kin thc c bn . - Ch trng vic rn luyn k nng gii ton.IV/ Tin trnh bi hc v cc hot ng: 1. Kim tra bi c: Hot ng 1 (10'): Kim tra bi c HGVHHV - Th no l hon v, chnh hp, t hp? -Tnh

- Ln bng tr li.- Tt c cc HV cn li tr li vo v nhp. - Nhn xt. 2. Bi mi: Hot ng 2 (10'): BT1/SGK/54 Hot ng ca GVHot ng ca HVNi dung BT1/sgk/54 a) l hon v no? b) S chn th s v ntn? C my cch chn? Cch chn cc ch s cn li ? Cc s cu a) b hn 432000? - Xem BT1/sgk/54. - HV trnh by bi lm.- Tt c cc HV cn li tr li vo v nhp. - Nhn xt. - Ghi nhn kt qu. BT1/SGK/54 :a) 6! b) S chn: 3.5! = 360 (s) S l: 3.5! = 360 (s) c) 3.5! + 2.4! + 1.3! = 414 (s) Hot ng 3 (10'): BT2/SGK/54 3 4 3 3 36 7 6 5 5 6, , , , ,P A A A C CHot ng ca GVHot ng ca HVNi dung BT2/sgk/54 ?- Th nol hon v? - Xem BT1/sgk/54. - HV trnh by bi lm.- Nhn xt. - Ghi nhn kt qu. BT2/SGK/54 : 10! cch sp xp Hot ng 4 (13'): BT3,4/SGK/54, 55 Hot ng ca GVHot ng ca HVNi dung BT3/sgk/54 ?- Th nol chnh hp? BT4/sgk/54 ? - Xem BT3,4/sgk/54. - HV trnh by bi lm.- Tt c cc HV cn li tr li vo v nhp. - Nhn xt. - Ghi nhn kt qu. BT3/SGK/54 : (cch) BT4/SGK/55 : (cch) V/Hng d hc v lm bi nh: (2')- Cng c: Ni dung c bn c hc? - Dn d: Xem bi tp gii v lm BT4-7 sgk T55 . Ngy son: 10/03/2011 Lp 11CB: 11/03/2011 Ngy dy:Lp 10B://2011 Lp 10D://2011 Tit 34LUYN TP 2 (Tip theo) I/ Mc tiu: Qua bi hc HV cn nm vng:1) Kin thc: - Khi nim hon v, s hon v, chnh hp, s chnh hp, t hp, s t hp, cc cng thc tnh. 2) K nng: - Vn dng hon v, chnh hp, t hp vo gii bi ton thc t. - Dng my tnh tnh hon v, chnh hp, t hp. 3) T duy:Hiu hon v, s hon v, chnh hp, s chnh hp, t hp, s t hp . 4) Thi : Cn thn trong tnh ton v trnh by. Qua bi hc HV bit c ton hc c ng dng trong thc tin. II. Chun b: 1. Gio vin: - Gio n, SGK, STK, phn mu. - Bng ph- Phiu tr li cu hi 2. Hc vin:- Hc bi, lm bi tp, c trc bi, my tnh b ti. - dng hc tp: Thc k, compa,... III/ Nhng iu cn lu : - Lm cho hc vin nm vng c cc kin thc c bn . - Ch trng vic rn luyn k nng gii ton.IV/ Tin trnh bi hc v cc hot ng: 1. Kim tra bi c: Xen ln trong bi dy 377!2104!A= =466!3602!A= =2. Bi mi: Hot ng 1 (13'): BT5/SGK/55 Hot ng ca GVHot ng ca HVNi dung - BT5/sgk/55 ?- Th nol t hp? - Xem BT5/sgk/55. - HV trnh by bi lm.- Tt c cc HV cn li tr li vo v nhp. - Nhn xt. - Ghi nhn kt qu. BT5/SGK/55 : a) (cch) b) (cch) Hot ng 2 (10'): BT6/SGK/55 Hot ng ca GVHot ng ca HVNi dung - BT6/sgk/55 ?- Th nol t hp? - Xem BT6/sgk/55. - HV trnh by bi lm.- Tt c cc HV cn li tr li vo v nhp. - Nhn xt. - Ghi nhn kt qu. BT6/SGK/55 : (tam gic) Hot ng 3 (10'): BT7/SGK/55 Hot ng ca GVHot ng ca HVNi dung - BT7/sgk/55 ?- Th no l hcn?- Cch chn hai ng thng song song? - Cch chn hai thng vung gc vi bn ng thng song song? - Xem BT7/sgk/55. - HV trnh by bi lm.- Tt c cc HV cn li tr li vo v nhp. - Nhn xt. - Ghi nhn kt qu. BT7/SGK/55: (hnh ch nht) Hot ng 4 (10): Bi tp lm thm Hot ng ca GVHot ng ca HVNi dung BT1:Mt ci khay trn ng bnh ko ngy tt c 6 ngn hnh qut mu khc nhau. Hi c bao nhiu cch by 6 loi bnh ko vo 6 ngn ? - Xem BT1. - HV trnh by bi lm.- Tt c cc HV cn li tr li vo v nhp. - Nhn xt. - Ghi nhn kt qu. BT1 Pn = 6! = 720 ( cch ) V/Hng d hc v lm bi nh: (2')- Cng c: Cu 1: Ni dung c bn c hc? Cu 2: Cng thc tnh hon v, chnh hp, t hp?- Dn d: Xem bi gii. Xem trc bi nh thc niu tn ..Ngy son: 10/03/2010 Lp 11CB: 11/03/2011 Ngy dy:Lp 10B://2011 Lp 10D://2011 Tit 353. NH THC NIU TN I/ Mc tiu: Qua bi hc HV cn nm:1) Kin thc: Bit: 355!602!A= =355!103!.2!C = =366!203!.3!C = =2 24 5. 60 C C =- Cng thc nh thc Niu tn (a + b)n - Tam giac Pa-xcan. 2) K nng: -Bit khai trin nh thc Niu -tn vi mt s m c th. 3) T duy:Hiu nm c cng thc nh thc Niu -tn, tam gic Pa xcan. 4) Thi : Cn thn trong tnh ton v trnh by. Qua bi hc HV bit c ton hc c ng dng trong thc tin. II. Chun b: 1. Gio vin: - Gio n, SGK, STK, phn mu. - Bng ph,- Phiu tr li cu hi 2. Hc vin:- Hc bi, lm bi tp, c trc bi, my tnh b ti. - dng hc tp: Thc k, compa,... III/ Nhng iu cn lu -Lm cho hc vin nm vng c cc kin thc c bn . -Ch trng vic rn luyn k nng gii ton. IV/ Tin trnh bi hc v cc hot ng: 1. Kim tra bi c: Hot ng 1 (10'): Kim tra bi c Hot ng ca GVHot ng ca HV - Tnh : - Nhc li cc hng ng thc : (a + b)2, (a + b)3 ? - Ln bng tr li.- Tt c cc HV cn li tr li vo v nhp. - Nhn xt. 2. Bi mi: Hot ng 2 (10'): Cng thc nh thc Niu - tn Hot ng ca GVHot ng ca HVNi dung - H1 sgk?- Khai trin : (a + b)n? - Cng thc nh thc Niu -tna = b = 1 suy c g t ct? a = 1 , b = -1 suy c g t ct? - Nhn xt s hng t VT, s m ca a v b, h s hng t cch u hai hng t u ? - c H1 sgk lm v nhp,nhnxt,ghi nhn. 1. Cng thc nh thc Niu-tn: (sgk) + H qu : (sgk) + Ch : (sgk) Hot ng 3 (10'): V d Hot ng ca GVHot ng ca HVNi dung - VD1 sgk ?- VD2 sgk ?- VD3 (sgk) ? - S dng cng thc nh thc Niu -tn gii. - c VD2 sgk, nhn xt, ghi nhn.- Trnh by bi gii.- Nhn xt.- Chnh sa hon thin. - Ghi nhn kin thc.V d 1: (sgk) V d 2: (sgk) V d 3: (sgk) Hot ng 4 (13'): Tam gic Pa -xcan Hot ng ca GVHot ng ca HVNi dung - nh ngha nh sgk.- Ch cho HV bit cch tnh cc - Xem sgk. - Nhn xt. 2) Tam gic Pa -xcan: (sgk) 0 1 2 0 1 2 32 2 2 3 3 3 3, , , , , , C C C C C C Ch s. - H2 sgk?- Da vo nhn xt, tam gic Pa xcan.- Ghi nhn kin thc. -LmH2sgk,nhnxt,ghi nhn. Nhn xt: (sgk) V/Hng dn hc v lm bi nh: (2') - Cng c: Cu 1: Ni dung c bn c hc? Cu 2: Cng thc nh thc Niu -tn, tam gic Pa -xcan ?- Dn d: Xem bi v VD gii. LmBT1->BT6/SGK/T57,58. Ngy son: 10/03/2011 Lp 11CB: 11/03/2011 Ngy dy:Lp 10B://2011 Lp 10D://2011 Tit 37LUYN TP 3. I. Mc tiu: Qua bi hc HV cn nm vng:1) Kin thc: - Cng thc nh thc Niu -tn. - Tam giac Pa - xcan. 2) K nng: - Bit cng thc nh thc Niu - tn, tam gic Pa - xcan. - Tnh cc ca khai trin nhanh chng bng cng thc Niu - tn hoc tam gic Pa - xcan3) T duy:Hiu nm c cng thc nh thc Niu - tn, tam gic Pa - xcan. 4) Thi : Cn thn trong tnh ton v trnh by. Qua bi hc HV bit c ton hc c ng dng trong thc tin II. Chun b: 1. Gio vin: - Gio n, SGK, STK, phn mu. - Bng ph- Phiu tr li cu hi 2. Hc vin:- Hc bi, lm bi tp, c trc bi, my tnh b ti. - dng hc tp: Thc k, compa,... III. Nhng iu cn lu : - Lm cho hc vin nm vng c cc kin thc c bn . - Ch trng vic rn luyn k nng gii ton.IV. Tin trnh bi hc v cc hot ng: 1. Kim tra bi c: Hot ng 1 (7'): Kim tra bi c Hot ng ca GVHot ng ca HVNi dung - BT1/SGK/57 ? - Cng thc nh thc Niu-tn ? - Ln bng tr li.- Tt c cc HV cn li tr li vo v nhp. - Nhn xt.1. BT1/SGK/57: c) 2. Bi mi: Hot ng 2 (7'): BT2/SGK/58 Hot ng ca GVHot ng ca HVNi dung - BT2/SGK/58 ?- Trnh by bi gii. 2. BT2/SGK/58: ( )131313 21301. 1 .kk kkx C xx=| | = |\ .- Cng thc nh thcNiu-tn ? -Khai trin

- H s ca x3 l phn no? - Nhn xt.- Chnh sa hon thin. - Ghi nhn kin thc. H s ca x3 l: Hot ng 3 (7'): BT3/SGK/58 Hot ng ca GVHot ng ca HVNi dung - BT3/SGK/58 ? - Cng thc nh thc Niu tn ? - Khai trin

- H s ca x2 l phn no? - Trnh by bi gii.- Nhn xt.- Chnh sa hon thin. - Ghi nhn kin thc. 3. BT3/SGK/58: H s ca x2 l: Hot ng 4 (7'): BT4/SGK/58 Hot ng ca GVHot ng ca HVNi dung - BT4/SGK/58 ? - Hng t khng cha x th x c s m bao nhiu? - Gi s hng l - Tm k ? - Tm C68? - Trnh by bi gii. - Nhn xt. - Chnh sa hon thin. - Ghi nhn kin thc. 4. BT4/SGK/58 Hot ng 5 (7'): BT5/SGK/58 Hot ng ca GVHot ng ca HVNi dung - BT5/SGK/58 ? - Cng thc nh thc Niu -tn? - Khai trin - x bao nhiu xut hin tng cc h s? (x = 1) - Trnh by bi gii.- Nhn xt.- Chnh sa hon thin. - Ghi nhn kin thc. 5. BT5/SGK/58 Hot ng 6 (8'): BT6/SGK/58 Hot ng ca GVHot ng ca HVNi dung - BT6/SGK/58 ? - Chia ht 100 l s pt ntn? - Cng thc nh thc Niu -tn? - Phn tch thnh tch c cha tha s 100? b) tng t cu a) c) phn tch - Trnh by bi gii.- Nhn xt.- Chnh sa hon thin. - Ghi nhn kin thc. 6. BT6/SGK/58 a) V. Hng dn hc v lm bi nh (2')- Cng c: Cu 1: Ni dung c bn c hc? 622xx| |+ |\ .162 12 C=( ) 1 3nx 29 90 5nC n = =( )8381kkkC xx| | |\ .( )173 4 x ( ) ( )17 173.1 4 1 1 = = ( ) ( )100 1001 10 , 1 10 + ( )10 1011 1 1 10 1 = + =( )2 2 2 9 9 1010 1010 10 ... 10 10 100 C C = + + + + Cu 2: Cng thc nh thc Niu -tn, tam gic Pa -xcan ?- Dn d:- Xem bi tp gii.-Lm BT cn li. -Xem trc bi php th v bin c Ngy son: 13/03/2011 Lp 11CB: 15/03/2011 Ngy dy:Lp 10B://2011 Lp 10D://2011 Tit 38 4. PHP TH V BIN C I/ Mc tiu: Qua bi hc HV cn nm: 1) Kin thc: Bit: -Php th ngu nhin; khng gian mu; bin c lin quan n php th ngu nhin. 2) K nng: - Xc nh c: Php th ngu nhin; khng gian mu; bin c lin quan n php th ngu nhin. 3) T duy: - Hiu th no l php th, kt qu ca php th, khng gian mu. - Hiu ngha xc sut ca bin c, cc php ton trn cc bin c. 4) Thi : - Cn thn trong tnh ton v trnh by. Tch cc hot ng tr li cu hi.- Qua bi hc HV bit c ton hc c ng dng trong thc tin. II. Chun b: 1. Gio vin: - Gio n, SGK, STK, phn mu. - Bng ph,- Phiu tr li cu hi 2. Hc vin:- Hc bi, lm bi tp, c trc bi, my tnh b ti. - dng hc tp: Thc k, compa,... III. Nhng iu cn lu - Lm cho hc vin nm vng c cc kin thc c bn . - Ch trng vic rn luyn k nng gii ton.IV. Tin trnh bi hc v cc hot ng: 1. Kim tra bi c: Xen ln trong bi dy 2. Bi mi: Hot ng 1 (10'): Php th, khng gian mu Hot ng ca GVHot ng ca HVNi dung - Gii thiu nh sgk - Nghe, suy ngh.- Tr li. I/ Php th, khng gian mu:- Php th ngu nhin?- Nhn xt.1) Php th: (sgk) Hot ng 2 (12'): Khng gian mu Hot ng ca GVHot ng ca HVNi dung - H1 sgk?- Khng gian mu? - Chnh sa hon thin- VD1 sgk ? - VD2 sgk ? - VD3 sgk ? - Kt qu c th xy ra? - c H1 sgk.- Tr li.- Nhn xt, ghi nhn.- Nghe, suy ngh.- Tr li, nhn xt.- Chnh sa hon thin. - Ghi nhn kin thc. 2) Khng gian mu:(sgk) K hiu: (c m ga) VD1 : (sgk) VD2 : (sgk) VD3 : (sgk) Hot ng 3 (8'): Bin c Hot ng ca GVHot ng ca HVNi dung - VD4 sgk ?- Bin c l g? - H2 (sgk) ? - Xem sgk. - Nghe, suy ngh. - Ghi nhn kin thc. II/Bin c: (sgk) Tp bin c khng thTp bin c chc chn Hot ng 4 (13'): Php ton trn cc bin c Hot ng ca GVHot ng ca HVNi dung - Nh sgk- Th no l bin c i? B =Akl g hai bc A, B ? - Hp, giao cc bin c? - Th no l bin c xung khc?( ) - VD5 sgk ? - Xem sgk, tr li. - Nhn xt. - Ghi nhn kin thc. - c VD5 sgk, nhn xt, ghi nhn. III/ Php ton trn cc bin c: (sgk) Bin c i ca bc A .K hiu:AK hiu: Ngn ng bin c

A l bin c A l bin c khngA l b.c chc chn C l bc :A hoc B C l bc: A v B A v B xung khc B =AA v B i nhu. V. Hng dn hc v lm bi nh: (2')- Cng c:Cu 1: Ni dung c bn c hc? Cu 2: Php th, khng gian mu, bin c i, bin c xung khc?- Dn d: - Xem bi v VD gii. - Lm BT1->BT7/SGK/63,64. ----------------------------------------------------------------------------------------------------------------- Ngy son: 13/03/2011 Lp 11CB: 15/03/2011 Ngy dy:Lp 10B://2011 Lp 10D://2011 Tit 39LUYN TP 4 OCOA B =CAcOA=CA=OC A B = C A B = A B =CI. Mc tiu: Qua bi hc HV cn nm vng: 1) Kin thc: Bit: -Php th ngu nhin; khng gian mu; bin c lin quan n php th ngu nhin.2) K nng: - Xc nh c: Php th ngu nhin; khng gian mu; bin c lin quan n php th ngu nhin; 3) T duy: - Hiu th no l php th, kt qu ca php th, khng gian mu. - Hiu ngha xc sut ca bin c, cc php ton trn cc bin c. 4) Thi :- Cn thn trong tnh ton v trnh by. Tch cc hot ng tr li cu hi.- Qua bi hc HV bit c ton hc c ng dng trong thc tin. II. Chun b: 1. Gio vin: - Gio n, SGK, STK, phn mu. - Bng ph,- Phiu tr li cu hi 2. Hc vin:- Hc bi, lm bi tp, c trc bi, my tnh b ti. - dng hc tp: Thc k, compa,... III. Nhng iu cn lu : - Lm cho hc vin nm vng c cc kin thc c bn . - Ch trng vic rn luyn k nng gii ton.IV. Tin trnh bi hc v cc hot ng: 1. Kim tra bi c: Hot ng 1 (6'): Kim tra bi c Hot ng ca GVHot ng ca HVNi dung Khng gian mu l g? Th no l bc i, bc xung khc? BT1/SGK/57 ? Chnh sa hon thin. - Ln bng tr li. - Tt c cc HV cn li tr li vo v nhp. - Nhn xt. 1. BT1/SGK/63 : a) b) 2. Bi mi: Hot ng 2 (6'): BT2/SGK/63 Hot ng ca GVHot ng ca HVNi dung BT2/SGK/63 ? Thc hin my h.ng c kq? b) A nhn xt kq ln gieo u?B nhn xt tng s chm hai ln? C nhn xt kq hai ln gieo? - Trnh by bi gii.- Nhn xt.- Chnh sa hon thin. - Ghi nhn kin thc. 2. BT2/SGK/63 : a)b) A l bin c: Ln u gieo xut hin mt 6 chm B l bin c: Tng s chm trong hai ln gieo l 8 C l bin c: Kt qu ca hai ln gieo nh nhau Hot ng 3 (6'): BT3/SGK/63 Hot ng ca GVHot ng ca HVNi dung BT3/SGK/63 ?- Trnh by bi gii. 3. BT3/SGK/63 : {}, , , , , , ,SSS SSN NSS SNSNNS NSN SNN NNNO ={ } , , , A SSS SSN SNS SNN ={ } , , B SNN NSN NNS ={ } \ C SSS = O( ) { }, /1 , 6 i j i j O= s sCc kq c th xy ra? Trng hp tng s trn hai th chn? Trng hp tach cc s trn hai th chn? - Nhn xt.- Chnh sa hon thin. - Ghi nhn kin thc. a) b) Hot ng 4 (6'): BT4/SGK/64 Hot ng ca GVHot ng ca HVNi dung BT4/SGK/64 ? Bin c i? Bin c xung khc? b) bin c: C hai bn trt nn B, C xung khc - Trnh by bi gii. - Nhn xt. - Chnh sa hon thin. - Ghi nhn kin thc. 4. BT4/SGK/64 a) Hot ng 5 (6'): BT5/SGK/64 Hot ng ca GVHot ng ca HVNi dung - BT5/SGK/64 ? - Khng gian mu ? - Kq ly th mu ? - Kq ly th mu trng? - Kq ly th ghi s chn? - Trnh by bi gii.- Nhn xt.- Chnh sa hon thin. - Ghi nhn kin thc. 5. BT5/SGK/64 a) b) Hot ng 6 (7): BT6/SGK/64 Hot ng ca GVHot ng ca HVNi dung BT6/SGK/64 ? Khng gian mu? S ln gieo khng qu 3? S ln gieo l 4? - Trnh by bi gii.- Nhn xt.- Chnh sa hon thin. - Ghi nhn kin thc. 6. BT6/SGK/64 a) b) Hot ng 7 (6'): BT7/SGK/64 Hot ng ca GVHot ng ca HVNi dung BT7/SGK/64 ? Khng gian mu? Kq ch s sau > hn ch s trc? Kq ch s trc gp i ch s sau?- Hai ch s bng nhu ? - Trnh by bi gii.- Nhn xt.- Chnh sa hon thin. - Ghi nhn kin thc.6. BT7/SGK/64 V. Hng dn hc v lm bi nh: (2')- Cng c: Php th, khng gian mu, bin c i, bin c xung khc?- Dn d: - Xem bi gii.- Xem trc bi xc xut ca bin c . ----------------------------------------------------------------------------------------------------------------- Ngy son: 17/03/2011 Lp 11CB: 18/03/2011 Ngy dy:Lp 10B://2011 ( ) { ( ) ( )( ) ( ) ( )}1, 2 , 1, 3 , 1, 4 , 2, 3 , 2, 4 , 3, 4O =( ) ( ) { }1,3 , 2, 4 A=( ) { }\ 1,3 B=OD1 2D A A A = =B C =C1 2 1 2A A A B A A = = ( ) ( )1 2 1 2C A A A A = 1 2D A A = { } 1, 2,...,10 O={ } 1, 2,3, 4,5 A={ } 7,8,9,10 B={ } 2, 4, 6,8,10 C ={ } , , , , S NS NNS NNNS NNNN O={ } , , A S NS NNS ={ } , B NNNS NNNN = Lp 10D://2011 Tit 41 5. XC XUT CA BIN C. I. Mc tiu: Qua bi hc HV cn nm vng: 1) Kin thc: Bit: - nh ngha xc sut ca bin c. - Bit tnh cht: P() = 0; P() = 1; 0 P(A) 1 2) K nng: - Tnh c xc sut ca bin c theo nh ngha. - Bit dng my tnh b ti h tr tnh xc sut. 3) T duy:- Hiu th no l xc sut ca bin c - Hiu c ngha ca xc sut . 4) Thi :- Cn thn trong tnh ton v trnh by. Tch cc hot ng tr li cu hi.- Qua bi hc HV bit c ton hc c ng dng trong thc tin. II. Chun b: 1. Gio vin: - Gio n, SGK, STK, phn mu. - Bng ph,- Phiu tr li cu hi 2. Hc vin:- Hc bi, lm bi tp, c trc bi, my tnh b ti. - dng hc tp: Thc k, compa,... III. Nhng iu cn lu :- Lm cho hc vin nm vng c cc kin thc c bn . - Ch trng vic rn luyn k nng gii ton.IV. Tin trnh bi hc v cc hot ng: 1. Kim tra bi c: Hot ng 1 (15'): Kim tra bi c Hot ng ca GVHot ng ca HV - Khng gian mu l g? - Gieo ng tin cn i v ng cht ngu nhin 2 ln . Xc nh khng gian mu?, bin c A: mt sp xut hin t nht mt ln? - Ln bng tr li.- Tt c cc HV cn li tr li vo v nhp. - Nhn xt.2. Bi mi: Hot ng 2 (28'): nh ngha c in ca xc sut Hot ng ca GVHot ng ca HVNi dung - VD1 sgk ?- H1 sgk?- Chnh sa hon thin. - nh ngha nh sgk. - Khng gian mu? s phn t khng gian mu? - Xc nh bin c A, B, C ? - S phn t cc bin c? - Tnh xc sut cc bin c? - c VD1 sgk, nhn xt, ghi nhn.- Trnh by bi gii.- Nhn xt.- Chnh sa hon thin. - Ghi nhn kin thc. - Tr li.- Nhn xt.- Chnh sa hon thin. - Ghi nhn kin thc . I. nh ngha c in ca xc sut1/ nh ngha: (sgk) Ch :(sgk) 2/ V d: VD2 : (sgk) VD3 : (sgk) VD4: (sgk) ( )( )( )n AP An=OV. Hng dn hc v lm bi nh: (2') - Cng c:Ni dung c bn c hc? - Bi tp: 1-4 sgk T74 Ngy son: 17/03/2011 Lp 11CB: 18/03/2011 Ngy dy:Lp 10B://2011 Lp 10D://2011 Tit42 5.XC XUT CA BIN C ( Tip theo) I. Mc tiu: Qua bi hc HV cn nm vng: 1) Kin thc: Bit: - nh ngha xc sut ca bin c. - Bit tnh cht: P() = 0; P() = 1; 0 P(A) 1 2) K nng: - Tnh c xc sut ca bin c theo nh ngha. - Bit dng my tnh b ti h tr tnh xc sut. 3) T duy:- Hiu th no l xc sut ca bin c. - Hiu c ngha ca xc sut. 4) Thi :Cn thn trong tnh ton v trnh by. Tch cc hot ng tr li cu hi.Qua bi hc HV bit c ton hc c ng dng trong thc tin. II. Chun b: 1. Gio vin: - Gio n, SGK, STK, phn mu. - Bng ph- Phiu tr li cu hi 2. Hc vin:- Hc bi, lm bi tp, c trc bi, my tnh b ti. - dng hc tp: Thc k, compa,... III. Nhng iu cn lu - Lm cho hc vin nm vng c cc kin thc c bn . - Ch trng vic rn luyn k nng gii ton. IV. Tin trnh bi hc v cc hot ng: 1. Kim tra bi c: Xen ln trong bi dy 2. Bi mi: Hot ng1 (25'): Tnh cht ca xc sut Hot ng ca GVHot ng ca HVNi dung - S phn tca bin c? - Xc sut cc bin c ny? - A, B xung khc s pt - H2 (sgk) ? - Chng minh h qu? - Khng gian mu? s phn t khng gian mu? - Xc nh cc bin c? - Xem sgk. - Nghe, suy ngh. - Tr li. - Ghi nhn k. thc. - H2 sgk. - Trnh by bi gii.- Nhn xt.- Chnh sa hon thin. II. Tnh cht ca xc sut:1/ nh l:(sgk) a)b), vi mi bin c A c) Nu A, B xung khc, th H qu: (sgk) , C O? A B ( ) ? P A B=( ) ( ) 0, 1 P P C= O=( ) 0 1 P A s s( ) ( ) ( ) P A B P A P B= +( )( ) 1 P A P A = - S phn t cc bin c? - Tnh xc sut cc bin c? - Ghi nhn k. thc. 2/ V d:VD5 : (sgk) VD6 : (sgk) Hot ng 2 (15'): K vng Hot ng ca GVHot ng ca HVNi dung - Khng gian mu? s phn t khng gian mu? - Xc nh cc bin c? - S phn t cc bin c? - Tnh xc sut cc bin c? c) Xc nh bin c A.B, s pt? - Xem sgk, tr li. - Nhn xt. - Ghi nhn kin thc. III. Cc bin c c lp, cng thc nhn xc sut: VD7 : (sgk) A v B l hai bin c c lp khi v ch khi P (A.B) = P(A) .P(B) V. Hng dn hc v lm bi nh: (5')- Cng c:Cu 1: Ni dung c bn c hc? Cu 2: Cch tnh xc sut ca bin c? th no l hai bin c c lp? - Dn d: - Xem bi v VD gii.- Lm BT1->BT7/SGK/74,75. - Xem trc bi lm bi luyn tp v n chng. Ngy son: 20/03/2011Ngy dy:Lp 11CB: 22/03/2011 Tit 43LUYN TP 5 I. Mc tiu: Qua bi hc Hv cn nm vng: 1) Kin thc: - Bin c, khng gian mu. - nh ngha c in ca xc sut. 2) K nng: Bit cch tnh xc sut ca bin c trong cc bi ton c th. 3) T duy: Hiu c ngha ca xc sut. 4) Thi :Cn thn trong tnh ton v trnh by. Tch cc hot ng tr li cu hi. Qua bi hc HV bit c ton hc c ng dng trong thc tin. II. Chun b: 1. Gio vin: - Gio n, SGK ,STK , phn mu. - Bng ph,- Phiu tr li cu hi 2. Hc vin:- Hc bi, lm bi tp, c trc bi, my tnh b ti. - dng hc tp: Thc k, compa,... III. Nhng iu cn lu :- Lm cho hc vin nm vng c cc kin thc c bn . - Ch trng vic rn luyn k nng gii ton.IV. Tin trnh bi hc v cc hot ng: 1. Kim tra bi c: Hot ng 1 (10'): Kim tra bi c Hot ng ca GVHot ng ca HVNi dung - Khng gian mu l g? - Xc sut ca bin c? - BT1/SGK/74 ? - Ln bng tr li.- Tt c cc HV cn li tr li vo v nhp. - Nhn xt. BT1/SGK/74 : a)b) 2. Bi mi: Hot ng 2 (10'): BT2/SGK/74 Hot ng ca GVHot ng ca HVNi dung - BT2/SGK/74 ? - Khng gian mu, s pt?- Xc nh bin c A, B? - S phn t cc bin c? - Tnh xc sut cc bin c? - Trnh by bi gii. - Nhn xt.- Chnh sa hon thin. - Ghi nhn kin thc. BT2/SGK/74 : a)b) Hot ng 3 (10'): BT3/SGK/74 Hot ng ca GVHot ng ca HVNi dung - BT3/SGK/74 ? - Khng gian mu, s pt?- Xc nh bin c A: Hai chic to thnh mt i, s pt? - Tnh xc sut cc bin c? - Trnh by bi gii.- Nhn xt.- Chnh sa hon thin. - Ghi nhn kin thc. BT3/SGK/74 : Hot ng 4 (13'): BT4/SGK/74 Hot ng ca GVHot ng ca HVNi dung - BT4/SGK/74 ? - Khng gian mu, s pt?- Phng trnh bc hai c nghim khi no? VN khi no? - Pt nghim nguyn l ntn? - Xc nh bin c A, B, C? - S phn t cc bin c? - Tnh xc sut cc bin c? - Trnh by bi gii.- Nhn xt . - Chnh sa hon thin. - Ghi nhn kin thc.b) c) BT4/SGK/74 : a) V/Hng dn hc v lm bi nh: (2')- Cng c: Ni dung c bn c hc ? ( )( )( )( )6 1 11;36 6 36n AP A P Bn= = = =O( ) { }, /1 , 6 i j i j O= s s( ) ( ) ( ) ( ) {( ) ( )}4, 6 , 6, 4 , 5, 5 , 5, 66, 5 , 6, 6A =( )( )( )( )1 1;4 2n AP A P Bn= = =O( ) { ( )( )( )} ( ) ( )341, 2, 3 , 1, 2, 4 , 1, 3, 4 ,2, 3, 44 n CO =O= =( ) { } ( ) 1,3, 4 ,1 A n A = =( ) ( ) { } ( ) 1, 2,3 , 2,3, 4 ,2 B n B = =( ) ( )( )2828; 44 128 7n C n AP AO= = == =( ) ( )113B A P B P A = = ={ } ( ) ( )13 , 16C n C P C = = ={ } ( ) ( )1, 2,3,..., 66 n O= O={ } { }( ) ( )2/ 8 0 = 3, 4, 5, 64 246 3A b bn A P A= eO >= = =- Bi tp: Lm BT 5 - 7 sgk T74. Ngy son: 21/03/2011 Ngy dy: Lp 11CB: 23/03/2011 Tit 45LUYN TP 5 (Tip theo) I. Mc tiu: Qua bi hc HV cn nm vng: 1) Kin thc: - Bin c, khng gian mu. - nh ngha c in ca xc sut. 2) K nng: - Bit cch tnh xc sut ca bin c trong cc bi ton c th. 3) T duy: - Hiu c ngha ca xc sut. 4) Thi : - Cn thn trong tnh ton v trnh by. Tch cc hot ng tr li cu hi.- Qua bi hc HV bit c ton hc c ng dng trong thc tin. II. Chun b: 1. Gio vin: - Gio n, SGK, STK, phn mu. - Bng ph- Phiu tr li cu hi 2. Hc vin:- Hc bi, lm bi tp, c trc bi, my tnh b ti. - dng hc tp: Thc k, compa,... III. Nhng iu cn lu : - Lm cho hc vin nm vng c cc kin thc c bn. - Ch trng vic rn luyn k nng gii ton. IV. Tin trnh bi hc v cc hot ng: 1.Kim tra bi c: Xenln trong bi dy 2.Bi mi: Hot ng 1 (13'): BT5/SGK/74 Hot ng ca GVHot ng ca HVNi dung - BT5/SGK/74 ? - Khng gian mu, s pt?- Xc nh bin c A, B, C? - S phn t cc bin c? - B l bc:t nht mt con t, i B nh th no? s pt? - Tnh xc sut cc bin c? - Trnh by bi gii.- Nhn xt.- Chnh sa hon thin. - Ghi nhn kin thc.c) BT5/SGK/74 : a) b) Hot ng 2 (15'): BT6/SGK/74 Hot ng ca GVHot ng ca HVNi dung - BT6/SGK/74 ? - Khng gian mu, s pt?- Xc nh bin c: A : Nam n ngi i din nhu - Trnh by bi gii.- Nhn xt.- Chnh sa hon thin. - Ghi nhn kin thc.BT6/SGK/74: a) ( ) ( )2 24 436. 36270725n C C C P C = = =( )452270725 n C O= =( ) ( )4411270725n A C P A = = =( )448194580 n B C = =( )( )( )1945801270725P B P B P B = = ( ) 4! 24 nO= =( ) 2.2.2.2 16 n A= =B : N ngi i din nam? - S phn t cc bin c? - Tnh xc sut cc bin c? b) Hot ng 3 (15'): BT7/SGK/75 Hot ng ca GVHot ng ca HVNi dung - BT7/SGK/75 ? - Khng gian mu, s pt?- Th no l hai bin c c lp? - Xc nh bin c A, B ? - S phn t cc bin c? C : Ly hai qu cng mu. Xc nh bc C? s pt? D : Ly hai qu khc mu. Xc nh bc D? D, C lin quan ntn? - Tnh xc sut cc bin c? - Trnh by bi gii. - Nhn xt. - Chnh sa hon thin. - Ghi nhn kin thc.b) .Do xung khc nn A, B c lp c) BT7/SGK/75: a) V/Hng dn hc v lm bi nh: (2')- Cng c: Cu 1: Ni dung c bn c hc? Cu 2: Cch tnh xc sut ca bin c? th no l hai bin c c lp? - Dn d: - Xem bi v VD giiBT1X ->BT7/SGK/T74,75. - Xem trc bi lm bi tp n chng.Ngy son: 24/03/2011Ngy ging: 25/03/2011Lp: 11CB Tit 46S DUNG MY TNH B TI TNH T, XC SUT I. Mc tiu: 1. V kin thc:- Gii thiu cho HS cch s dng my tnh FX500 MS tnh s hon v, s chnh hp, s t hp 2. V k nng: - Rn k nng s dng my tnh FX500 MS tnh s hon v, s chnh hp, s t hp - Vn dng vo vic gii mt s cc bi ton 3 . V t duy, thi : -Thi cn thn, chnh xc -Thy c nhng ng dng cng ngh trong ton hc, thy c nhng tin ch, tit kim thi gian v cng sc trong vic tnh ton khi s dng my tnh II. Chun b: 1. Gio vin: My tnhFX500 MS, dng ging dy 2. Hc sinh: My tnhFX500 MS, dng hc tp III/ Nhng iu cn lu : Ch trng vic rn luyn k nng gii ton bng my tnh.( ) ( )113B A P B P A = = = ( )16 224 3P A = =. . C AB AB = . , . A B A B( ) ( ) ( )24 24 48 12100 100 100 25P C P AB P AB === + = =( ) ( )13125D C P D P C = = =( ) { }, /1 6;1 10 A i j i j = s s s s( ) { }, /1 10;1 4 B i j i j = s s s s( ) ( )6.10 6 10.4 4;10.10 10 10.10 10P A P B = = = =( ) { }. , /1 6;1 4 AB i j i j = s s s s( ) ( ) ( )6.4.10.10P AB P A P B = =IV. Tin trnh bi ging: 1. Kim tra bi c: Khng 2. Bi mi: Hot ng 1: Tnh cc s hon v Bi 1: Tnh 4!, 7!, 10! , 15! GV: Lm mu tnh 4! n=14 SHIFT X KQ: 24 HS: Lm tng t vi cc cn li Hot ng 2: Tnh cc s chnh hp Bi 1: Tnh 5 10 2 37 13 9 11A , A , A , A GV: lm mu tnh 57An 7 SHIFT nCr 5=KQ: 2 520 HS: Lm tng t vi cc cn li Hot ng 3: Tnh cc s t hp Bi 3: Tnh 5 10 2 37 13 9 11C , C ,C ,C GV: lm mu tnh 57C n 7 nCr 5=KQ: 21 HS: Lm tng t vi cc cn li Hot ng 4: Vn dng vo vic gii cc bi ton thc t v mt s bi ton khc Bi 4: Trong mt phng c bao nhiu hnh bnh hnh c to thnh t 5 ng thng song song v 7 ng thng song song vi 5 ng thng trn Gii: C 2 25 7C C .hnh bnh hnh Dng my tnh n5 nCr . 7 nCr22=KQ: 210 (hnh bnh hnh) Bi 5: Dng my tnh v cng thc nh thc Niutn khai trin biu thc ( )52a 3 +Gii:p dng cng thc nh thc Niutn ta c 5 0 5 1 4 1 2 3 2 3 2 3 4 4 5 55 5 5 5 5 5(2a+3)=C (2a) +C (2a) 3 +C (2a) 3 +C (2a) 3 +C 2a3 +C 3n5 nCr . 2 ^ 50= nh 32 5 nCr ^1. 24. 3 = nh 240 5 nCr ^ ^2. 23 . 32= nh 720 5 nCr ^ ^3. 22 . 33= nh 1080 5 nCr ^ ^4 . 21 . 34 = nh 810 5 nCr ^5. 35 = nh 1215 Kt qu: 5 5 4 3 2(2a+3)=32a +240 a +720 a +1080 a +810 a+1215 V. Hng dn hc sinh hc v lm bi nh: -Thnh tho vic s dng my tnh b ti tnh s hon v, chnh hp, t hp - n tp chng II Ngy son://2011Ngy dy://2011Lp: 11CB Tit 49N TP CHNG II. I. Mc tiu: Qua bi hc HV cn nm vng: 1) Kin thc: - Qui tc cng, qui tc nhn, hon v, chnh hp, t hp, nh thc Niu tn. - Php th, bin c, khng gian mu. - nh ngha c in ca xc sut, t/c ca xc sut. 2) K nng: - Bit cch tnh s phn t ca tp hp da vo qui tc cng, nhn. Phn bit hon v, chnh hp, t hp. Bit khi no dng chng tnh s phn t tp hp. Bit cch biu din bin c bng li v tp hp. - Bit cch xc nh khng gian mu, s pt, tnh xc sut ca bin c trong cc bi ton c th. 3) T duy: - Hiu c hon v, chnh hp, t hp. Bit khi no dng chng tnh s phn t tp hp. 4) Thi : - Cn thn trong tnh ton v trnh by. Tch cc hot ng tr li cu hi. Qua bi hc HV bit c ton hc c ng dng trong thc tin. II. Chun b: 1. Gio vin: - Gio n, SGK ,STK , phn mu. - Bng ph,- Phiu tr li cu hi 2. Hc vin:- Hc bi, lm bi tp, c trc bi, my tnh b ti. - dng hc tp: Thc k, compa,... III. Nhng iu cn lu :- Lm cho hc vin nm vng c cc kin thc c bn . - Ch trng vic rn luyn k nng gii ton.IV.Tin trnh bi hc v cc hot ng: 1. Kim tra bai c: Hot ng 1 (10'): Kim tra bi c Hot ng ca GVHot ng ca HVNi dung - Pht biu qt cng, nhn, cho vd? - Khng gian mu l g? - Xc sut ca bin c? - BT4/SGK/76 ? - G. s s to thnh tm s cch chn a, b, c, d ? - Ln bng tr li.- Tt c cc HV cn li tr li vo v nhp. - Nhn xt.Vy s chn c 4 ch s khc nhu: 120 + 300 = 420 (s) . BT4/SGK/76 : a) 6.7.7.4 = 1176 (s) b) d = 0 : : d c 3 cch chn, a c 5 cch chn, bc c cch chn . S cch: 3.5.20 = 300 2. Bai mi: Hot ng 2 (10'): BT5/SGK/76 abcd36120 A=0 d =2520 A =Hot ng ca GVHot ng ca HVNi dung - BT5/SGK/76 ? - Khng gian mu, s pt?- Xc nh bin c A, B? - S phn t cc bin c? - Tnh xc sut cc bin c? b) Ba nam ngi cnh nhu th c th xp v tr no ? my cch? - S cch xp n vo cc ch cn li? Theo qui tc nhn s cch? - Trnh by bi gii.- Nhn xt.- Chnh sa hon thin. - Ghi nhn kin thc. BT5/SGK/76 : a) Nam ngi gh 1 c 3!.3! cch N ngi gh 1 c 3!.3! cch Theo qui tc cng:b) Hot ng 3 (10'): BT6/SGK/76 Hot ng ca GVHot ng ca HVNi dung - BT6/SGK/76 ? - Khng gian mu, s pt?- Xc nh bin c A, B ? - Cng mu ln tn? t nht 1 qu trng l ntn? B : t nht 1 qu trng, th bc i l ntn? s pt?- S phn t cc bin c? - Tnh x.sut cc bin c? - Trnh by bi gii.- Nhn xt.- Chnh sa hon thin. - Ghi nhn kin thc. BT6/SGK/76 : a) b)B : 4 qu ly ra t nht 1 qu trng : C 4 qu u en, Hot ng 4 (13'): BT7/SGK/77 Hot ng ca GVHot ng ca HVNi dung - BT7/SGK/77 ? - Khng gian mu, s pt?- Xc nh bin c A? bin c i bin c A ntn? - S phn t cc bin c? - Tnh xc sut cc bin c? - Trnh by bi gii.- Nhn xt.- Chnh sa hon thin. - Ghi nhn kin thc. BT7/SGK/77 : V. Hng dn hc v lm bi nh: (2') - Cng c: Ni dung c bn c hc? - Bi tp: Lm BT 8-9 sgk T77. Ngy son://2011Ngy dy: //2011Lp: 11CB Tit 50 ( )( )( )110n AP An= =O( ) 6! nO=( ) ( )23. 3! n A=( ) ( )14.3!.3!5n B P B = =( )( )1 2091 1210 210P B P B = = =( ) ( )( )( )( )4 4 410 6 4210; 1616 8210 105n C n A C Cn AP AnO= = = + == = =OB( )441 n B C = =( )( )351 16P A P A| |= = |\ .( ) { }, , /1 , , 6 a b c a b c O= s s( )36 216 nO= =( ) ( )3355 6n A P A| |= = |\ .N TP CHNG II ( tip theo ) I. Mc tiu: Qua bi hc HV cn nm vng: 1) Kin thc: - Qui tc cng, qui tc nhn, hon v, chnh hp, t hp, nh thc Niu-tn. - Php th, bin c, khng gian mu . - nh ngha c in ca xc sut, t/c ca xc sut. 2) K nng: - Bit cch tnh s phn t ca tp hp da vo qui tc cng, nhn. Phn bit hon v, chnh hp, t hp. Bit khi no dng chng tnh s phn t tp hp . - Bit cch biu din bin c bng li v tp hp . - Bit cch xc nh khng gian mu, s pt, tnh xc sut ca bin c trong cc bi ton 3) T duy: Hiu c hon v, chnh hp, t hp.Bit khi no dng chng tnh s phn t tp hp. 4) Thi :Cn thn trong tnh ton v trnh by. Tch cc hot ng tr li cu hi Qua bi hc HV bit c ton hc c ng dng trong thc tin II. Chun b: 1. Gio vin: - Gio n, SGK ,STK , phn mu. - Bng ph,- Phiu tr li cu hi 2. Hc vin:- Hc bi, lm bi tp, c trc bi, my tnh b ti. - dng hc tp: Thc k, compa,... III. Nhng iu cn lu : - Lm cho hc vin nm vng c cc kin thc c bn . - Ch trng vic rn luyn k nng gii ton.IV. Tin trnh bi hc v cc hot ng: 1.Kim tra bai c: Xen ln trong bai dy 2.Bi mi: Hot ng 1 (15'): BT8/SGK/77 Hot ng ca GVHot ng ca HVNi dung - BT8/SGK/77 ? - Lc gic c bao nhiu cnh, ng cho? khng gian mu, s pt?- Xc nh bin c A, B, C? - S phn t cc bin c? - Tnh xc sut cc bin c? - Trnh by bi gii.- Nhn xt.- Chnh sa hon thin. - Ghi nhn kin thc.c) BT8/SGK/77 : a) b) Hot ng 2 (13'): BT9/SGK/77 Hot ng ca GVHot ng ca HVNi dung - BT9/SGK/77 ? - Khng gian mu, s pt?- Xc nh bin c A, B ? - S phn t cc bin c? - Tnh xc sut cc bin c? - Trnh by bi gii.- Nhn xt.- Chnh sa hon thin. - Ghi nhn kin thc.b)BT9/SGK/77 : a) ( ) ( )135n C P C = =( )2615 n C O= =( ) ( )6 2615 5n A P A = = =( ) ( )2636 95n B C P A = = =( ) ( )194n B P B = =( ) { }( ), / 1 , 636i j i jnO = s s O=( ) { }, / , 2, 4, 6 A i j i j = = Hot ng 3 (15'): BTTN/SGK/76 Hot ng ca GV Hot ng ca HVNi dung -BTTN/SGK/76 ? - Trnh by bi gii.- Nhn xt.- Chnh sa hon thin. - Ghi nhn kin thc. BTTN/SGK/76 101112131415 BD B A C C V. Hng dn hc v lm bi nh: (2')- Cng c: Ni dung c bn c hc ? - Dn d: Xem bi tp gii n tp chun b cho kim tra ht chng.

Ngy son: //2011Ngy kim tra: Tit 53KIM TRA 45 I. Mc tiu:1. Kin thc- nh gi qu trnh nhn thc ca hc sinh - Kim tra k nng gii ton v kin thc c bn ca chng II. ( ) ( )9 1936 4n A P A = = =2. K nng- Tnh -c xc sut ca cc bin c- Tnh -c s cc: hon v, t hp, chnh hp. Phn bit -c t hp v chnh hp - Rn luyn t duy l gc khi gii ton. 3. Thi - H/v t nh gi kt qu hc tp, tch cc hn trong hc tp.- T gic, tch cc, nghim tc lm bi II. Chun b: 1, Gio vin: kim tra, p n biu im 2, Hc vin: My tnh b ti, giy nhp, giy kim tra, n tp chng II. III. Tin trnh bi mi: 1. n nh lp: 2. kim tra: A. Phn trc nghim khch quan: 4 1 ) Mt lp hc c 40 hc vin: 26 nam v 14 n. S cch chn 3 hc vin tham gia lao ng tng v sinh ton trng l: a) A340 b) C340c) A326 d) C314 2) S cc s t nhin c hai ch s l :a)50 b)45c)90d) 35 3) S cch xp ch ngi cho 6 hc vin trn mt dy bn gm 6 ch l: a)12b)120 c)36d) 720 4) Gieo mt ng tin kim loi cn i v ng cht hai ln ta c: a) n( O) = 2b) n( O ) = 3c) n( O ) = 4d) n( O) = 6 B. Phn t lun: 6 Mt hp ng 7 qu cu, trong c 3 qu mu trng v 4 qu mu en. Ly ngu nhin 2 qu. 1) Tnh n( O). 2) Tnh xc sut sao cho hai qu ly ra cng mt mu. 3. p n v biu im: A. Phn trc nghim khch quan: Cu hi 1234 p n bbdc Biu im1111 B. Phn t lun: 1) Ta c :n( O) =C27 =212 2) GiA l bin c : " Hai qu cu ly ra cng mt mu"1 Ta c:n(A)=C23 + C24=92 Vy: P(A)= ( ) 9 3( ) 21 7nnA= =O 1 CHNG III:DY S. CP S CNG. CP S NHN. Ngy son: //2011 Ngy dy: Tit 54 1. PHNG PHP QUY NP TON HC I. Mc tiu: Qua bi hc HV cn nm vng: 1) Kin thc: Hiu th no l phng php quy np ton hc, trnh t gii bi ton. 2) K nng:Bit cch la chn v s dng phng php 1uy np ton hc gii cc bi ton mt cch hp l . 3) T duy: Hiu th no l phng php quy np ton hc. 4) Thi : Cn thn trong tnh ton v trnh by. Tch cc hot ng tr li cu hi. Qua bi hc HV bit c ton hc c ng dng trong thc tin. II. Chun b: - Gio n, SGK, STK, phn mu. - Bng ph. Phiu tr li cu hi. III. Nhng iu cn lu : - Lm cho hc vin nm vng c cc kin thc c bn . - Ch trng vic rn luyn k nng gii ton.IV. Tin trnh bi hc v cc hot ng: 1. Kim tra bi c: Hot ng 1 (10'): Kim tra bi c Hot ng ca GVHot ng ca HV - Mnh l g? - Cho vd vi m cha bin? - Ln bng tr li. - Tt c cc HV cn li tr li vo v nhp. - Nhn xt.2. Bi mi: Hot ng 2 (15'): Phng php quy np ton hc Hot ng ca GVHot ng ca HVNi dung - H1/SGK/ ?- Chng t KL ng, ta CM ng vi mi trng hp? - Chng t KL sai, ta ch ra mt trng hp sai? - Trnh by nh sgk. - Xem sgk . - Nghe, suy ngh. - Tr li.- Nhn xt.- Chnh sa hon thin. - Ghi nhn kin thc. I. Phng php quy np ton hc: (sgk) B1 : Kim tra m ng vi n = 1 B2 : Gi thit m ng vi n = k Ta cm m ng vi n = k + 1 Kt lun m ng n e N* Hot ng 3 (18'): V d p dng Hot ng ca GVHot ng ca HVNi dung - VD1 sgk ?- H2/SGK ? - VD2 sgk ? - Ktra vin = 1 lm ntn? - Gi s ng vi n = k ta c g? - Ta cn chng minh g? - H3/SGK ? - Xem sgk. - Trnh by bi gii.- Nhn xt.- Chnh sa hon thin. - Ghi nhn kin thc. II. V d p dng: V d 1: (sgk) V d 2: (sgk) Ch : (sgk) V. Hng dn hc v lm bi nh: (2') - Cng c: Ni dung c bn c hc ? - Dn d: Xem bi v VD gii.BT->BT/SGK. Xem trc bi lm bi luyn tp. Ngy son: //2011Ngy dy: Tit 55 LUYN TP 1. I. Mc tiu: Qua bi hc HV cn nm vng: 1) Kin thc: - Hiu th no l phng php quy np ton hc, trnh t gii bi ton . 2) K nng: - Bit cch la chn v s dng phng php 1uy np ton hc gii cc bi ton mt cch hp l . 3) T duy: Hiu th no l phng php quy np ton hc . 4) Thi :Cn thn trong tnh ton v trnh by. Tch cc hot ng tr li cu hiQua bi hc HV bit c ton hc c ng dng trong thc tin. II. Chun b: - Gio n, SGK, STK, phn mu. - Bng ph.Phiu tr li cu hi. III. Nhng iu cn lu - Lm cho hc vin nm vng c cc kin thc c bn . - Ch trng vic rn luyn k nng gii ton.IV. Tin trnh bi hc v cc hot ng: 1. Kim tra bi c: Hot ng 1 (7'): Kim tra bi c Hot ng ca GVHot ng ca HVNi dung -Trnh by Phng php quy np ton hc? - Ln bng tr li. - Tt c cc HV cn li tr li vo v nhp. - Nhn xt. Phng php quy np ton hc: (sgk) B1 : Kim tra m ng vi n = 1 B2 : Gi thit m ng vi n = k . Ta cm m ng vi n = k + 1 Kt lun m ng n e N* 2. Bi mi: Hot ng 2 (8'): BT1/82/SGK Hot ng ca GVHot ng ca HVNi dung - BT1/82/SGK ? - nh ngha nh sgk- Cho HV bit c ngha ca k vng- Trnh by bi gii.- Nhn xt. - Chnh sa hon thin. - Ghi nhn kin thc.BT1/82/SGK Hot ng 3 (13'): BT2/82/SGK Hot ng ca GVHot ng ca HVNi dung - BT1/82/SGK ? - Hng dn hc vin s dng phng php quy np ton hc vo gii bi tp.- Trnh by bi gii.- Nhn xt.- Chnh sa hon thin. - Ghi nhn kin thc. BT2/82/SGK:a) t Sn=n3+3n2+5n Vi n =1 ,s1=9 + 3 Gi s: Sk=k3+3k2+5k +3 Ta c/m: Sk+1=(k+1)3+3(k+1)2+5(k+1) + 3 Tht vy:Sk+1=k3+3k2+5k+3k2+9k+9 = Sk+3(k2+3k+3) +3 Nn Sk+1 chia ht cho 3 vi mi n eN* Hot ng 4 (15'): BT3/82/SGK Hot ng ca GVHot ng ca HVNi dung - BT3/82/SGK ? - Hng dn hc vin s dng phng php quy np ton hc vo gii bi tp.- Trnh by bi gii.- Nhn xt.- Chnh sa hon thin. - Ghi nhn kin thc. BT3/82/SGK:b) Vi n=1 ,VT=8, VP=7, vy btng vi n =1. Gi s: btng vi n =k (k=1) 2k+1>2k+3 (*) Ta c/m: btng vi n =k+1 2k+2>2(k+1)+3Hay : 2k+2>2k+5 Nhn(*)vi 2 v, ta c : 2k+2>4k+6>2k+5+2k+1 Nn 2k+2>2k+5 Vy 2n+1>2n+3 vi n e N* V. Hng dn hc v lm bi nh: (2')- Cng c:Ni dung c bn c hc? - Dn d:- Xem bi v VD gii.- BT/SGK. - Chun b bi "Dy s". Ngy son: //2011 Ngy dy: Tit 582. DY S. I. Mc tiu: Qua bi hc HV cn nm vng: 1) Kin thc: - Hiu th no l dy s. - Nm chc khi nim dy s, cch cho dy s, dy s hu hn, v hn. - Cc tnh cht tng, gim v b chn ca dy s. 2) K nng: - Bit cch gii cc bi tp v dy s nh: Tm s hng tng qut, Xt tnhtng, gim v b chn ca dy s. - Vit c dy s cho bng 3 cch. 3) T duy: Hiu v vn dng thnh tho cch tnh dy s. 4) Thi : Cn thn, chnh xc trong tnh ton v trnh by. Qua bi hc HV bit c ton hc c ng dng trong thc tin. II. Chun b: - Gio n, SGK, STK, phn mu. - Bng ph. Phiu tr li cu hi. III. Nhng iu cn lu :- Lm cho hc vin nm vng c cc kin thc c bn . - Ch trng vic rn luyn k nng gii tonIV. Tin trnh bi hc v cc hot ng: 1.Kim tra bi c: Xen ln trong bi dy 2.Bi mi: Hot ng 1 (8'): nh ngha Hot ng ca GVHot ng ca HVNi dung - H 1: sgk- Qua hot ng trn cc em c nhn xt g v hm s - HV suy ngh, tr li. - Mt HV ln bng trnh by. -TtcccHVcnlinhn 1/ nh ngha dy s: ( sgk) cho? - Chnh sa hon thin.- VD1:sgk. xt. - HV suy ngh, tr li. - Xem sgk. - HV suy ngh tr li. - Ghi nhn kin thc. Hot ng 2 (8'): nh ngha dy s hu hn Hot ng ca GVHot ng ca HVNi dung - Dy s nh th no c gi l dy s hu hn?- GV nu nh ngha sgk. - VD2:sgk. - HV lng nghe tr li. - Ghi nhn kin thc. - HV ln