www.MATHVN.com Dnh cho hc sinh lp 12 n thi i hc- Cao ngB thi t lun- mn Ton 1 Vn Ph Quc- GV. Trng i hc Qung Nam D: 0982 333 443;0934 825 925 MATHVN.COM - Ton hc Vit Nam 1 A- PHN CHUNG(7,0 im) Cu I (2,0 im) . Cho hm s: y = x4 2x2 3(C). 1. Kho st s bin thin v v th (C) ca hm s. 2. Tm m ng thngy = mct th (C) ti bn im phn bit M, N, P, Q ( sp th t t tri sang phi) sao cho di cc on thng MN, NP, PQ c gi s l di 3 cnh ca mt tam gic bt k. Cu II (2,0 im) 1. Gii phng trnh:sin x.sin 4 x = 2 2 cos | t x | 4 3 cos2 x.sin x.cos 2 x 6 | 2. Gii h phng trnh: 2 x2 + 3 y \. y 2 + 8x= 1 ( x,y e ) . x ( x + 8) + y ( y + 3) = 13 4 1x + ex Cu III (1,0 im) . Tnh tch phn: I = } 1 + 4xxe2 x dx . Cu IV (1,0 im).Tnh th tch khi t din ABCD bit AB = a, AC = b, AD = c vBAC = CAD = DAB = 600 . Cu V (1,0 im). Chng minh phng trnh: duy nht. x x+1 = ( x + 1)xlun cnghim thc dng B- PHN RING(3,0 im). Th sinh ch c chn mt trong hai phn B.1. CHNG TRNH CHUN Cu VI a(2,0 im) 1.TrongmtphngOxy,chongthngd : x y + 1 = 0vngtrn (C ) : x2 + y2 + 2x 4 y = 0 . Tm ta im M thuc ng thng d m qua k c hai ng thng tip xc vi ng trn(C ) ti A v B sao choAMB = 600 . 2. Trong khng gian Oxyz, cho3 imA (a; 0; 0 ) ,B (0; b; 0),C (0; 0; c )vi a,b, cl cc sdng thay ivthamna2 + b2 + c2= 3 .Xcnh a,b,csaochokhong cch t gc to O (0; 0; 0)n mt phng( ABC )t gi tr ln nht. CuVIIa(1,0 im). Tma,b ephngtrnhz 2 + az + b = 0cnhnsphc z = 1 + ilm nghim. B.2. CHNG TRNH NNG CAO Cu VI b(2,0 im) 1. Trong mt phng Oxy, cho prabol( P ) : y = x2 . Vit phng trnh ng thng d i qua M(1; 3) sao cho din tch hnh phng gii hn bi (P) v d t gi tr nh nht. 2. Trong khng gian vi h to Oxyz, cho hai imA (1; 5; 0 ) ,B (3; 3; 6 )v ng thng d: x + 1 = y 1 = z . Xc nh v tr ca im C trn ng thng d din tch tam 212 gic ABC t gi tr nh nht. www.MATHVN.com Dnh cho hc sinh lp 12 n thi i hc- Cao ngB thi t lun- mn Ton 2 Vn Ph Quc- GV. Trng i hc Qung Nam D: 0982 333 443;0934 825 925 MATHVN.COM - Ton hc Vit Nam 2 () Cu VII b (1,0 im).Gii phng trnh: log 2 x2 + x + 1 log x2 x + 1= 1 log 3 x4 + x2 + 1+ log x4 x2 + 1 . 4()=1() 2 3 2() BI GII A- PHN CHUNG Cu I 1. Hc sinh t gii. 2.Phng trnh honh giao im ca ng thngy = m v th (C): x 4 2x2 3 = m x 4 2x 2 3 m = 0 (1) tt = x 2 ( t > 0 ) .Phng trnh trn thnh:t 2 2t 3 m = 0 (2 ) Gix1 , x 2 , x3 , x 4 ln lt l honh cc giao im M, N, P, Q. Khi x1 , x 2 , x3 , x 4l nghim ca phng trnh (1). Davoth,tathyviiukin:4 < m < 3thphngtrnh(2)chainghim l: t1 = 1 m + 4; t 2= 1 + m + 4 . Suy rax1 = t 2 ,x 2= t1 ,x3=t1 ,x 4=t2 Ta cMN = PQ = x 4 x3, NP = x3 x 2= 2x3 a + b > c rng: iu kin ba s dng a, b, c l di 3 cnh ca mtAl: b + c > a . c + a > b VMN = PQnn MN, NP, PQ l di 3 cnh ca mt tam gic bt k nn ta ch cn: MN + PQ > NP 2PQ > NP 2 ( x4 x3 ) > 2x3 x4> 2x3 t > 2 t t > 4t 1 + m + 4 > 41 m + 4 m + 4 > 3 m > 91 hay 2121 525 Kt hp vi iu kin : Cu II 1.Ta c: 4 < m < 3ta c: 91 < m < 3. 25 sin x.sin 4 x = 2 2 cos | t x | 3 cos x.sin 4 x 6 | sin 4x (sin x + \. 3 cos x ) = 2 2 cos | t x | 6 | \. sin 4x.| sin t sin x + cos t cos x | = 2 cos | t x | (sin 4 x 2 )cos | t x | = 0 66 | 6 | 6 | \.\.\. www.MATHVN.com Dnh cho hc sinh lp 12 n thi i hc- Cao ngB thi t lun- mn Ton 3 Vn Ph Quc- GV. Trng i hc Qung Nam D: 0982 333 443;0934 825 925 MATHVN.COM - Ton hc Vit Nam } . 4 } aH B cos | t x | = 0 ( vsin 4x 2 < 0 ) t x = t + kt x = 2t + kt ( k e ) . 6 | 623 \. 2. iu kin:x 2 + 3y > 0 , y2 + 8x > 0 tu =x2 + 3y, v =y2 + 8x(u, v > 0 ) H phng trnh thnh: 2u v = 1 v = 2u 1 v = 2u 1 u2 + v2= 13 u 2 + v2= 13 u2 + ( 2u 1)2= 13 v = 2u 1 v = 2u 1 5u 2 4u 12 = 0 u = 2

6 u = 2 . v = 3

u = 5 Khi : x 2 + 3y = 2 x 2 + 3y = 4 y = 4 x2 3 2 y2 + 8x= 3 y2 + 8x = 9 | 4 x2| | + 8x = 9 y = 4 x2 3 y = \ 3 . 4 x2 3 y = 4 x2 3

x = 1

y = 1 =

= x4 8x2 + 72x 65 = 0 ( x 1) ( x + 5) ( x2 4x + 13) = 0 x1

x5 x = 5 y = 7 Kthpviiukintautathuctphpnghimcahphngtrnhl: S = {(1;1), (5; 7)} . Cu III 4 1x + ex 4 111 2 |11| Ta c:I = } 4x + xe2 x dx = } ++ 4xxex dx e2 x =

\ 2 x +x| dx e. 111 4 |11| 4 11 A =

\ 2 x + ex| dx = x e x 1 = 1 + ee4 Cu IV Gi s 1 c a = min {a, b, c} Trn cnh AC ly im E, AD ly im F sao cho F AB = AE = AF = a . T din ABEF c bn mt l cc tam gic u bng D nhau nn l t din u cnh bng a.E b C www.MATHVN.com Dnh cho hc sinh lp 12 n thi i hc- Cao ngB thi t lun- mn Ton 4 Vn Ph Quc- GV. Trng i hc Qung Nam D: 0982 333 443;0934 825 925 MATHVN.COM - Ton hc Vit Nam BH.S 2 Ta d dng tnh c VABEF= a 3 2 . 12 Gi H l chn ng cao h t B. 1 V AAEF 1 AEAF sin 600 a Ta c: ABEF = 3= 2=. Suy ra VABCD VABCD= 1 BH.S 3 abc 2 12 AACD . 1 ACAD sin 600bc 2 Cu V Ta c:x x +1 = (x + 1)x (x > 0) (x + 1) ln x x ln (x + 1) = 0 Xt hm s:f ( x ) = ( x + 1) ln x x ln (x + 1)(x > 0) f ' ( x ) = ln |1 + =1 | + 1 + 1 x | xx + 1 \. Chng minh bt ng thc c bnln (1 + t ) < t t > 0 ta suy ra ln |1 + =1 | > 1 x | x Do f ' ( x ) > 1 + 1 + 1 = 1 > 0 f ( x ) \. ng bin trn (0; + ) (1) xxx + 1x + 1 Mt khc: dof ( x )lin tc trn(0; + )vf (2 ) f (3) = (ln 8 ln 9 ) (ln 81 ln 64) < 0suy ra tn tix 0 e (2; 3) c (0, + )sao chof ( x 0 ) = 0 T (1) v (2) iu phi chng minh. B- PHN RING B.1. CHNG TRNH CHUN Cu Via 1.Vit li(C )di dng( x + 1)2+ ( y 2)2= 5 (2) A I - Vy(C )c tm I ( 1; 2 ) ,bn knhR =5 . M M e d M (t; t + 1) B Theo gi thitAMB = 60o Suy raAMI= 30o MI = 2IA = 2R = 2 5 hay t = 3 M (3; 4) MI 2= 20 (1 t )2 + (1 t )2= 20 t 2= 9

t = 3 M ( 3; 2) 2. Phng trnh mt phng (ABC): x + y + z = 1 abc d = dO; (ABC)= 1 1 + 1 + 1 a 2 b2 c2 www.MATHVN.com Dnh cho hc sinh lp 12 n thi i hc- Cao ngB thi t lun- mn Ton 5 Vn Ph Quc- GV. Trng i hc Qung Nam D: 0982 333 443;0934 825 925 MATHVN.COM - Ton hc Vit Nam 2 9 = | a. 1 + b. 1 + c. 1 | s (a 2 + b2 + c2 )|1 + 1 + 1| = 3|1 + 1 + 1| . abc | a 2 b2 c2 | a 2 b2 c2 | \.\.\. Suy ra: 1 + 1 + 1 a 2 b2 c2 > 3 . Do :d s1. 3 Du bng xy ra 1=1 a 2 b2 = 1 = 1 a = b = c = 1 . c2 VyMaxd = 1 khia = b = c = 1 . 3 Cu VIIa Theo , ta c:(1 + i )2+ a (1 + i ) + b = 0 1 + 2i + i2 + a + ai + b = 0 (a + 2 ) i + (a + b ) = 0 a + 2 = 0 a = 2. a + b = 0 b = 2 B.2. CHNG TRNH NNG CAO Cu VIb 1. Gi s d ct (P) ti hai im phn bit A (a; a 2 ),B (b; b2 ) PT ng thng d: ( b > a) y = (a + b ) x ab Gi S l din tch hnh phng gii hn bi (P) v t d . Ta c: bbb S = } (a + b ) x ab x 2 dx = } ( x a ) ( x b ) dx= } ( x a ) ( x b ) dx a b = | 1 x3 a + b x 2 + abx | aa = 1 (b a )3 . 32 | 6 \. a DoM (1; 3) e d a + b = ab + 3 Suy raS 2=1 3

(b a )2 (= 1

(a + b )2 3 4ab(= 1

(ab + 3)2 4ab( 36 36 36 12 3 83 1288 2 =

(ab + 1) + 8( >= S > 36 3693 MinS = 8 2 ab + 1 = 0 ab = 1 a + b = 2 . 3 Vy ta lp c phng trnh ng thngd : y = 2x + 1 . 2. Ta c:C e d C (1 + 2t;1 t; 2t ) AB = ( 2; 2; 6), AC = ( 2t 2; t 4; 2t ) |

AB; AC( =26622 ;;2| = ( 2t + 24; 8t 12; 2t 12) t 42t2t2t 2 2t 2 t 4 | \. Gi S l din tch tam gic ABC. www.MATHVN.com Dnh cho hc sinh lp 12 n thi i hc- Cao ngB thi t lun- mn Ton 6 Vn Ph Quc- GV. Trng i hc Qung Nam D: 0982 333 443;0934 825 925 MATHVN.COM - Ton hc Vit Nam ( Ta c: S = 1

1 AB; AC= ( 2t + 24 )2+ (8t 12 )2 + (2t 12)2 2 2 = 1 72t 2 144t + 864 = 2 1 72 ( t 1)2+ 792 > 3 22 . 2 Du = xy ra khit = 1 C (1; 0; 2) Vy khiC (1; 0; 2)thMinS = 3 22 Cu VIIb Phng trnh cho tng ng vi: (2 )(2 )(42 )(42 ) log2x + x + 1+ log2 x x + 1= log2 x + x+ 1+ log2 x x + 1 (2 )(2 )(42 )(42 ) log2 x + x + 1 x x + 1= log2 x + x+ 1+ log 2x x + 1 (42 )(42 )(42 ) log2 log2 x + x (x 4 x 2 + 1 + 1) = log2x = 0 + x + 1+ log2x x + 1 x = 0 x 4 x 2 + 1 = 1 x 4 x 2= 0

x = 1 x = 1 Vy tp hp nghim ca phng trnh l:S = {1; 0;1} 2 A- PHN CHUNG(7,0 im) Cu I (2,0 im) . Cho hm s:y = 2x 3 x 2 (C). 3. Kho st s bin thin v v th (C) ca hm s. 4. Gi I l giao im ca hai tim cn. Tm im M thuc (C). Bit tip tuyn ca (C) ti M ct cc ng tim cn ti J v K sao cho ng trn ngoi tip tam gic IJK c din tch nh nht. Cu II (2,0 im) 1. Tm nghim x e | 0; =t | ca phng trnh sau y : 2 | 4 sin2| t =x | \. 3 sin | t 2 x | = 1 + 2 cos2 | x = 3t | . 2 | 2 | 4 | \. 2. Gii h phng trnh: \.\. 8x3 y + 27 = 18 y3 . 4x2 y + 6x =y2 t 2 Cu III (1,0 im) .Tnh tch phn: I =I= } 10 1 cos5 x.sin x. cos9 xdx . 0 www.MATHVN.com Dnh cho hc sinh lp 12 n thi i hc- Cao ngB thi t lun- mn Ton 7 Vn Ph Quc- GV. Trng i hc Qung Nam D: 0982 333 443;0934 825 925 MATHVN.COM - Ton hc Vit Nam CuIV(1,0 im). Chohnh chpS.ABC cyltam gicABCvung cn ti nhB, BA = BC = 2a, hnh chiu vung gc ca S trn mt phng y (ABC) l trung im E ca AB v SE = 2a. Gi I, J ln lt l trung im ca EC, SC ; M l im di ng trn tia i ca tia BA sao cho ECM= o (0 < o < 900 ) v H l hnh chiu vung gc ca S trn MC. Tnh th tch ca khi t din EHIJ theoa,ov tm o th tch ln nht. x1 Cu V(1,0 im).Chng minh rng: B- PHN RING(3,0 im) B.1. CHNG TRNH CHUN Cu VI a(2,0 im) x1 x+ x1 x> 2 e x e (0;1) . 1. Trong mt phng Oxy, cho hnh thoi ABCD c phng trnh hai cnh AB, AD th t l:x + 2 y 2 = 0 ;2x + y + 1= 0 .CnhBDchaim M (1; 2) .Tm toccnhca hnh thoi. 2. Trong khng gian Oxyz, cho ng thng d : x 1 = y + 2 = z . Vit phng trnh mt 122 phng (P) bitrng (P) cha ng thng d v to vi mt phng (xOy) mt gc nh nht. Cu VII a (1,0 im). Tm tp hp im M m ta phc ca n tha mn iu kin: z 2 + i= 1 . B.2. CHNG TRNH NNG CAO Cu VI b(2,0 im) 1. Trong mt phng Oxy, cho tam gic ABC cn ti Be Ox, phng trnh cnh AB c dng: 3x y 2 3 = 0 ; tm ng trn ngoi tip tam gic l tam gic. I (0; 2 ) . Tm to cc nh ca 2. Trong khng gian Oxyz, cho hai imA (2; 0; 0 )vJ ( 2; 0; 0) . Gi s(o )l mt phng thay i,nhnglun i quang thngAJ vctcc trcOy,Ozlnltti ccim B (0; b; 0) ,C (0; 0; c ) vib, c > 0 . Chng minh rng:b + c = bc 2 v tm b, c saochodin tch tam gic ABC nh nht. Cu VII b (1,0 im). 001 1223320102010 Tnh P = 2C2010 2 C2010+ 2C2010 2C2010+ ... + 2C2010 1.22.33.44.52011.2012 BI GII A- PHN CHUNG Cu I 1. Hc sinh t gii. www.MATHVN.com Dnh cho hc sinh lp 12 n thi i hc- Cao ngB thi t lun- mn Ton 8 Vn Ph Quc- GV. Trng i hc Qung Nam D: 0982 333 443;0934 825 925 MATHVN.COM - Ton hc Vit Nam 0 ( ) 0.0 yy \ M 0 |2x 3 | 2. Ta c:M x;eC, x = 2 ,y' (x) = 1 0 x 2 | 00 (x 2)2 Phng trnh tip tuyn A vi ( C) ti M : A : y = (x0 1 2)2 (x x0 ) + 2x0 3 x0 2 |2x 2 | To giao im J, K ca (A) v hai tim cn l: J 2; 0 | ; K ( 2 x0 2; 2) \ x0 2 . Ta c: xJ+ xK = 2 + 2 x0 2 = x = x ,yJ+ yK = 2x0 3 =y M l trung im JK. 22 0M 2x0 2 Mt khc I(2; 2) v AIJK vung ti I nn ng trn ngoi tip AIJK c din tch: 2 S =t IM 2= t (x 2)2+ | 2 x0 3 2 |(

= t ( x 2)2+ = 1 ( > 2t | ( ( 0

\ x0 20 . ( ( x0 2)2 1 x0= 1 M (1;1) Du = xy ra khi( x 2)2= . Cu II ( x0 2)2 x0= 3 M (3; 3) 1. Ta c:4 sin2| t =x | 3 sin | t 2 x | = 1 + 2 cos2 | x = 3t | 2 | 2 | 4 | \.\.\. 2 1 cos ( 2t x )( 3 cos 2 x = 1 + 1 + cos | 2 x = 3t | 2 | \. 2 2 cos x 3 cos 2 x = 2 sin 2 x sin 2 x 3 cos 2 x = 2 cos x 1 sin 2 x 3 cos 2 x = cos x sin | 2 x t | = cos | t x | 22

3 | 2 |

2x t = t x + k 2t \.\. x = 5t + k 2t 32 183 ( k e ) .

2x t = t + x + k 2t x = 5t + k 2t 326 Vx e | 0; =t | nn ta chn c nghim x = 5t . 2 | 18 2.Vi \. y = 0h phng trnh cho thnh: | 3 |3 (2x)3 + | 27 = 0 6x = 0 = 18 ( V l). Suy ra y = 0 8x3 y + 27 = 18 y3 \ y . Khi : 22 . 4xy + 6x =y3 |3 |2x. 2x + \ | = 3 . tu = 2x,v = 3 .Hphngtrnhtrnthnh: y www.MATHVN.com Dnh cho hc sinh lp 12 n thi i hc- Cao ngB thi t lun- mn Ton 9 Vn Ph Quc- GV. Trng i hc Qung Nam D: 0982 333 443;0934 825 925 MATHVN.COM - Ton hc Vit Nam vv | 4 . 42 | 22 A J H u3 + v3 = 18 (u + v )3 3uv (u + v ) = 18 (u + v )3 = 27 u + v = 3 += = .Suyrau,vl uv (uv )3 uv (u + v ) = 3 uv (u + v ) = 3 uv1 3 +5

t = nghim ca phng trnh:t 2 3t + 1 = 0 2 . 3 5 t = 3 +5 2x = 3 5 2x = 2 3 +5 x = 3 5 x = Do : 2 33 5 2 33 +5 3(3 +5 ) 4 3(3 5 ) = y2 = y2 y = 2 y = 2 Vytpnghimcahphngtrnhchol: | 3 + 53(3 + 5 ) || 3 53(3 5 ) | S =

; | ;

; | . ` Cu III \.\.) tt = 10 1 cos5 x t10= 1 cos5 x 10t9 dt = 5 sin x. cos4 xdx sin x. cos4 xdx = 2t 9 dt t 211 Ta c: I = } 10 1 cos5 x.cos5 x.sin x cos4 xdx = 2} t (1 t10 )t9 dt = 2} (t10 t 20 ) dt 000 | t11 = 2

1 t 21 | | = 20 . \ 1121 . 0 231 Cu IV ABC vungcntiBnn S AC =AB2 + BC 2=4a2 + 4a 2= 2 2a EBC vung cn ti B nn C =BE 2 + BC 2=a 2 + 4a 2= a 5 . Theo nh l ba ng vung gc ta c: EH HC . Suy ra:EH= EC sin o EI= 1 EC = a 5 . E 22 M B S= 1 EH .EI sinHEI= 1 a 5 sin o . a 5 cos o AHEI 222 I = 5 a2 sin 2o . www.MATHVN.com Dnh cho hc sinh lp 12 n thi i hc- Cao ngB thi t lun- mn Ton 10 Vn Ph Quc- GV. Trng i hc Qung Nam D: 0982 333 443;0934 825 925 MATHVN.COM - Ton hc Vit Nam 8 C www.MATHVN.com Dnh cho hc sinh lp 12 n thi i hc- Cao ngB thi t lun- mn Ton 11 Vn Ph Quc- GV. Trng i hc Qung Nam D: 0982 333 443;0934 825 925 MATHVN.COM - Ton hc Vit Nam 2 3 Trong ASCE , IJl on trung bnh nnIJ= 1 SE = a 2 vIJ ( HEI ) . Do V = 1 IJ .S = 1 a 5 a2 sin 2o = 5a sin 2o. EHIJ 3 HEI 3824 |t |t VEHIJ t gi tr ln nht khisin 2o = 1 0 < o < | \. tc o =. 4 Cu V x1xxx Xt hm s:f ( x ) = x1 x+ x1 x= x1 x+ x.x1 x= (1 + x ) x1 x x x e (0;1) . ( ) (1 + x ) x1 x| 1 x| f 'x = (1 x )2 2. \ 1 + x + ln x | . . Xt:g ( x ) = 2.1 x + ln x 1 + x 2 g' ( x ) = (1 x ) x (1 + x )2 > 0 g ( x ) ng bin trn (0;1) g ( x ) < g (1) = 0 f ' ( x ) < 0 x e (0;1) f ( x )nghch bin trn(0;1) 1 ||1 x 1 1 | f ( x ) > lim f ( x ) = lim (1 + x ) x 1.x1 x= 2 lim1 + = 2 x e (0;1) | . x1x 1x 1 1 | e B- PHN RING B.1. CHNG TRNH CHUN Cu VIa \1 x . 1.Tac:A = AB AD toimAlnghimcahphngtrnh: = 4 A x + 2 y 2 = 0 2x + y + 1 = 0 x 3 . N y = 5 M Do 3 D B A | 4 ; =5 | . 3 3 | C \. GiN ( x; y ) e AC ( AC l tia phn gicBAD). Hn na M v Nnm cng pha i vi x + 2 y 22x + y + 1 = 55 x + 2 y 2=2x + y + 1 ng thng AB nn ta c: ( x + 2 y 2) (1 + 4 2) > 0 x + 2 y 2 > 0 ( 2x + y + 1) ( 2 + 2 2) > 0 2x + y 2 > 0 www.MATHVN.com Dnh cho hc sinh lp 12 n thi i hc- Cao ngB thi t lun- mn Ton 12 Vn Ph Quc- GV. Trng i hc Qung Nam D: 0982 333 443;0934 825 925 MATHVN.COM - Ton hc Vit Nam 3 3 | x + 2 y 2 = 2x + y + 1hayx y + 3 = 0 .Vy phng trnh ng thng cha cnhAC l:x y + 3 = 0 . ngchoBDquaM (1; 2 ) nhnu BD= n AC = (1; 1) lmvectchphngnnc phng trnh l: x 1 = y 2 x + y 3 = 0 . 11 B = AB BD toimBlnghimcahphngtrnh: x + 2 y 2 = 0 x + y 3 = 0 x = 4 y = 1 . Do B (4; 1) . D = AD BD to im D l nghim ca h phng trnh: 2x + y + 1 = 0 x + y 3 = 0 x = 4 y = 7 . Do D ( 4; 7 ) . GiI= AC BD to im I l nghim ca h phng trnh: x y + 3 = 0 x + y 3 = 0 x = 0 y = 3 . Do I (0; 3) . V I l trung im AC nnC | 4 ; 1=3 | . \. 2.Gi vect php tuyn ca mt phng (P) l: n p = (a; b; c )(a2+ b2 + c2> 0) . vect php tuyn ca mt phng (xOy) l:nxOy = (0; 0;1) vect ch phng ca ng thng d l:ud= (1;1; 2) d c ( P ) ud .nP= 0 a + b + 2c = 0 b = a 2c . Gi l gc gia hai mt phng (Q) v mt phng (xOy) | 0 s s =t | . 2 | \. ccc Ta c:cos ===. a2 + b2 + c2 a2 + ( a 2c )2+ c2 2a2 4ac + 5c2 + Nuc = 0thcos = 0 = t . 2 + Nuc = 0th ta chnc = 1 . Khi :cos = 1 = 1 s1. Du = xy ra khia = 1 . 2a2 4a + 5 2 ( a 1)2+ 3 3 nh nht cos ln nht. Do so snh hai trng hp ny ta c maxcos = t ( e

0; ( 1 khia = c = 1 ; b= 3 1. 2 Mt phng( P ) quaA (1; 2; 0) e dvnP= (1; 1;1)lm vect php tuyn nn c phng trnh: 1.( x 1) 1.( y + 2) + 1.( z 0 ) = 0 x y + z 3 = 0 . Cu VIIa Hai s phc lin hp c moun bng nhau, ta suy ra www.MATHVN.com Dnh cho hc sinh lp 12 n thi i hc- Cao ngB thi t lun- mn Ton 13 Vn Ph Quc- GV. Trng i hc Qung Nam D: 0982 333 443;0934 825 925 MATHVN.COM - Ton hc Vit Nam C z 2 + i=z 2 i (vz 2 + i = z + ( 2 + i ) = z 2 i). T ta c:z 2 i= 1 . t z = x + iy ( x, y e ) . Suyra : z 2 i= 1 ( x 2) + ( y 1) i = 1 ( x 2)2 + ( y 1)2 = 1 ( x 2)2 + ( y 1)2= 1 Tp hp cc im M l ng trn tmI (2;1) , bn knhR = 1. B.2. CHNG TRNH NNG CAO Cu VI b 1. Giphngtrnhngthng ABiqua imA MB M(4;5) l:a ( x 4) + b ( y 5) = 0 (a2+ b2> 0) . Khi phng trnh ng thngBC i quaN(6;5) vQN vung gc vi AB l:b ( x 6 ) a ( y 5) = 0 . Din tch ca hnh ch nht ABCD l: DP d ( P; AB ).d (Q; BC ) = a 3b a2 + b2 4b + 4a . a 2 + b2 4 ( a 3b ) (a b ) =. a2 + b2 Theo , ta c: 4 (a 3b ) ( a b ) a2 + b2 = 16 ( a 3b) ( a b)= 4 (a2+ b2 ) . Cho b = 1, ta c:(a 3)(a 1) = 4 (a2+ 1) a2 4a + 3= 4a2 + 4 a2 4a + 3 = 4a2 + 4

a2 4a + 3 = (4a2+ 4 ) 3a2 + 4a + 1 = 0

5a2 4a + 7 = 0 a = 1 1

a = 3 Vy phng trnh cnh AB l:x y + 1 = 0 ;x 3 y + 11 = 0. 2.mp (o ) quaA (2; 0; 0) , B (0; b; 0) , C (0; 0; c )(b,c> 0) x + y + z = 1 2bc nn c phng trnh l: J (1;1;1) e (o ) 1 + 1 + 1 = 1 b + c = bc 2bc2 (1) | b0 0 AB = ( 2; b; 0), AC = ( 2; 0; c )

AB; AC( =; 2 2b | ;= ( bc; 2c; 2b ) 0cc 2 20 | 1

1 S=

AB; AC( = b2c2 + 4c2 + 4b2= 1 \. b2c2 + 4 (b2+ c2 ) . ABC 2 22 p dng bt ng thc Bunhiacopski, ta c: (1.b + 1.c)2s 2 (b2+ c2 ) 4 (b2+ c2 ) > 2 ( b + c )2 . Do :S > 1 b2c2 + 2 ( b + c)2 = 1 4 (b + c )2+ 2 (b + c)2 = 6 (b + c ) . ABC 222 www.MATHVN.com Dnh cho hc sinh lp 12 n thi i hc- Cao ngB thi t lun- mn Ton 14 Vn Ph Quc- GV. Trng i hc Qung Nam D: 0982 333 443;0934 825 925 MATHVN.COM - Ton hc Vit Nam Du bng xy ra khi v ch khib = c. www.MATHVN.com Dnh cho hc sinh lp 12 n thi i hc- Cao ngB thi t lun- mn Ton 15 Vn Ph Quc- GV. Trng i hc Qung Nam D: 0982 333 443;0934 825 925 MATHVN.COM - Ton hc Vit Nam 12 2 T (1) suy rac = 2b b 2 > 0 b > 2. 6 |2b |6 | 4|6 4( Suy raS ABC> b + | = b + 2 + | =

(b 2) ++ 4( 2\ b 2 . 2\b 2 . 2b 2 p dng bt ng thc Cauchy cho 2 s dng:( b 2) + b = c > 0 4 b 2 > 4 S ABC > 4 6 . Do MinSABC = 4 6 Cu VIIb Ta c: t c khi b + c = bc b = c = 4 . 2 2k Ck ( 2)k2010! ( 2 )k2010! k 2010 ( 1)( k + 1) == k !( 2010 k )!(k + 1)(k + 1)!(2010 k )! k 1 ( 2 ) 2011!== 1 (2)k +1 Ck +1 2011 (k + 1)!(2011 k 1)!4022 2011 P = 1

( 2)1 C1 + (2)2 C2 + ... + (2)2011 C2011 ( 4022 201120112011 = 1

( 2 + 1)2011 ( 2)0C0 ( = 1 . 4022 2011 2011 3 A- PHN CHUNG(7,0 im) Cu I (2,0 im) . Cho hm s: y = 1 x3 5 mx2 4mx 4(C). 32 5. Kho st s bin thin v v th (C) ca hm s khim = 0 . 6. Tmmhmstcctrtix1, x2 saochobiuthc: A = m x2 + 5mx+ 12m + 21 t gi tr nh nht. x2 + 5mx + 12mm2 Cu II (2,0 im) 3. Gii phng trnh: tan x ( tan x + 2 sin x + 1) 6 cos x = 3 + sin x |1 + tan x tan =x | . 2 | 4. Gii h phng trnh: x6 + y6 + 2xy 5x2 2x + 33 2xy 5x2 2 y + 33 = x2 + y6 =y2 + x6 \. ( x,y e ) . www.MATHVN.com Dnh cho hc sinh lp 12 n thi i hc- Cao ngB thi t lun- mn Ton 16 Vn Ph Quc- GV. Trng i hc Qung Nam D: 0982 333 443;0934 825 925 MATHVN.COM - Ton hc Vit Nam Cu III (1,0 im) .Tnh tch phn: I= ln 5 } dx . ln 2(10e x 1) ex 1 CuIV(1,0 im). Chohnh chp tgicS.ABCD cy ABCDlhnhvung tm O, cnh bng a. Cnh bn SA vung gc vi y hnh chp vSA = a 2 . Gi H v K ln lt l hnh chiu ca A trn SB, SD. Chng minhSC (AHK )v tnh th tch O.AHK. CuV(1,0im).Tmmphngtrnhsaucnghim: (4m 3)x + 3 + (3m 4)1 x + m 1 = 0 B- PHN RING(3,0 im). Th sinh ch c chn mt trong hai phn B.1. CHNG TRNH CHUN Cu VI a(2,0 im) 3. TrongmtphngOxy,chohaingtrn: (C1 ) : x 2 + y2 = 9 ;(C2 ) : ( x 1)2 + ( y 1)2= 25 .Gi A,Blccgiaoim ca(C1 ) v(C2 ) .VitphngtrnhngthngAB.Hy chngminh rngnuK e AB th KI < KJvi I, J ln lt l tm ca(C1 )v(C2 ) . 4. Trong khng gianOxyz,choim A (5; 5; 0) vng thng d : x + 1 = y + 1 = z 7 . 234 TmtoccimB,CthucdsaochotamgicABCvungcntiAv BC = 2 17 . Cu VII a (1,0 im). Gii phng trnh:z2 + 2011 = 0trn tp s phc . B.2. CHNG TRNH NNG CAO Cu VI b(2,0 im) 1. TrongmtphngOxy,xcnhtoccimBvCcatam gicuABCbit A (3; 5) v trng tmG (1;1) . 2. Trong khng gian Oxyz, cho hai imM (0; 0; 3) ,N (2; 0; 1)v mt phng (o ) : 3x 8 y + 7 z 1 = 0 .TmtaPnm trnmtphng(o ) u. saochotamgicMNP Cu VII b (1,0 im).Gii hphng trnh: x log3 y + 2ylog3 x= 27 . log3 y log3 x = 1 Cu I 1. Hc sinh t gii. 2. TX: D = y' = x2 5mx 4m . BI GII Hm s t cc tr tix1 ,x2Phng trnhy' = 0c hai nghim phn bitx1 ,x2 www.MATHVN.com Dnh cho hc sinh lp 12 n thi i hc- Cao ngB thi t lun- mn Ton MATHVN.COM - Ton hc Vit Nam 21 | 2 < 16 A = 25m2 + 16m > 0

m 25 m > 0 (1) x+ x = 5m Theo nh l Viet, ta c: 12 x1 x2= 4m 22 Vx1 l nghim ca phng trnh nn:x1 5mx1 4m = 0 x1 = 5mx1 + 4m . Do : x2 + 5mx + 12m = 5mx + 4m + 5mx + 12m = 5m ( x + x) + 16m = 15m2 + 16m > 0 . 121212 Tng t, ta cng tnh c: 2 x2 + 5mx 2 + 12m = 25m2 + 16m > 0 . 22 Khi :A = m + x2 + 5mx1 + 12m = m + 25m + 16m > 2 2222 x1 + 5mx2 + 12mm 25m+ 16mm Du = xy ra ( Bt ng thc Cauchy cho 2 s dng). m25m2 + 16m 2 = 2 m4 = (25m2 + 16m)2 m2 = 25m2 + 16m 25m+ 16mm m = 0 24m2 + 16m = 0

m = Vymin A = 2khim = 2 . 3 2 2 . Kt hp vi (1) ta chnm = 3 . 3 Cu II 1.iu kin: cos x = 0 cos x = 0 cos x = 0 2 cos2 x = 0 cos x = 0 1 + cos x = 0 cos x = 0 . cos x = 1 22 Ta c:tan x ( tan x + 2 sin x + 1) 6 cos x = 3 + sin x |1 + tan x tan =x | 2 | \. | x |

tan x ( tan x + 2 sin x + 1) 6 cos x = 3 + sin x

1 + sin x sin 2 x | cos x cos | \2 . cos x cos x + sin x sin x tan x ( tan x + 2 sin x + 1) 6 cos x = 3 + sin x 2 2 cos x cos x 2 cos x tan x ( tan x + 2 sin x + 1) 6 cos x = 3 + sin x 2 cos x cos x 2 Vn Ph Quc- GV. Trng i hc Qung Nam D: 0982 333 443;0934 825 925 15 www.MATHVN.com Dnh cho hc sinh lp 12 n thi i hc- Cao ngB thi t lun- mn Ton MATHVN.COM - Ton hc Vit Nam x . tan x ( tan x + 2 sin x ) + tan x 6 cos x = 3 + tan x tan x ( tan x + 2 sin x ) = 6 cos x + 3 tan 2 x (1 + 2 cos x ) = 3 (1 + 2 cos x ) (1 + 2 cos x )( tan2 x 3) = 0 1 11

cos x = 1

cos x = 2

cos x = 2

cos x = 2 2

tan 2 x = 3

sin 2 x = 3cos2 x

1 cos2 x = 3 cos2 x

cos2 x = 1

4 cos2 x = 1 cos2x = 1 2x = 2t + k 2t x = t + kt 4233 2. Cng v theo v ca hai phng trnh trn ta c: ( k e ) (tho iu kin) | 1 2xy+ 1 | = x2 + y2 | 52 \ 2x + 33 5y2 2 y + 33 | || 2 xy 1 + = 1 | = x2 + y2 (*) 5 ( x 1)2+ 32 5 ( y 1)2+ 32 | \. Nhn xt:VT (*)s 2xys x2 + y2= VP (*) . x=y VT (*) = VP (*) x =y = 1 x =y = 0 .

x =y = 0 x =y = 1 Vy tp hp nghim ca h phng trnh cho l:S = {(0; 0 ) , Cu III (1;1)} . Ta c: ln 5 dx I= } ln 5 = } ex dx . ln 2(10e x 1) ex 1 ln 2(10 ex ) ex 1 tt = i cn: ex 1 t 2= ex 1 2tdt = e x dx x = ln 2 t = 1 ; x = ln 5 t = 2 Khi : 2 2tdt 2 dt1 2 | 11 | 2 13 + t15 } (9 t 2 ) t } (3 t ) (3 + t ) 3 } \ 3 t3 + t . 33 t32 I == 2= + | dt =ln=ln. 1111 Cu IV + Chng minhSC ( AHK)S BC AB Ta c: BC ( SAB ) BC AH . BC SA I Mt khcAH SBnn AH SC . AH ( SBC ) , do K M G www.MATHVN.com Dnh cho hc sinh lp 12 n thi i hc- Cao ngB thi t lun- mn Ton MATHVN.COM - Ton hc Vit Nam O C DH Vn Ph Quc- GV. Trng i hc Qung Nam D: 0982 333 443;0934 825 925 16 A B www.MATHVN.com Dnh cho hc sinh lp 12 n thi i hc- Cao ngB thi t lun- mn Ton 17 Vn Ph Quc- GV. Trng i hc Qung Nam D: 0982 333 443;0934 825 925 MATHVN.COM - Ton hc Vit Nam Chng minh tng t ta cng c:AK SC . VySC ( AHK ) . + Tnh th tch khi chp O.AHK GiG = SO HKvI= AG SC . VSC ( AHK) nnSC AI . Mt khc SA = AC = a 2nn tam gic SAC l tam gic vung cn ti A. Suy ra I l trung im SC v G l trng tm tam gic SAC. Ta c:SC = SA 2 = 2a nnAI= 1 SC = av 2 AG = 2 AI= 2a . 33 Ngoi ra: BD AC BD ( SAC ) BD SC . BD SA ( AHIK ) SC T : BD SC BD c ( AHIK ) BD / / ( AHIK ) . Do :( AHK ) ( SBD ) = HK / / BD. Ta c: HK = SG = 2 HK= 2 BD = 2a 2 . BDSO333 11 2a 2 2a 2a2 2 Din tch ca tam gic AHK l:S =HKAG =..=. AHK 22339 Gi M l trung im ca AI thOM / / ICvOM= 1 IC = a . 22 MSC ( AHK) nnOM ( AHK ) . 112a2 2 aa3 2 Th tch hnh chp O.AHK l:V=S .OM=..=. Cu V iu kin : 3 s x s 1. 3 AHK39227 Khi :( 4m 3) x + 3 + (3m 4 )1 x + m 1 = 0 m = 3 4 x + 3 + 4 1 x + 1 x + 3 + 31 x + 1 (*) 22 | x + 3 || V+ 1 x | = 1 nn ta tx + 3 = 2 sin o;1 x= 2coso 2 | 2 | \.\. tt = tan o , 0 s o s t ,0 s t s 1 . 22 Phng trinh (*) tr thnh: 7t 2 + 12t + 9 m = 5t 2 + 16t + 7 (**) Phng trnh cho c nghim khi v ch khi phng trnh (**) c nghim trn |0;1|. 7t 2 + 12t + 9 Xt hm s:f (t ) = 5t 2 + 16t + 7 www.MATHVN.com Dnh cho hc sinh lp 12 n thi i hc- Cao ngB thi t lun- mn Ton 18 Vn Ph Quc- GV. Trng i hc Qung Nam D: 0982 333 443;0934 825 925 MATHVN.COM - Ton hc Vit Nam

t = 3 f ( t )lin tc trn |0;1| 2 f ' (t ) = 52t 8t 60 < 0 f ( t )l hm nghch bin trn |0;1| . (5t 2 + 16t + 7 )2 Bng bin thin: t0 1 f ' ( t ) f ( t ) 9 7 9 7 Da vo bng bin thin ta suy ra phng trnh cho c nghim khi 7 s m s 9 . 97 B- PHN RING B.1. CHNG TRNH CHUN Cu VI a x2 + y2= 9 1.Ta c:M ( x; y ) e (C1 ) (C1 ) 22 x2 + y2= 9 ( x 1)+ ( y 1)= 25 x2 + y2 2x 2 y 23 = 0 x + y + 7 = 0l phng trnh ng thng AB. K e AB K (t; t 7 ) . ng trn(C1 ) c tm l:I (0; 0) , ng trn(C2) c tm l:J (1;1) . KJ 2 KI 2= (1 a )2+ (a + 8)2 a2 (a + 7 )2= 16 > 0 KI< KJ . 2. V tam gic ABC vung cn ti A nnAB2= 1 BC 2= 34 . 2 B, C e dnn c to l( 1 + 2t; 1 + 3t; 7 4t ) . Khi AB2= 34 (2t 6 )2 + (3t 6 )2 + ( 4t + 7 )2= 34 29t 2 116t + 87 = 0 t = 1 . VyB (1; 2; 3) ,C (5; 8; 5) hocB (5; 8; 5) ,C (1; 2; 3) . Cu VIIa tz = a + bi (a,b e ) . Khi z 2= a2 b2 + 2abi z 2= (a2 b2 ) 2abi z 2 + 2011 = (a2 b2 + 2011) 2abi Do z 2 + 2011 = 0 (a2 b2 + 2011) 2abi = 0 a 2 b2 + 2011 = 0 . 2ab = 0 www.MATHVN.com Dnh cho hc sinh lp 12 n thi i hc- Cao ngB thi t lun- mn Ton 19 Vn Ph Quc- GV. Trng i hc Qung Nam D: 0982 333 443;0934 825 925 MATHVN.COM - Ton hc Vit Nam y00 000 x2 + y2 + ( z Nub = 0tha2 + 2011 = 0( v l) . Do b = 0 a = 0 . Dn n:b = 2011 . Vy s phc z cn tm l:2011.i B.2. CHNG TRNH NNG CAO Cu VI b(2,0 im)A 3 1.Gi M l trung im BC, suy ra MA =GA M (0; 4) 2 Trong tam ABM vung ti M ta c: cos 300= AM AB AB = AM cos 300 = 310 = 2 30 . 3 2 G BMC Suy ra:BM= CM= 2 30 =30 2 22 2 Gi sB ( x0 ; y0 ) . Khi MB= 30 x0+ ( y0 4) = 30(1) Mt khcAM BM AM .BM= 0 x0 3y0 + 12 = 0 x0= 3 y0 12 (2) Thay (2) vo (1) ta c:(3 y 12)2+ ( y 4)2= 30 ( y 4)2= 3 000 y0= 7 x0= 9 = 1 x = 9 VyB (9; 7 ) ,C (9;1) hocB (9;1) ,C (9; 7 ) . 2. ng thng MN quaM (0; 0; 3) nhnMN = (2; 0; 2)lm vect ch phng nn c phng trnh: x = 2t y = 0 z = 3 + 2t ( t e ). Gi( | )l mt phng trung trc ca on thng MN. Gi I l trung im MN I (1; 0; 2 ) . Mt phng( | )quaI (1; 0; 2 )nhnMN= (2; 0; 2) lm vect php tuyn nn c phng trnh:2.( x 1) + 2.(z + 2 ) = 0 x + z + 1 = 0 . P e (o )sao choAMNP u P e (o ) ( | ) MN2= NP2 Gi s ta im N l( x0 ; y0 ; z0 ) , ta c: 3x 8 y+ 7 z1 = 0 x0 + z0 + 1 = 0. 000 + 3)2= 8 Gii h phng trnh ny ta tm c P ( 2; 2; 3), P | 2 ; 2 ; =1 | . 333 | Cu VIIb iu kin : \. x, y > 0 . www.MATHVN.com Dnh cho hc sinh lp 12 n thi i hc- Cao ngB thi t lun- mn Ton 20 Vn Ph Quc- GV. Trng i hc Qung Nam D: 0982 333 443;0934 825 925 MATHVN.COM - Ton hc Vit Nam (

x = 9 3 |` b ) a ta = log3 x x = 3a b b = log3yy = 3 3a = HPT cho thnh: 2 (3b )= 27 3ab= 9 (a ) b = 2 b a = 1a + b = 1 a + b = 1 a ; bl nghim ca phng trnh:t 2 t 2 = 0 t = 1

t = 2

a = 1

a = 1

x = 3 y = 9 Khi :

b = 2

a = 2

b = 2

a = 2

1

9 ( tho iu kin).

b = 1b = 1

y = 1 3 Vy tp hp nghim ca h phng trnh cho l:S = (3; 9) , | 1 ; =1 | . \.) 4 A- PHN CHUNG(7,0 im) Cu I (2,0 im) . Cho hm s: y = x 1 x + 1 (C). 7. Kho st s bin thin v v th (C) ca hm s. 8. Tm im M thuc (C) tng khong cch t M n hai trc to l nh nht. Cu II (2,0 im) 1. Gii phng trnh: tan | + x | 3 tan 2 x = cos 2x 1 . 2 | cos2 x 2. Gii h phng trnh: \. 3y3 1 + x = 3 x2 + y3= 82 t 4 Cu III (1,0 im) .Tnh tch phn:I= } (tan 2 x tan x ) e x dx . 3t 4 CuIV(1,0 im).ChohnhchpS.ABCcSA ( ABC ) ,tam gicABCvungcn nh C vSC = a . Tnh gcgia hai mt phng( SBC )v( ABC ) th tch khi chp ln nht. CuV(1,0 im). Choa,b,c,dlccsthcdngsaocho:a2 + b2 + c2 + d 2= 4 . Chng minh:a3 + b3 + c3 + d 3 s 8 . B- PHN RING(3,0 im) www.MATHVN.com Dnh cho hc sinh lp 12 n thi i hc- Cao ngB thi t lun- mn Ton 21 Vn Ph Quc- GV. Trng i hc Qung Nam D: 0982 333 443;0934 825 925 MATHVN.COM - Ton hc Vit Nam 55 x 1 0.\ B.1. CHNG TRNH CHUN Cu VI a(2,0 im) 1. Trong mt phng vi h to Oxy, cho tam gic ABC vi AB = 5,C ( 1; 1) , ng thngAB cphng trnhx + 2 y 3 = 0vtrng tm Gcatam gicABC thuc ng thngx + y 2 = 0.Hy tm to cc im A v B. 2.Trong khng gianvi hto Oxyz, choccimA (3;1;1) ,B (7; 3; 9 ),C (2; 2; 2 )v mt phng (P) c phng trnh:x + y + z + 3 = 0 . Tm im M thuc mt phng (P) sao cho Cu VII a (1,0 im) MA + 2MB + 3MCnh nht. Gi A, B theo th t l cc im ca mt phng phc biu din s z khc 0 v Chng minh tam gic OAB vung cn. B.2. CHNG TRNH NNG CAO Cu VI b(2,0 im) z' = 1 + i z . 2 1.TrongmtphngvihtoOxy,chongthngd :2x + my + 1 2 = 0 v ng trn(C ) : x2 + y2 2x + 4 y 4 = 0 . Gi I l tm ng trn (C). Tm m sao cho d ct (C) ti hai im phn bit A v B. Vi gi tr no ca m th din tch tam gic IAB ln nht v tnh din tch . 2.TrongkhnggianOxyz,chotamgicABCc trung tuyn : d: x 3 = y 6 = z 1 ; d : x 4 = y 2 = z 2 A (1; 2; 5) vphngtrnhhaing 1 221 2 141 Vit phng trnh chnh tc cc cnh ca tam gic ABC. Cu VII b (1,0 im). 22 y x + 2 y= 2x+1 Gii h phng trnh sau: log( x2 + 3 y + 1) log . y = 2 x2 + 4 y 1 BI GII A- PHN CHUNG Cu I 1. Hc sinh t gii 2. Ta c:|| M x;0 e (C ) 0x+ 1 | GidM l khong cch t M n hai trc ta , khi : dM= d ( M ; Ox ) + d (M ; Oy ) =x0+ x0 1 x0 + 1 ChnM (1; 0) e ( H ) d M= 1 . Do , tmMindMta ch cn xt: www.MATHVN.com Dnh cho hc sinh lp 12 n thi i hc- Cao ngB thi t lun- mn Ton 22 Vn Ph Quc- GV. Trng i hc Qung Nam D: 0982 333 443;0934 825 925 MATHVN.COM - Ton hc Vit Nam x \ 2 x0 < 1 x 1 < x0< 1 0 < x < 1 . 0 1 < 1 1 x< 1 + x 0 x0 + 1 00 Vi0 < x0< 1 th d= x + 1 x0= x 1 + 2 = ( x + 1) + 2 2 M000 1 + x0 x0 + 1 x0 + 1 > 2x+ 1 2 2 = 2 2 2 = 22 1 ( Bt ng thc Cauchy). (0 )( ) x0 + 1 0 < x0< 1 Du bng xy ra 0 + 1 = 2 x0 + 1 x0= 2 1 VymindM Cu II = 2 ( 2 1) khiM ( 2 1;1 2 ) . cos x = 0 = 1. iu kin: | | cos x 0 sin 2x = 0 x = k (k e ) co s + x | = 0 . sin x = 02 Phng trnh tng ng vi 2 cot x 3 tan 2 x = 2 sin cos2 x x 1 tan x tan 2 x = 0 tan3 x = 1 tan x = 1 x = + k 4 ( k e) . x = u2 x2= u4 2. tu =x,u > 0 ;v =y3 1 . Suy ra: y3 1 = v3 y3= v3 + 1 H cho tr thnh:u + v = 3 v = 3 u u4 + (3 u )3= 81 u 4 + v3= 81u 4 + v3= 81 Dou > 0 u4 u3 + 9u 2 27u 54 = 0 (u 3) (u3+ 2u 2 + 15u + 18) = 0 . nnu3 + 2u 2 + 15u + 18 > 0 . Suy rau = 3,v = 0 . Khi : x = 3 x = 9 = . 3y3 1 = 0 y1 x = 9 Vy y = 1 Cu III l nghim ca h phng trnh cho. ttt I =(tan 2 x tan x ) e x dx = tan 2 x.e x dx t anx.e x dx = I I www.MATHVN.com Dnh cho hc sinh lp 12 n thi i hc- Cao ngB thi t lun- mn Ton 23 Vn Ph Quc- GV. Trng i hc Qung Nam D: 0982 333 443;0934 825 925 MATHVN.COM - Ton hc Vit Nam }}} 12 3t3t3t 444 S dng tch phn tng phn i viI2ta c: www.MATHVN.com Dnh cho hc sinh lp 12 n thi i hc- Cao ngB thi t lun- mn Ton 24 Vn Ph Quc- GV. Trng i hc Qung Nam D: 0982 333 443;0934 825 925 MATHVN.COM - Ton hc Vit Nam t e| AC I = t anx.e x t + (1 + tan 2 x ).e x dx = et+ I I= I I = et . 23t } 112 43t 4 Cu IV BC = ( SBC ) ( ABC ) Ta c:SC BC S = SCA, e | 0; =t | 2 | CA BC \. TrongSAC vung ti A, ta c:SA = SC sin = a sin . vAC = SC cos = a cos . V ABC vung cn ti C nn: S ABC = 1 AB.AC = 1 AC 2= 1 a 2 cos2 .B 222 Th tch hnh chpS .ABCl: 2 V= 1 SA.S = 1 a sin . 1 a2 cos2 = 1 a3 sin cos2 = a (sin sin3 ) S . ABC 3 ABC 3266 Xt hm s y =f ( ) = sin sin 3 , e | 0; t=| . 2 | tt = sin \. , e | 0; t=| t e (0;1) . 2 | Khi xt hm s g ' (t ) = 1 3t 2 \. g (t ) = t t 3, t e (0;1)

t = g' (t ) = 0 1 3t 2= 0

1e (0;1) 3 Lp bng bin thin t g' ( t )

t =1 3 0 e (0;1) 1 3 1 0 + g ( t ) 00 2 3 3 Da vo bng bin thin ta suy ra a3 a3 a3 2 a3 3 1|t | max VS . ABC= max f( ) = max g (t ) =. =khisin =, e 0; | 6 | 0;t |6 te(0;1)63 3 27 3\ 2 . Cu V \ 2 . www.MATHVN.com Dnh cho hc sinh lp 12 n thi i hc- Cao ngB thi t lun- mn Ton 25 Vn Ph Quc- GV. Trng i hc Qung Nam D: 0982 333 443;0934 825 925 MATHVN.COM - Ton hc Vit Nam xBB x vv yy Ta2 + b2 + c2 + d 2= 4 suy raa2s 4 a s 2.Khi a2 ( a 2 ) s 0 a3 s 2a 2 . Tng t ta cng chng minh c:b3 s 2b2,c3 s 2c2 ,d 3s 2d 2 . Cng v theo v ca bn bt ng thc trn ta c a3 + b3 + c3 + d 3s 2 (a2+ b2 + c2 + d 2 ) = 8. B- PHN RING B.1. CHNG TRNH CHUN Cu VI a x + x + x = x + x 1 = 3x 1. V G l trng tm tam gic ABC nn : ABCABG Suy ra yA + yB + yC=yA + yB 1 = 3 yG xA + yA + xB + yB 2 = 3 ( xG + yG ) . V G e ng thngx + y 2 = 0nnxG + yG 2 = 0 xG + yG= 2 . Do :xA + yA + xB + yB 2 = 6 xA + yA + xB + yB= 8 Li v A, B thuc ng thng x + 2 y 3 = 0nn: xA + 2 yA 3 = 0 + 2 y 3 = 0 (1) . 8 = x + y + x + y = 3 2 y + y + 3 2 y + y = 6 ( y + y) yA + yB= 2 AABBAABBAB A + xB = 10 Mt khc AB = 5 ( x x)2+ ( y y)2= 5 ABAB Nhng theo (1)xA= 3 2 yA, x B= 3 2 yB nnxA xB= 2 ( yB yA ) . Suy ra( x x)2+ ( y y)2= 5 ( y y)2= 5 ABABAB y = 1 y = 1 Vy yA yB= 1 yA + yB= 2 yA yB= 1 yA + yB= 2 A B 2 = =3 2 A 2 = 3 B 2 Cui cng ta thu c: A| 4; =1 | , B| 6; =3 | hoc A| 6; =3 | ,B | 4; =1 | . 2 | 2 | 2 | 2 | \.\.\.\. 2. Gi s I l mt im tho IA + 2IB + 3IC = 0 . Khi ta tnh c I | 23 ; 13 ; =25 | . 666 | \. Ta c:MA + 2MB + 3MC=MI + IA + 2 (MI + IB ) + 3 (MI + IC ) = 6 MI MA + 2MB + 3MC phng (P). nh nht MI nh nht M l hnh chiu vung gc ca I ln mt www.MATHVN.com Dnh cho hc sinh lp 12 n thi i hc- Cao ngB thi t lun- mn Ton 26 Vn Ph Quc- GV. Trng i hc Qung Nam D: 0982 333 443;0934 825 925 MATHVN.COM - Ton hc Vit Nam 6 6 GiAl ng thng qua I nhnu

= nP= (1;1;1) lm vect ch phng nn c phng x = 23 + t 6 trnh tham s : y = 13 + t (t e ) . z = 25 + t Ta cM= A ( P ) to im M ng vi tham s t l nghim ca phng trnh : 23 + t + 13 + t + 25 + t + 3 = 0 t = 79 M | 5 ; 20 ; =2 | . 66618 999 | Vy MA + 2MB + 3MC nh nht khiM | \. 5 ; 20 ; =2 | 999 | Cu VII a Gi sz = x + yi th ta c \. A ( x; y ) . Vz = 0nnx 2 + y2> 0 . Ta cz' = 1 + i z = 1 (1 + i ) ( x + yi ) = x y + x + y i. 2222 Vy B c ta : B | x y ; =x + y | . 22 | \. 22 22 Ta li c:OA2= x2 + y2;OB2= | x y | + | x + y | = x + y. 2 | 2 | 2 \.\. 2222 22 AB2= | x x y | + | y x + y | = | x + y | + | y x | = x + y. 2 | 2 | 2 | 2 | 2 \.\.\.\. OB = AB T , suy ra : OA2= OB2 + AB2 .Vy tam gic OAB vung cn ti B. B.2. CHNG TRNH NNG CAO Cu VI b 1.(C )c tm I (1; 2) v bn knh R = 3. d ct(C )ti 2 im phn bit A, BB d (I , d ) < R d A 2 2m + 1 2< 3 2 + m2 I 1 4m + 4m2 < 18 + 9m2 5m2 + 4m + 17 > 0 ( lun ngm e ). Ta c:S IAB = 1 IA.IB sin AIB s 1 IA.IB = 9 222 Vy:SIAB ln nht l 9khi 2 AIB = 900 AB = R 2 = 3 2d ( I , d ) = 3 2 2 www.MATHVN.com Dnh cho hc sinh lp 12 n thi i hc- Cao ngB thi t lun- mn Ton 27 Vn Ph Quc- GV. Trng i hc Qung Nam D: 0982 333 443;0934 825 925 MATHVN.COM - Ton hc Vit Nam 55 33 1 2m= 3 2 2 2 + m2 16m2 16m + 4 = 36 + 18m2 2m2 + 16m + 32 = 0 m = 4 2. Thay ta A (1; 2; 5) vo hai phng trnhd1 ,d2 thy khng tha. Do A e d1 ,A e d 2 . Gi sd1l trung tuyn v t B,d2l trung tuyn v t C B e d1 B (3 2b; 6 + 2b;1 + b ), C e d 2 C (4 + c; 2 4c; 2 + c ) . G = d1 d2 Ta im G l nghim ca h : x 3 = y 6 = z 1 221 .Gii h ny ta thu cx = 3; y = 6; z = 1. x 4 y 2 z 2 == 141 Do G (3; 6;1) G l trng tm ca tam gic ABC nn x= x A + x B + xC 3 = 1 + 3 2b + 4 + c G 3 3 2b c = 1 y = yA + yB + yC 6 = 2 + 6 + 2b + 2 4c b 2c = 4 G z + z + z 5 + 1 + b + 2 + c b + c = 5 zG = A B C 3 1 = 3 b = 2 B (7; 2; 1) c = 3 C (1;14; 1) T ta lp c phng trnh cc cnh ca tam gic ABC l: AB : x 1 = y 2 = z 5 101 BC : x 7 = y 2 = z + 1 120 CA : x 1 = y 14 = z + 1 . 0 Cu VII b 21 22 y x + 2y= 2x +1 (1) log ( x2 + 3 y + 1) log y = 2 x2 + 4 y 1 (2 ) iu kin :y > 0 . Chia c hai v ca (1) cho2x> 0ta c: y x 22( y x ) + 2y x= 2 22( y x ) + 2y x 2 = 0 2= 1

2 y x= 2 < 0 www.MATHVN.com Dnh cho hc sinh lp 12 n thi i hc- Cao ngB thi t lun- mn Ton 28 Vn Ph Quc- GV. Trng i hc Qung Nam D: 0982 333 443;0934 825 925 MATHVN.COM - Ton hc Vit Nam \.\ .> 2. Ta c:2y x= 1 x =y . Thayy = xvo (2) ta c: log ( x2 + 3x + 1) log x = 2x2 + 4x 1 log | x + 1 + 3 | = 1 2 ( x 1)2 (3) 555 x | \. |1| p dng bt ng thc Cauchy: VT (3) = log5 x ++ 3 | > log5 ( 2 + 3) = 1 . \ x . VP (3) s 1 . x = 1 Vy VT (3) = VP (3) = 1 x x = 1 y = 1 (tha iu kiny > 0 ) Vy ( x 1)2= 0 x =y = 1l nghim duy nht ca h phng trnh. 5 A- PHN CHUNG(7,0 im) Cu I (2,0 im) . Cho hm s: y = x3 (3m + 1) x2 + (5m + 4) x 8 (Cm) 9. Kho st s bin thin v v th(Cm)ca hm s khim = 0 . 10. Tm m (Cm) Cu II (2,0 im) ct trc honh ti 3 im phn bit lp thnh mt cp s nhn. 1.Gii phng trnh: 8 sin x = 3 cos x + 1 . sin x 2. Gii phng trnh:x + 4x (1 x )2+ 4 (1 x )3 = 0 1 x + 4x3+ 4x2 (1 x ). dx Cu III (1,0 im) .Tnh tch phn:I= } . 1 1+ x (1+ x ) 2 Cu IV (1,0 im). Cho hnh chp S.ABCD c y ABCD l hnh thoi ; hai ng cho AC =2 3a , BD = 2a v ct nhau ti O; hai mt phng (SAC) v (SBD) cng vung gc vimtphng(ABCD).BitkhongcchtimOnmtphng(SAB)bng Tnh th tch khi chp S.ABCD theo a. Cu V(1,0 im). Tm m bt phng trnh sau v nghim: 2 a 3 . 4 2 | s inx + =1 | | s inx =1 | 7

s inx | s inx | 2 3| s inx + =1 |+ | s inx =1 | + m 12 s inx | s inx | \.\. B- PHN RING(3,0 im). Th sinh ch c chn mt trong hai phn www.MATHVN.com Dnh cho hc sinh lp 12 n thi i hc- Cao ngB thi t lun- mn Ton 29 Vn Ph Quc- GV. Trng i hc Qung Nam D: 0982 333 443;0934 825 925 MATHVN.COM - Ton hc Vit Nam 21 3 3 2 B.1. CHNG TRNH CHUN Cu VI a(2,0 im) 1.TrongmtphngvihtoOxy,choim A (2;1) .Ly imBthuctrcOxc honh b > 0v im C thuc trc Oy c tung c > 0sao cho tam gic ABC vung ti A. Tm B, C sao cho din tch tam gic ABC ln nht. 2.Trong khng gianOxyzchoccimA ( 2; 0; 0 ) ,M (0; 3; 6) . Vitphng trnhmt phng( P ) VOABC= 3. chaA,Mvctcctrc Oy,Oz ticcimtngngB,Csaocho Cu VII a (1,0 im). Xt s phc:z = i m 1 m ( m 2i ) . Tm m z.z = 1 . 2 B.2. CHNG TRNH NNG CAO Cu VI b(2,0 im) 1.TrongmtphngvihtoOxy,chongthng A : x + 2 y 2 = 0 vhaiim A (1; 3) ,B (3; 2 ) . Tm M trnAsao choMA MBt gi tr ln nht. 2. Trong khng gian Oxyz, cho hai im x = t A (2; 3; 0) ,B (0; 2; 0) v ng thng A : y = 0 z = 2 t . TmC e Asao cho chu vi tam gic ABC nh nht. Cu VII b (1,0 im). Tm min xc nh ca hm s: y = ln ( 82+ lg x 3 42lg x ) Cu I BI GII 1. Hc sinh t gii 2.+iu kincn:Gis(Cm) cttrchonh ti 3im phn bit x1 , x2 , x3 lpthnh mtcpsnhn.Khix3 (3m + 1) x2 + (5m + 4 ) x 8 = 0 c3nghimlx1 , x2 , x3 x3 (3m + 1) x2 + (5m + 4) x 8 = ( x x) ( x x ) ( x x) x e 123 x3 (3m + 1) x 2 + (5m + 4) x 8 = x3 ( x1 + x2 + x3 ) x+ ( x1 x2 + x2 x3 + x3 x1 ) x x1 x2 x3 Suyra:x1 x2 x3= 8 .V x1 , x2 , x3 lpthnhmtcpsnhnnnx2= x x.Do x2= 8 x2= 2. Thayx2= 2 vophngtrnhx3 (3m + 1) x2+ (5m + 4) x 8 = 0 ta thu c4 2m = 0 m = 2. www.MATHVN.com Dnh cho hc sinh lp 12 n thi i hc- Cao ngB thi t lun- mn Ton 30 Vn Ph Quc- GV. Trng i hc Qung Nam D: 0982 333 443;0934 825 925 MATHVN.COM - Ton hc Vit Nam += + iu kin : Vim = 2 thay vo phng trnhx3 (3m + 1) x2 + (5m + 4 ) x 8 = 0ta thuc x3 7 x2 + 14 x 8 = 0 ( x 1) ( x 2 ) ( x 4 ) = 0 x = 1 , x = 2,x= 4 lp thnh mt cp s nhn . 123 Vym = 2l gi tr cn tm. Cu II 1.iu kin: sin x = 0t sin 2x = 0 x = k cos x = 02 ( k e ). Phng trnh cho tng ng vi: 8 sin2 x cos x =3 sin x + cos x 4 (1 cos 2x ) cos x =3 sin x + cos x 4 cos x 4 cos 2 x cos x = 3 sin x + cos x 4 cos x 2 (cos 3x + cos x ) =3 sin x + cos x 1 cos x 3 sin x = cos 3x cos | x + t | = cos 3x 22 3 | x + t = 3x + k 2t \. x = t + kt 3 6 ( k e ) tho iu kin ban u.

x + t = 3x + k 2t x = t + k t 3122 2. iu kin:0 s x s 1 t u = 4x v = 4 1 x u > 0 v > 0 44 uv1 T phng trnh cho ta c : u 2 + uv2 + v3= v2 + u 2 + u 2v (u v )

u + v (u 2 + uv + v2 )( = 0 (u v ) (u + v ) (1 u v ) = 0 (u v ) (1 u v ) = 0 ( do u , v khng m v u , v khng ng thibng 0 nnu + v > 0). Ta c cc h : u4= 1 x = 1 u v = 0 a) u4 + v4= 1 2 .Do 1 2 x = 1 . 12 v4= 2 1 x = 2 www.MATHVN.com Dnh cho hc sinh lp 12 n thi i hc- Cao ngB thi t lun- mn Ton 31 Vn Ph Quc- GV. Trng i hc Qung Nam D: 0982 333 443;0934 825 925 MATHVN.COM - Ton hc Vit Nam 2 } 0 0 } 2 u + v = 1 u + v = 1 b) u 4 + v4= 1 u + v 2 2uv ( 2u 2 v2= 1 ( ) u + v = 1 u + v = 1

u + v = 1

uv = 0

u = 0

v = 1

x = 0

1 x = 1 x = 0 2u2v2 4uv = 0 uv = 0

u + v = 1

u = 1 . Do

x = 1 . x = 1 uv = 2

uv = 2

v = 0

1 x = 0 Vy tp hp nghim ca phng trnh cho l: S = 0; 1 ;1 . 2 ` ) Cu III 0 dx 0 dx Ta c:I== 1 1 + x (1 + x )1 1| 1| 22 1 +

+ x | 4 \ 2 . t 1 + x = 1 sin t dx = 1 costdt; t e

t ; t=( 2222 2 t 1 tt 2 cos tdt 2 2 cos t2 2(t Lc : I= } = } dt = } 1 ( dt = 2J . 0 1 + t 2 1 1 sin2 t 44 dt 02 + cos t x 0 2dt 2 + cos t 2 Cn tnh J = } 2 + cos t . tt = tan 2dt dx = 21 + t 2 1 1 + t 2 1 dt Ta c:J= } 0 2 + 1 t 2 1+ t 2 = 2} 3 + t 2 tt =3 tan u dt =3 (1 + tan 2 u ) du tt 6 3 (1+ tan2 u ) du 16 t } 3 + 3 tan 2 u 3 } 6 3 J== 00 du =. Suy ra: I = t 2.t (9 4 3 )t =. 26 318 www.MATHVN.com Dnh cho hc sinh lp 12 n thi i hc- Cao ngB thi t lun- mn Ton 32 Vn Ph Quc- GV. Trng i hc Qung Nam D: 0982 333 443;0934 825 925 MATHVN.COM - Ton hc Vit Nam \.\ .> 2 . S Cu IV Trong ABO vung ti O ta c: tan ABO = OA = AC = 2 3a =3 OBBD2a ABO = 600hay ABD u.I D ( SAC ) ( ABCD ) A ( SBD ) ( ABCD ) SO = ( SAC ) ( SBD ) SO ( ABCD ) . O H a K Do ABD u nn vi H, KC B ln lt l trung im ca AB, HB nnDHABv DH =a 3 ; OK // DHvOK= 1 DH= a 3 22 OK AB AB (SOK) Gi I l hnh chiu caO ln SK ta c OI SK; AB OI OI (SAB) , hayOI= d (O; (SAB )) SOK vung ti O ta c: 1 OI 2 = 1 OK 2 + 1 SO2 SO = a . 2 Din tch y:S ABCD= 4S ABO = 2OA.OB = 2a3a = 2 3a2 . 3 Th tch khi chp S.ABCD: Cu V VS . ABCD = 1 SO.S 3 ABCD = 1 a 2 3a2= a3 . 3 23 Bt phng trnh cho tng ng vi: 2 2 | sin x =1 | | sin x =1 | + 1

sin x | sin x | 2 3| sin x =1 |+ | sin x =1 | + m sin x | sin x | \.\. tt = sin x 1 sin x ( x = kt , t e ) . Bi ton tng ng vi vic tm m bt phng trnh sau ng vi mi t: 2t 2 t + 1 3t 2 + t + m s 2 . V mu thc xc nht ennA = 1 12t < 0 m > 2 1 . Khi 3t 2 + t + m > 0 12 t e . V2t 2 t + 1 > 0 t enn 2t t + 1 s 2 t e 4t 2 + 3t + 2m 1 > 0 t e 3t 2 + t + m A = 9 16 ( 2m 1) s 0 m > 35 . 12 www.MATHVN.com Dnh cho hc sinh lp 12 n thi i hc- Cao ngB thi t lun- mn Ton 33 Vn Ph Quc- GV. Trng i hc Qung Nam D: 0982 333 443;0934 825 925 MATHVN.COM - Ton hc Vit Nam Vym > 35 12 l gi tr cn tm tha mn yu cu bi ton. B- PHN RING B.1. CHNG TRNH CHUN Cu VI a 1. Ta c:B (b; 0) e Ox , C (0; c ) e Oyvib,c > 0 . AB = (b 2; 1) ,AC = (2; c 1) . Do tam gic ABC vung ti A nn AB AC = 0 2 (b 2) (c 1) = 0 c 1 = 2 (b 2) . S ABC = 1 ABAC = 1 22 (b 2)2+ 1. 4 + (c 1)2 = 1 (b 2)2+ 1 4 + 4 (b 2 )2 = (b 2)2+ 1 > 1 . 2 Doc = 2b + 5 > 0 0 s b s 5 . 2 Du = xy ra khib = 2 ,c = 1 . Vymin S ABC= 1 khi B ( 2; 0 ),C (0;1) . 2.GisB (0; b; 0 ) ,C (0; 0; c ) .Mtphng( P )quaA,B,Cnncphngtrnhl: x + y + z = 1. 2bc M (0; 3; 6) e ( P ) 3 + 6 = 1 6b 3c = bc. bc Li c : V = 1 .OA.S = 2 . 1 bc= bc = 3 bc= 9. OABC 3 OBC 323 T ta c c cc h phng trnh : b = 3 c = 2b 3 = 6b 3c = 9 c = 2b 3 c = 2b 3 b = 3

c3 (a) . bc = 9 b ( 2b 3) = 9

2b2 3b 9 = 0

3 . 3b =

b = 2

2 c = 6 (ii ).6b 3c = 9 bc = 9 c = 2b + 3 b ( 2b + 3) = 9 c = 2b + 3 2b2 + 3b + 9 = 0 ( H v nghim). Vy phng trnh mt phng( P ) cn tm l : www.MATHVN.com Dnh cho hc sinh lp 12 n thi i hc- Cao ngB thi t lun- mn Ton 34 Vn Ph Quc- GV. Trng i hc Qung Nam D: 0982 333 443;0934 825 925 MATHVN.COM - Ton hc Vit Nam 7| x + y + z = x +y+z= 1, 2332 Cu VII a = 3 6 1. 2 Ta c:z = i m = (1 m2 ) + 2mi ( m + i ) (1 m2 2mi ) (1 m2 )2+ 4m2 m (1 m2 ) + 2m + i (1 m2 + 2m2 ) == (1 + m2 )2 m (1+ m2 ) + i (1+ m2 ) (1 + m2 )2 = m 1 + m2 + 1 i 1 + m2 z = 2 m 1 + m2 1 i 1+ m2 Do z.z = 1 m + 1 = 1 1 = 1 m2 + 1 = 2 m = 1. 2 (m2+ 1)2 2 m2 + 12 B.2. CHNG TRNH NNG CAO Cu VI b 1.tF ( x; y ) = x + 2 y 2 .DoF (1; 3).F (3; 2 ) = 15 < 0nnAvBnmvhaipha khc nhau i viA . Gi d l ng thng qua A vung gc viA . Khi d nhnud= nA= (1; 2) lm vect ch phng nn c phng trnh: x 1 = y 3 2x y + 1 = 0 . 12 GiH= d A To im H l nghim ca h: 2x y + 1 = 0 x = 0 H (0;1) . x + 2 y 2 = 0 y = 1 Gi C l im i xng ca A quaAnn H l trung im ca AC. Suy raC ( 1; 1) . Ta c: MA MB=MC MBs BC = 17. Du = xy ra khi v ch khiM= BC A. ngthngBCquaB (3; 2 )nhnBC = (4;1)lmvectchphngnncphng trnh: x 3 = y + 2 x + 4 y + 5 = 0 . 41 x + 2 y 2 = 0 x = 9 7 To im M cn tm l nghim ca h: M | 9; | . x + 4 y + 5 = 0y = 2 2 \. VyMax MA MB= 17khiM | 9; =7 | . 2 | \. 2. Ta c:C e A C (t; 0; 2 t ) Gi P l chu vi ca tam gic ABC, ta c: P = AB + AC + BC= 15 + 6 2 + 2t 2 8t + 17 + 2t 2 4 y + 6 www.MATHVN.com Dnh cho hc sinh lp 12 n thi i hc- Cao ngB thi t lun- mn Ton 35 Vn Ph Quc- GV. Trng i hc Qung Nam D: 0982 333 443;0934 825 925 MATHVN.COM - Ton hc Vit Nam 55 | = 15 + 6 2 +2 (t 2 )2 + 32+2 (t 1)2+ 22 Trong mt phng Oxy, tu = ( AC + BC =u+v>u + v= 3 3 2 (t 2); 3) , v = ( 2 (t 1); 2 ) Du bng xy ra khi v ch khiu vvcng phng 2 (t 2) 2 (t 1) = 3 t = 7 . 25 MinP = 15 + 6 3 + 3 3khiC | 7 ; 0; =3 | . \. Cu VII b Hm s xc nh khi v ch khi 2+ lg x 32lg xx > 0 x > 0 84> 0 82+lg x > 3 42lg x (82+lg x )3> (42lg x )2 x > 0 x > 0 9(2+lg x ) 2 24( 2lg x ) 9 ( 2lg x )4 (2 lg x ) x > 0> x > 0 +> lg x > 2 x > 100 x > 100 . Vy min xc nh ca hm s cho l:D = (100; + ) . 6 A- PHN CHUNG(7,0 im) Cu I (2,0 im) . Cho hm s: y = x3 3x2 + 2 (C ) 11. Kho st s bin thin v v th(C )ca hm s. 12. Tmm (C )cimccivcctiunmvhaiphaivingtrn (Cm ) : x2 + y2 2mx 4my + 5m2 1 = 0 . Cu II (2,0 im) 1. Tm nghim thuc khong(0; t ) ca phng trnh: 7 | sin 3x cos 3x cos x | = 4 cos 2x . 2 sin 2x 1 | \. 2.Tm m h bt phng trnh sau c nghim: 75 x + x +1 75+ x+1 + 2012 x s 2012 x2 (m + 2) x + 2m + 3 > 0 1 dx Cu III (1,0 im) .Tnh tch phn: I = } 242 . 1 1 + x + x +x + 3x + 1 CuIV(1,0im).ChohnhchpS.ABCcyltamgicABCvungtiC, AC = a,AB = 2a, SAvung gc vi y. Gc giahai mtphng (SAB) v (SAC) bng www.MATHVN.com Dnh cho hc sinh lp 12 n thi i hc- Cao ngB thi t lun- mn Ton 36 Vn Ph Quc- GV. Trng i hc Qung Nam D: 0982 333 443;0934 825 925 MATHVN.COM - Ton hc Vit Nam 600 . Gi H, K ln lt l hnh chiu ca A ln SB, SC. Chng minhAK HK tch khi chp S.ABC. v tnh th Cu V(1,0 im). Chox,y, z e (0,1) . Chng minh rngxyz +(1 x ) (1 y ) (1 z ) < 1 . B- PHN RING(3,0 im). Th sinh ch c chn mt trong hai phn B.1. CHNG TRNH CHUN Cu VI a(2,0 im) 5. TrongmtphngOxy,choimM ( 2; 3)vngthng A : ( m 2 ) x + ( m 1) y + 2m 1 = 0 .TmthamsthcmkhongcchtMn ng thngAl ln nht. 6. Trong khng gian Oxyz, cho hai ng thng x = 2 2t d: x 2 = y 1 = z vd : y = 3 (t e ) .Chngminhhaingthngtrncho 1 212 2 z = t nhau. Hy vitphng trnh mtcu (S) bit(S) cng knh l on vung gc chung cad1 ,d2 . Cu VII a (1,0 im). Cho M, N lhai im trong mtphng phcbiu din theoth t 22 cc s phcz1 ,z2 khc 0 tha mn ng thcz1+ z2= z1 z2 . Chng minh tam gic OMN l tam gic u. B.2. CHNG TRNH NNG CAO Cu VI b(2,0 im) 1. Trong mt phng vi h ta Oxy, cho hnh ch nht ABCD c din tch bng 12, tm Ilgiaoimcahaingthngd1 : x y 3 = 0,d 2 : x + y 6 = 0 .Trung im Mca cnh AD l giao im ca ng thng nht. d1 vi trc Ox. Tm to cc nh cahnh ch 2.Trong khng gian vi h to Oxyz, chomtphng(o )vmtcu( S )ln ltc phngtrnh:2x y + 2z 3 = 0 ;( x 1)2+ ( y + 2)2+ ( z 4)2= 25. Xtvtrtngi gia mt cu( S )v mt phng(o ) . Vit phng trnh mt cu(V) mt phng(o ) . i xng vi( S )qua Cu VII b (1,0 im). Gii bt phng trnh:log2(3x + 1 + 6) 1 > log2(7 10 x ) . Cu I 1. Hc sinh t gii 2. Hm s xc nh v lin tc trn . BI GII Ta c: y' = 3x2 6x;y' = 0 3x2 6x = 0 x = 0 x = 2 Ta cc im cc tr:A (0; 2) ,B ( 2; 2) . Cch 1: (C) c im cc tr nm v hai pha i vi ng trn(Cm) khi v ch khi www.MATHVN.com Dnh cho hc sinh lp 12 n thi i hc- Cao ngB thi t lun- mn Ton 37 Vn Ph Quc- GV. Trng i hc Qung Nam D: 0982 333 443;0934 825 925 MATHVN.COM - Ton hc Vit Nam ( 2 ) (4 8m + 5m2 1)(4 + 4 4m + 8m + 5m2 1) < 0 (5m2 8m + 3)(5m2+ 4m + 7 ) < 0 5m2 8m + 3 < 0 3 < m < 1 . 5 ( v5m2 + 4m + 7 > 0 m e ). Cch 2: ng trn(Cm) : ( x m)2+ ( y 2m )2= 1c tm 2 I ( m; 2m ) , bn knh R = 1 . Tac: IB = 5m2 + 4m + 8 = 5 | m + 2 | + 36 >6> 1 = R imBnmphangoi 5 | 55 \. ng trn (Cm ) . Do im A nm pha trong ng trn (Cm ) , tc l: IA < 1 = R Cu II 5m2 8m + 4 < 1 5m2 8m + 3 < 0 3 < m < 1. 5 x = 5t + k 2t 1. iu kin:sin 2x = 1 12 (k e ) . x = t 12 + k 2t Phng trnh cho tng ng vi: | 3sin x 4 sin 3 x 4 cos3 x + 3 cos x| 7

\ 2 sin 2 x 1 cos x | = 4 cos 2x . | 3(sin x + cos x ) 4 (sin x + cos x ) (1 sin x cos x ) | 7 | = 4 cos 2 x \ 2 sin 2 x 1 . | (sin x + cos x ) (3 4 + 4 sin x cos x ) 7

| cos x | = 4 cos 2 x \ 2 sin 2 x 1 . | (sin x + cos x ) (2 sin 2x 1) | 7 cos x = 4 cos 2 x 7 sin x = 4 1 2 sin x 2 sin 2 x 1 | \. = t +t sin x = =1 xk 2 2 sin2 x 7 sin x + 3 = 0 2 6

sin x = 3 x = 5t + k 2t

6 Vx e (0;t )nn ta c c hai nghim sau y: x = t , x = 5t . 2. iu kin: 66 x > 1 Ta c:75 x + x +1 75+x+1 + 2012x s 2012 7 x +1 (75 x 75 ) s 2012 (1 x ) (*) Nux > 1th VP (*) < 0 < VT (*)( V l). Bt phng trnh (*)lun ng khi1 s x s 1 . Hbtphngtrnhchocnghim Btphng trnh x2 (m + 2) x + 2m + 3 > 0c nghimx e |1;1| Bt phng trnh:m ( x 2 ) s x2 2x + 3c nghimx e |1;1| www.MATHVN.com Dnh cho hc sinh lp 12 n thi i hc- Cao ngB thi t lun- mn Ton 38 Vn Ph Quc- GV. Trng i hc Qung Nam D: 0982 333 443;0934 825 925 MATHVN.COM - Ton hc Vit Nam m > x2 2x + 3 x 2 c nghim x e |1;1| m > min xe|1;1| f( x ) vi f ( x ) = x2 2x + 3 . x 2 Hm s ny c vit li : 2 f ( x ) = x + 3 x 2 ,x e |1;1| f ' ( x ) = 1 3 = x 4x + 1 f ' ( x ) ( x 2)2 x = 2 + = 0 ( x 2)2 3 e |1;1| x = 2 3 e |1;1| f (2 3 ) = 2 2 3; f ( 1) =f (1) = 2 ; Do min xe|1;1| f( x ) =f( 1) =f(1) = 2 . Vym > 2tho mn yu cu bi ton. Cu III tx = t dx = dt . 1 dt 1 dx Khi :I= } 242 = } 242 . 11 t + t + 1 t + 3t + 11 1 x + x +x + 3x + 1 Suy ra:2I= } ( g ( x ) + g ( x )) dx vig ( x ) = 1 . 242 Ta c: 1 g ( x ) + g ( x ) = 1+ x + x +x 1 + 1 + 3x + 1 1 + x + x2 +x4 + 3x21 x + x2 +x4 + 3x2 1 x + x2 + = x4 + 3x2 + 1 + 1 + x + x2 + x4 + 3x2 + 1 = 2 (1 + x2 + x4 + 3x2 + 1 ) (1 + x + x2 + x4 + 3x2 + 1)(1 x + x2 + x4 + 3x2 + 1) ( 242 )2 2 2 (1 + x2 + = x4 + 3x2 + 1) 1 + x +x + 3x + 1 x 1 + x4 + x4 + 3x2 + 1 + 2 x2 + 2x4 + 3x2 + 1 + 2x2x4 + 3x2 + 1 x2 2 (1 + x2 + = x4 + 3x2 + 1) 2 + 2x4 + 4x2 + 2x4 + 3x2 + 1 + 2x2x4 + 3x2 + 1 1 + x2 + = x4 + 3x2 + 1 1 + x4 + 2x2 +x4 + 3x2 + 1 + x2x4 + 3x2 + 1 1 + x2 + x4 + 3x2 + 11 ==. (1 + x2 + x4 + 3x2 + 1) + (x2+ x4 + x2 x4 + 3x2 + 1 ) 1+ x2 1 dx 1 dx 1 dx Do :2I= } 1 + x2= 2 } 1 + x2 I= } 1 + x2 t 100 x = tan u dx = (1 + tan 2 u ) du www.MATHVN.com Dnh cho hc sinh lp 12 n thi i hc- Cao ngB thi t lun- mn Ton 39 Vn Ph Quc- GV. Trng i hc Qung Nam D: 0982 333 443;0934 825 925 MATHVN.COM - Ton hc Vit Nam . tt 4 (1 + tan2 u ) du 4 t t } 1 + tan 2 u }0 4 VyI== 00 du = u 4=. Cu IV. S . SA BCH Ta c: BC (SAC ) BC AK AC BC MAK SCnnAK (SBC) AK HK . Hn naAK (SBC) AK SB Suy ra:SB ( AHK ) SB HK A K B MSB = ( SAB ) ( SBC )nn AHK= 600 . a 2 3 D dng tnh cSABC= 2 C Trong tam vung AKH c AK = AH sin 600= 3 AH 2 Cc tam gic SAB, SAC vung ti A nn 1 AH2 = 1 SA2 + 1 AB2 = 1 SA2 +1 4a 2 (1) 1 AK2 = 1 SA2 + 1 AC2 4 3AH2 = 1 + 1 SA2 a 2 1 AH2 = 3 4SA2 +3 4a 2 (2) T (1) v (2) suy ra: 1 4SA2 = 1 2a 2 SA = a 2 . 2 VyVS.ABC= a3 6 . 12 Cu V Cch 1:Do x, y, z e (0;1) nn xyz e (0;1) (1 x ) (1 y ) (1 z ) e (0;1) Khi : xyz + (1 x ) (1 y ) (1 z ) < 3xyz + 3 (1 x ) (1 y ) (1 z ) p dng bt ng thc Cauchy ta c: 3xyz + 3 (1 x ) (1 y ) (1 z ) s x + y + z + 1 x + 1 y + 1 z = 1 33 Suy ra: Cch 2: xyz +(1 x ) (1 y ) (1 z ) < 1 . Ta cz e (0;1) 1 z e (0;1) Do : xyz + (1 x )(1 y )(1 z ) < xy + (1 x )(1 y ) p dng bt ng thc Bunhiacopski ta c: www.MATHVN.com Dnh cho hc sinh lp 12 n thi i hc- Cao ngB thi t lun- mn Ton 40 Vn Ph Quc- GV. Trng i hc Qung Nam D: 0982 333 443;0934 825 925 MATHVN.COM - Ton hc Vit Nam ( xy + 1 x1 y )2s ( x + 1 x ) (1 y + y ) = 1 xy + (1 x )(1 y ) s 1 Suy ra: xyz + (1 x ) (1 y ) (1 z ) < 1 . B.1. CHNG TRNH CHUN Cu VI a 1. Ta c:d = d ( M , A ) = 2 2 (m 2) + 3 (m 1) + 2m 1 = (m 2)2 + ( m 1)2 7m 8 2m2 6m + 5 Suy ra: d 2= 49m 112m + 64 Cch 1: 2m2 6m + 5 2 Ta c:d 2= 49m 112m + 64 2d 2 m2 6d 2 m + 5d 2= 49m2 112m + 64 2m2 6m + 5 hay (2d 2 49) m2 + 2 (56 3d 2 ) m + 5d 2 64 = 0 . + Nu2d 2 49 = 0 d= 7 2 2 ( dod > 0 ) th35m + 117 = 0 m = 117 . 270 +Nu2d 2 49 = 0 d = 7 2 2 th phngtrnhbchai n theonm trn cnghim khi v ch khiA' = (56 3d 2 )2 (2d 2 49)(5d 2 64 ) > 0 d 4 37d 2s 0 d 2 37 s 0 ( dod > 0 ) d 2s 37 d s37 Vid =37ta c:25m2 110m + 121 = 0 m = 11 . 5 Vymax d= Cch 2: 37khim = 11 5 49m2 112m + 64 Xt hm s:f (m) = 2m2 6m + 5 Tp xc nhD = 2 f ' ( m) = 70m + 234m 176 (2m2 6m + 5)2

m = 8 f ' ( m) = 0 70m2 + 234m 176 = 0 7 lim f ( m) = 49 , lim f ( m) = 49 .

m = 11

5 x 2 x + 2 www.MATHVN.com Dnh cho hc sinh lp 12 n thi i hc- Cao ngB thi t lun- mn Ton 41 Vn Ph Quc- GV. Trng i hc Qung Nam D: 0982 333 443;0934 825 925 MATHVN.COM - Ton hc Vit Nam 5 3 33 | \. z \. 2 3 Bng bin thin: 8 t 7 f ' ( m ) 0 f( m )49 2 0 11 5+ +0 37 49 2 Da vo bng bin thin ta suy ra maxf ( m) =f | 1=1 | = 37 . Vymax d= 37khim = 11 . 5 me| 2. ng thngd1i qua M ( 2;1; 0)v c vect ch phngu1 = (1; 1; 2) . ng thng d2i qua N (2; 3; 0 )v c vect ch phngu1 = (2; 0;1) .

Ta c:

u1; u2 ( MN = 10 = 0 d1 ,d2 cho nhau. A (2 + a;1 a; 2a ) e d1; B (1 2b; 3; b ) e d2 AB.u= 0 a = 1 AB l on vung gc chung ca d1 ,d2 1 3 . Do : A | 5 ; 4 ; =2 | ,B (2; 3; 0 ) \. AB.u2= 0 b = 0 Mt cu (S) c tm I | 11 ; 13 ; =1 | , bn knh R = AB = 30 663 |26 222 Vy phng trnh ca mt cu (S) l : | x 11 | + | y 13 | + | z + 1 | = 5 . 6 | 6 | 3 | 6 Cu VII a \.\.\. z 2= z( z z) z2=z z z Ta c: z 2 + z 2 = z z 2121 2121 (-) 121 2 z 2= z ( z z) zzzz V z1 ,z2= 0 nn z1,z2 1212 = 0 . 22 1 = 22 1 T(-) ta c: z z =z2 =z1 z 3=zz =z 2122 z12 1212 Do :z2 z1=z1 =z2 MOM=z1 ; ON = z2 ; MN =z2 z1. Vy tam gic OMN u. B.2. CHNG TRNH NNG CAO Cu VI b(2,0 im) www.MATHVN.com Dnh cho hc sinh lp 12 n thi i hc- Cao ngB thi t lun- mn Ton 42 Vn Ph Quc- GV. Trng i hc Qung Nam D: 0982 333 443;0934 825 925 MATHVN.COM - Ton hc Vit Nam 2 2 | 1

t = 4 M = 9 x y 3 = 0 x 2 1. Ta c :I= d1 d2to imI l nghim ca h : x + y 6 = 0 3 . y = 2 Do I | 9 ; =3 | . A B \. M= d1 Ox M (3; 0) . I Ta c :AB = 2 I M= 3 2 D C Gi thit S ABCD = 12 AB.AD = 12 AD =12 3 2 = 2 2 VI ,M e d1nnd1 AD . ngthngADquaM (3; 0) vnhnuAD= nd= (1; 1)lmvectchphngnnc phng trnh: x 3 = y 0 x + y 3 = 0 . 11 To A, D c dng:(t; 3 t ) . Tac: MA = MD =2 MA2= MD2= 2 (t 3)2+ (3 t )2= 2 t 2 6t + 8 = 0 t = 2 . Do :A ( 2;1) ,D ( 4; 1)hocA ( 4; 1) ,D ( 2;1) . + ViA ( 2;1) ,D ( 4; 1)suy raC (7; 2 ),B (5; 4) + ViA ( 4; 1) ,D ( 2;1)suy raC (5; 4 ),B (7; 2) 2. +Mt cu( S )c tmI (1; 2; 4 )v bn knhR = 5 . d ( I ; (o )) = 2.1 ( 2 ) + 2.4 3 22 + (1)2 + 22 = 3 < 5 = R . Vy mt phng(o )ct mt cu( S ) . + Gi J l im i xng ca I qua mt phng(o )th(V) R = 5 . l mt cu c tm J, bn knh Gi dlngthngiquaIvvunggcvi(o ) .Khidnhn vect ch phng nn c phng trnh tham s: x = 1 + 2t y = 2 t z = 4 + 2t no= ( 2; 1; 2) lm Gi H l trung im IJ. Khi H l hnh chiu vung gc ca I ln mt phng(o ) . To im H ng vi tham s t l nghim ca phng trnh: 2 (1 + 2t ) ( 2 t ) + 2 (4 + 2t ) 3 = 0 t = 1 . www.MATHVN.com Dnh cho hc sinh lp 12 n thi i hc- Cao ngB thi t lun- mn Ton 43 Vn Ph Quc- GV. Trng i hc Qung Nam D: 0982 333 443;0934 825 925 MATHVN.COM - Ton hc Vit Nam 22 Suy raH ( 1; 1; 2) . V H l trung im IJ nn d dng suy ra cJ ( 3; 0; 0 ) . Vy phng trnh mt cu (V) l:( x + 3)2+ y2 + z 2= 25. Cu VII b(1,0 im) iu kin: 1 s x s 10. 3 Bt phng trnh cho tng ng vi: log 3x + 1 + 6 > log7 10 x 3x + 1 + 6 > 7 10 x 22() 3x + 1 + 2 10 x > 8 3x + 1 + 4 (3x + 1) (10 x ) + 4 (10 x ) > 64 4 (3x + 1) (10 x ) > x + 23 16 (3x + 1) (10 x ) > ( x + 23)2 49 x2 418 x + 369 s 0 1 s x s 369 49 Kt hp vi iu kin ban u ta thu c tp hp nghim ca bt phng trnh trn l: S =1; = 369 ( .

49 ( 7 A- PHN CHUNG(7,0 im) Cu I (2,0 im) . Cho hm s: y = 3x 2 x + 1 (C ) 13. Kho st s bin thin v v th(C )ca hm s. 2.Gi I l giao im hai tim cn. Vit phng trnh tip tuyn d ca th hm s bit d ct tim cn ng v tim cn ngang ln lt ti A v B thocos BAI= 5 26 . 26 Cu II (2,0 im) 1.Gii phng trnh: 1cot x + 2 sin 2 x sin x + cos x = 2 sin( x + t ) 2 2.Gii bt phng trnh sau:x2 3x + 2 +x2 4 x + 3 > 2x2 5x + 4 Cu III (1,0 im)Tnh din tch hnh phng gii hn bi ba ng sau: 2 Elip (E): x 4 + y2= 1 , ng thng d: x 2 3 y 4 = 0v trc honh. Cu IV (1,0 im). Cho hnh chp SABCD c y ABCD l hnh thang vung ti A v D, AB = AD = 2a,CD = a ,gcgiahaimtphng(SBC)l(ABCD)bng600 .GiIl trung im ca cnh AD. Bit hai mt phng (SBI) v (SCI) cng vung gc vi mt phng (ABCD). Tnh th tch khi chp S.ABCD. www.MATHVN.com Dnh cho hc sinh lp 12 n thi i hc- Cao ngB thi t lun- mn Ton 44 Vn Ph Quc- GV. Trng i hc Qung Nam D: 0982 333 443;0934 825 925 MATHVN.COM - Ton hc Vit Nam C+ C+ C+ ... + C= 2 3x 2 Cu V(1,0 im). Tm m phng trnh: bit trong on

0; =t ( . mx2 + 2 cos x = 2c ng 2 nghim thc phn 2 B- PHN RING(3,0 im). Th sinh ch c chn mt trong hai phn B.1. CHNG TRNH CHUN Cu VI a(2,0 im) 1.Trong mtphng vi h ta Oxycho ABC ctrng tmG ( 2; 0 ).Bitphng trnh cc cnh AB,AC theoth t l4x + y + 14 = 0 ,2x + 5 y 2 = 0. A,B,C . Tm ta cc nh 2. Trong khng gian Oxy cho cc imA ( 3; 5; 5) ,B (5; 3; 7 )v mt phng ( P ) : x + y + z = 0 .Tm im Cu VII a (1,0 im) M e ( P )sao cho (MA2+ MB2 ) nh nht. n Trongkhaitrinsauycbaonhiushnghut( 3 + 4 5 ) bitnthamn 1232n 4n+14n+14n +14 n+1 496 1 . B.2. CHNG TRNH NNG CAO Cu VI b(2,0 im) 1. Cho paraboly = x2 .Mt gc vung nh O ct Parabol tiA1v A2 . Hnh chiu ca A1, A2 ln Ox lB1 ,B2 . Chng minh rng: OB1.OB2= const . 2. Cho mt cu:( S ) : x2 + y 2 + z 2 2x + 2z 2 = 0v cc imA (0;1;1) ,B ( 1; 2; 3) C (1; 0; 3) . Tm im D thuc mt cu (S) sao cho th tch t din ABCD ln nht. n | Cu VII b (1,0 im) Tm s nguyn dng n b nht 3 + i | l s thc

1 i| \. BI GII A- PHN CHUNG Cu I 1. Hc sinh t gii || 2. GiM x; 0 e (C ) ( x = 1) 0|0 \ x0 + 1 . Tip tuyn d vi (C) ti M c phng trnh: y 3x0 2 = 5 ( x x) . x+ 1 ( x+ 1)2 0 0 0 Do d ct tim cn ng, tim cn ngang ln lt ti A, B vIABccos BAI= 5 26 26 nn tan BAI=5 . www.MATHVN.com Dnh cho hc sinh lp 12 n thi i hc- Cao ngB thi t lun- mn Ton 45 Vn Ph Quc- GV. Trng i hc Qung Nam D: 0982 333 443;0934 825 925 MATHVN.COM - Ton hc Vit Nam 0 4 Li c tan BAIl h s gc ca tip tuyn d m y' ( x) = 5 > 0 nn 5 = 5 ( x + 1)2= 1 x0= 0. ( x0 + 1)2 ( x1)20 x = 2 0 + 0 Vy c hai tip tuyn tho mn yu cu bi ton: Cu II 1. iu kin:sin x = 0, sin x + cos x = 0. Ta c: y = 5x 2;y = 5x + 2 . 1cot x + 2 sin 2x sin x + cos x = 2 sin( x + t ) 2 cos x 2 sin x + 2 sin x cos x 2 cos x = 0 sin x + cos x cos x 2 cos2 x

cos x = 0 | t |

= 0 cos x sin( x + ) sin 2 x | = 0

|=t | 2 sin xsin x + cos x \4. sin x + \ | = sin 2 x . x = t + kt 2 x = t + kt x = t + k 2t 2 ( k e ) ( tho iu kin ban u) 4 x = t + k 2t x = t + =k 2t 43

43 x = t + kt 2 Vy phng trnh cho c hai h nghim: (k e ) x = t + k 2t

43 x2 3x + 2 > 0 x s 1v x > 2 2.iu kin: x2 4x + 3 > 0 x s 1v x > 3 x s 1v x > 4 . x2 5x + 4 > 0 x s 1v x > 4 x2 3x + 2 +x2 4 x + 3 > 2x2 5x + 4 ( x2 3x + 2 x2 5x + 4 ) + ( x2 4x + 3 x2 5x + 4 ) > 0 2 ( x 1) x 1 +> 0 x2 3x + 2 +x2 5x + 4x2 4x + 3 +x2 5x + 4 www.MATHVN.com Dnh cho hc sinh lp 12 n thi i hc- Cao ngB thi t lun- mn Ton 46 Vn Ph Quc- GV. Trng i hc Qung Nam D: 0982 333 443;0934 825 925 MATHVN.COM - Ton hc Vit Nam D C ( x 1) | 2 + x 1 | > 0 | \x2 3x + 2 +x2 5x + 4x2 4 x + 3 +x2 5x + 4 . x 1 > 0 x > 1 . Kt hp vi iu kin ban u, ta c:x = 1 v x > 4 . Cu III + Giao im ca d v (E) l | 3 | A1; 2| y \. + d ct trc honh tiB ( 4; 0) . Gi A l chn ng vung gc h t A xung Ox. Khi din tch hnh phng cn tm A l phn gch xin nh hnh v. OB x Ta c:S = SAKB SAKC . KC SAKB = 1 AK.KB = 13 .3 = 3 3 22 24 Nhngimthuchnhphngc tung y > 0nnt phngtrnh(E) suyra: y = 1 2 4 x 2. 2 Do :S 1 AKC= } 4 x 2 dx t 2 1 x = 2 sin t dx = 2 cos tdt Khi : ttt t 1 222 |1| 2 t3 S= } 4 4 sin 2 t .2 cos tdt = 2 } cos2 tdt = } (1 + cos 2t ) dt = t + sin 2t | = AKC 22 t 34 ttt\. 666 6 Vy S = 3 3 | t 3 | =3 t ( 4 34| 3 \. vdt).S Cu IV Ta c: ( SBI ) ( ABCD ) ( SCI ) ( ABCD ) SI= ( SBI ) ( SCI ) SI ( ABCD ) . Gi I l hnh chiu ca I ln BC. A E B Theo nh l 3 ng vung gc suy SH BC .I MBC = ( SBC ) ( ABCD )nnSHI= 600 H www.MATHVN.com Dnh cho hc sinh lp 12 n thi i hc- Cao ngB thi t lun- mn Ton 47 Vn Ph Quc- GV. Trng i hc Qung Nam D: 0982 333 443;0934 825 925 MATHVN.COM - Ton hc Vit Nam 4 ( 3 2 2 l gc gia hai mt phng ( SBC )v (ABCD). Ta c :S ABCD = 1 AD ( AB + CD ) = 1 2a (2a + a ) = 3a 2 . 22 2 S IBC= S ABCD S ICD S IAB = 3a2 1 ID.CD 1 IA.AB = 3a 2 1 a 2 a2= 3a. 2222 3a2 2S 2.2 3a2 3a 5 KCE AB , ta c: IH= IBC === BC CE 2 + BE 24a 2 + a 2 5 Trong tam gic SIH vung ti I, ta c:SI= IH tan 600= 3a 5 . 3 = 3a 15 . Do : V= 1 SI .S = 1 . 3a 15 .3a2= 3a15 55 ( vtt). Cu V 3 ABCD355 + R rng:x = 0l mt nghim ca PT. 4 sin 2x + Vi x e | 0; =t ( ta c:m = 2 . tt = x ,x e | 0; =t ( t e | 0; =t ( 2 ( x2 2 2 ( 4 ( \ Ta c phng trnh sau: m = | sin t | , t e | 0; t ( . \\ t | 4 ( Xt hm s: f (t ) = | sin t | \.\ , t e | 0; t ( . t | 4 ( ' sin 2t \.\ |=t ( |t=( f(t ) = t 3 (1 tan t ) < 0 t e 0; \ 4 ( suy ra f(t) nghch bin trn 0;. \ limf (t ) = 1 . t 0+ Phng trnh trn c 2 nghim thc phn bit tong on nghim thc trong | 0; t=( .

0; =t ( PT

2 ( f (t ) = mc ng 1 4 ( t \ Bng bin thin : t 0 4 f ' (t ) f(t ) 1 8 t 2 Da vo bng bin thin ta tm c : 8s m < 1 . t 2 B- PHN RING(3,0 im). Th sinh ch c chn mt trong hai phn B.1. CHNG TRNH CHUN www.MATHVN.com Dnh cho hc sinh lp 12 n thi i hc- Cao ngB thi t lun- mn Ton 48 Vn Ph Quc- GV. Trng i hc Qung Nam D: 0982 333 443;0934 825 925 MATHVN.COM - Ton hc Vit Nam 2=C+ C+ C+ C+ ... + C CC = k= k 2 Cu VI a(2,0 im) 1. Ta c :A = AB AC Ta im A l nghim ca h 4x + y + 14 = 0 2x + 5 y 2 = 0 x = 4 y = 2 hayA ( 4; 2 ) B e AB B (b; 4b 14 ); C e AC C | c; 2 1 c | 55| \. 4 + b + c = 6 b = 3 B (3; 2) V G l trng tm ABC nn 21 2 4b 14 + 55 c = 0 c = 1 C (1; 0) VyA ( 4; 2 ) ,B ( 3; 2 ) ,C (1; 0) . 2. Gi H l trung im ca on thng AB. Theo cng thc di trung tuyn MH ca tam 2 gic MAB ta c:MA2 + MB 2= 2MH 2 + AB. 2 MA2 + MB2nh nht MHnh nht. Ta c:H (1;1;1)c nh. Suy ra MH nh nht Ml chn ca ng vung gc h t H n (P). rngOH= (1;1;1)l vect php tuyn ca mp (P) vO e ( P ) nn M O (0; 0; 0 ). VyMA2 + MB2nhnhtkhivchkhiMtrngvigctav Min ( MA2 + MB2 ) = Min (OA2+ OB2 ) = 142 . Cu VII a(1,0 im) Ta c:()4 n +1 0122334 n +14 n +1 1 + x= C4 n +1 + C4 n +1x + C4 n +1x + C4n +1x + ... + C4 n +1x Chnx = 1 4n +101234 n +1 2= C4 n +1 + C4n +1 + C4n +1 + C4n +1 + ... + C4 n +1 ( 01232 n ) Suy ra = 2C4 n +1 + C4n +1 + C4n +1 + C4n +1 + ... + C4 n +1 4n01232 n 4n +14n +14n +14n +14 n +1 Hay24n= 2496 4n = 496 n = 124. 124 124 124 kk 124 124k k (3 + 4 5 ) 124 ( 3 )( 4 5 ) 124 3 5 4 . k =0k =0 124 k 2 Trong khai trin c s hng hu t k 4 0 s k s 124 k4 0 s k s 124 k = 4t 0 s 4t s 124 0 s t s 31 C 32 gi tr ca tsuy ra c 32 gi tr ca k. Vy trong khai trin trn c 32 s hng hu t. www.MATHVN.com Dnh cho hc sinh lp 12 n thi i hc- Cao ngB thi t lun- mn Ton 49 Vn Ph Quc- GV. Trng i hc Qung Nam D: 0982 333 443;0934 825 925 MATHVN.COM - Ton hc Vit Nam 100 2 B.2. CHNG TRNH NNG CAO Cu VI b(2,0 im) 1. Gi sA ( x; x2 ) e ( P ) .Khi : A2- B1l hnh chiu ca A1ln Ox nn c ta -A1 - lB1 ( x0 ; 0 ) OB1 =x0 . B2 OB1 Phng trnh ng thngOA1 : y = xx0 . DoOA2 OA1 1 nn phng trnh ng thngOA2 : y = x . x0 y = x2 | 11 | Ta A2 lnghim cahphng trnh: y = 1 A xx ; | . B2 x2 lhnhchiu x0 | 1 | 1 \00 . caA2 ln Ox nn c ta lB2 x; 0 | OB2=x.V thOA1.OB2= 1 . \0.0 2.( S )c tmI (1; 0; 1), bn knhR = 2 . AB = (1; 3; 4),AB = (1; 1; 4) . Gi(o ) l mt phng cha 3 im A, B, C nhnn = AB, AC ( = (8; 8; 4) lm vect php tuyn nn c phng trnh:8 ( x + 1) 8 ( y + 2) + 4 ( x + 3) = 0 2 x 2 y + z + 1 = 0 . d ( I ,o ) = 2 0 1 + 1 22 + (2)2+ 12 = 2 < 2 = R ( S ) (o ) = C . 3 Ta c: V = 1 h.S nn Vln nht h ln nht. ABCD 3D ABC ABCDD GiD1D2 l ng knh ca( S ) vung gc vi mt phng(o ) . V D l im bt k thuc( S )nn d ( D,o ) s max {d ( D1 ,o ) , d ( D2 ,o )} . Du bng xy ra khi v ch khi D trng vi mt trong hai imD1hocD2 . www.MATHVN.com Dnh cho hc sinh lp 12 n thi i hc- Cao ngB thi t lun- mn Ton 50 Vn Ph Quc- GV. Trng i hc Qung Nam D: 0982 333 443;0934 825 925 MATHVN.COM - Ton hc Vit Nam 3| \. D1D2quaInhnvectphptuynca(o ) x = 1+ 2t lmvectchphngnncphngtrnh tham s:D1 D2 : y = 2t z = 1 + t t e . GiD (1 + 2d0 ; 2d0 ; 1 + d0 ) e D1D2 limcntm.KhiDlnghimcaphng trnh:(1 + 2d )2+ 4d 2 + (1 + d )2 2 (1 + 2d ) + 2 (1 + d ) 2 = 0 d = 2 . 000000 3 Ta c:d ( D,o ) = 9d0 + 2 . V 9. 2 + 2 39.| =2 | + 2 > \. nn D phi ng vid = 2 . 333 0 3 VyD | 7 ; 4 ; =1 | 333 | l im cn tm. Cu VIIb (1,0 im) Ta c:3 + i = 2 | cos t + i sin t=| ; 1 i =2 | cos | =t | + i sin | =t | | . 66 | 4 | 4 | | 3 + i = \. 2 | cos 5t + i sin 5t | . \\.\. . 1 i 1212 | \. n n nn | Do :3 + 1 | = 2 2 | cos 5 t + i sin 5 t | .

1 i| 1212| \. \. S l thc khi v ch khisin 5nt = 0 5nt = kt 5n = k ( k e ). 121212 S nguyn dng b nht cn tm l:n = 12 . 8 A- PHN CHUNG(7,0 im) Cu I (2,0 im) . Cho hm s: y = 1 x3 x2 3x + 8 33 (C). 14. Kho st s bin thin v v th(C )ca hm s. 2.Lp phng trnh ng thng d song song vi trc honh v ct th (C) ti hai im phn bit A, B sao cho tam gic OAB cn ti O ( O l gc to ). Cu II (2,0 im) 1.Gii phng trnh: (1 4 sin 2 x )sin 3x = 1 . 2 www.MATHVN.com Dnh cho hc sinh lp 12 n thi i hc- Cao ngB thi t lun- mn Ton 51 Vn Ph Quc- GV. Trng i hc Qung Nam D: 0982 333 443;0934 825 925 MATHVN.COM - Ton hc Vit Nam } 2 2.Gii phng trnh :4x x2 1 +x + t 3 x2 + 1 = 2 . dx Cu III (1,0 im). Tnh tch phn:I= 0 1 + s inx + cos x Cu IV (1,0 im). Cho lng tr tam gicABC.A'B'C'c y ABC l tam gic u cnh a v nhA'cch u cc nh A, B, C. CnhAA'to vi y gc lng tr. 600 . Tnh th tch khi Cu V(1,0 im). Cho cc s thc x, y, z tha: x2 + xy + y2= 3 . y 2 + yz + z 2= 16 Chng minh rng:xy + yz + zx s 8 . B- PHN RING(3,0 im). Th sinh ch c chn mt trong hai phn B.1. CHNG TRNH CHUN Cu VI a(2,0 im) 1.Trong mt phng Oxy, cho imP ( 7; 8) v hai ng thng:d1 : 2 x + 5 y + 3 = 0, d2 : 5x 2 y 7 = 0ctnhautiA.Vitphngtrnhngthngdi quaPvtovi 29 d1 , d2thnh tam gic cn ti A v c din tch bng2. 2. Trong khng gian Oxyz, choimH (4; 5; 6 ).Vitphng trnh mtphng (P) qua H, ct cc trc to Ox, Oy, Oz ln lt ti A, B, C sao cho H l trc tm ca tam gic ABC. Cu VII a (1,0 im) . Tnhinvin e . B.2. CHNG TRNH NNG CAO Cu VI b(2,0 im) 1.Trong mtphng Oxy,choParabol( P ) : y2= 64 x v ng thng A : 4x + 3 y + 46 = 0 . Tm A thuc (P) sao cho khong cch t A nAnh nht. Tnh khong cch nh nht . 2. Trong khng gian Oxyz, cho mt phng (P) ct Ox, Oy, Oz ln lt ti A (a; 0; 0 ) , B (0; b; 0), C (0; 0; c ) Gio , | , ln ltl cc gc ca cc mt phng (OAB), (OBC) , (OCA) vi mt phng (ABC). Chng minh rng: cos2o + cos2 | + cos2= 1. Cu VII b (1,0 im) Gii h phng trnh: 2 log1 x ( xy 2x + y + 2) + log 2+ y ( x 2x + 1) = 6 log1 x ( y + 5) log 2+ y ( x + 4) = 1 A- PHN CHUNG Cu I 1. Hc sinh t gii BI GII 2. V d song song vi trc honh nn phng trnh ca d l: Phng trnh honh giao im ca (C) v d: y = m ( m = 0) . www.MATHVN.com Dnh cho hc sinh lp 12 n thi i hc- Cao ngB thi t lun- mn Ton 52 Vn Ph Quc- GV. Trng i hc Qung Nam D: 0982 333 443;0934 825 925 MATHVN.COM - Ton hc Vit Nam 1 x3 x2 3x + 8 = m x3 3x2 9x + 8 3m = 0 33 (1) d ct (C) ti hai im phn bit A, B sao cho tam gic OAB cn th phng trnh (1) phi c cc nghim: x1 , x1 ,x2 ( vix1 , x1l honh ca A,B ). Khi x1 , x2 l nghim ca phng trnh: (22 )( ) 3222 x x1 x x2 = 0 x x2 x x1 x + x1 x2= 0 (2) x2= 3 ng nht (1) v (2) ta c: x2= 9 9.3 = 8 3mhaym = 19 . 1 x2 x 3 = 8 3m Vyd : y = 19. 3 1 2 Cu II 1.Nhn xt:cos x = 0khng phi l nghim ca phng trnh . Do , nhn c hai v ca phng trnh chocos x = 0ta c: (cos x 4 cos x sin 2 x )sin 3x = 1 cos x

cos x 4 cos x (1 cos2 x )( sin 3x = 1 cos x 2 2 2 sin 3x (4 cos3 x 3 cos x ) = cos x 2 sin 3x cos 3x = cos x sin 6 x = cos x x = t + k 2t sin 6x = sin | t x | 147 ( k e ) . |

\ 2 . x = t + k 2t 15. iu kin: x2 1 > 0 105 x > 1 . x >x2 1 Khi :x +x2 + 1 >x +x2 1 > 4x +x2 1( dox > 1). Suy ra:4x x2 1 +x + x2 + 1 > 4x x2 1 + 4x + x2 1 p dng bt ng thc Cauchy ta c: 4x x2 1 + 4x +x2 1 > 2 4x x2 1.4x +x2 1= 2 Do : 4x Cu III x2 1 +x +x2 + 1 > 1 . Vy phng trnh cho v nghim. tt = tan x dx = 2dt 21 + t 2 x = 0 t = 0 ; x = t t =1 33 11 3 2dt3dt1 |1| I= } = } = ln 1 + t3= ln

1 + | . 20 0 (1 + t 2 ) |1 + 2t + 1 t| 01 + t \3 . 1 + t 2 1 + t 2| \. www.MATHVN.com Dnh cho hc sinh lp 12 n thi i hc- Cao ngB thi t lun- mn Ton 53 Vn Ph Quc- GV. Trng i hc Qung Nam D: 0982 333 443;0934 825 925 MATHVN.COM - Ton hc Vit Nam ( 2 Cu IV A' Gi H l hnh chiu caA'ln mp (ABC) vM l trung im ca BC. C' Do ABC l tam gic u v A'A = A'B = A'C gic ABC. nn H l trng tm tam B' Ta c: AH= 2 AM= 2 a 3 = a 3 . 3323 Mt khcAHl hnh chiu caAA'ln A mt phng (ABC) nnA'AH= 600 hp bi cnh bnAA'v( ABC ) . l gcC H Trong tam gicA'HAvung ti H, ta c : M A'H= AH tan 600= a a2 3 3 . 3 = a . B 3 S ABC= Vy V . 4 = A'H .S = a. a3 = 3 a3 . ABC . A'B'C ' Cu V ABC 44 p dng bt ng thc Bunhiacopski, ta c: 2 3 ( xy + yz + zx )2=

| y + x | 3 z + 3 x | y + z |( 4

2 |22

2 |( \.\. 22

|x | sy + + 3 x2 (3 z 2 + | y + z | = ( x2 + xy + y 2 )( y2 + yz + z 2 ) = 48

| (| (

\ 2 . 4 4 \ 2 . ( xy + yz + zx )2s 64 xy + yz + zx s 8 ( pcm). B- PHN RING B.1. CHNG TRNH CHUN Cu VI a 1. Ta c: ca h : A = d1 d2to im A lnghim d1 d2 2x + 5 y + 3 = 0 5x 2 y 7 = 0 x = 1 y = 1 hay A A (1; 1) . Phng trnh cc ng phn ca gc to bi d1 , d2l:A1 : 7 x + 3 y 4 = 0 , A 2 : 3x 7 y 10 = 0 B C d Vdtovid1 , d2 mttamgicvungcnnn P H www.MATHVN.com Dnh cho hc sinh lp 12 n thi i hc- Cao ngB thi t lun- mn Ton 54 Vn Ph Quc- GV. Trng i hc Qung Nam D: 0982 333 443;0934 825 925 MATHVN.COM - Ton hc Vit Nam d A1 3x 7 y + C1 = 0

d A27 x + 3 y + C2= 0 Mt khcP ( 7; 8) e dnnC1 = 77,C2= 25 . Suy ra: d : 3x 7 y + 77 = 0 d : 7 x + 3 y + 25 = 0 Gi B = d1 d ,C = d2 d V tam gic ABC vung cn ti A nnS ABC = 1 ABAC = 1 AB2= 29 AB = 222 29. vBC = AB 2 =58 2. 29 Suy ra:AH= 2S ABC BC = 2= 58 58 . 2 Vid : 3x 7 y + 77 = 0 , ta c:d ( A; d ) = 3.1 7.( 1) + 77 = 32 + (7 )2 87=AH= 58 58 ( loi). 2 Vid : 7 x + 3 y + 25 = 0 , ta c :d ( A; d ) = Vyd : 7 x + 3 y + 25 = 0 . 7.1 + 3.( 1) + 25 = 72 + 32 29= 58 58 = AH 2 ( tho). 2. Gi s : ( P ) Ox = A ( a; 0; 0 ) ,( P ) Oy = B (0; b; 0 ) ,( P ) Oz = C (0; 0; c ) Khi (P) c phng trnh : x + y + z = 1 . abc Ta c :H (4; 5; 6) e ( P ) 4 + 5 + 6 = 1 abc AH= (4 a; 5; 6),BH= (4; 5 b; 6),BC = (0; b; c ), AC = (a; 0; c ) V H l trc tm tam gic ABC nn AH .BC = 0 5b + 6c = 0 += 456 BH .AC = 0 a = 77 4b6c0 ++= 1 4 Gii h phng trnh : abc 5b + 6c = 0 77 b =. 5 4b + 6c = 0 77 c = 6 Vy phng trnh mt phng (P) l : x+y+z= 1 . 777777 456 Cu VIIa + Nun = 4k ( k e )thi n= i 4k= (i 4 )k = 1 www.MATHVN.com Dnh cho hc sinh lp 12 n thi i hc- Cao ngB thi t lun- mn Ton 55 Vn Ph Quc- GV. Trng i hc Qung Nam D: 0982 333 443;0934 825 925 MATHVN.COM - Ton hc Vit Nam \. 2 2 + Nun = 4k + 1( k e )thin= i4k i = 1.i = i + Nun = 4k + 2 ( k e ) thi n= i 4k i 2= 1.( 1) = 1 + Nun = 4k + 3( k e ) thi n= i 4k i3= 1(i ) = i . B.2. CHNG TRNH NNG CAO Cu VI b 1. Ta c: A e ( P ) : y 2= 64x A | a; a | 64 | \. 4. a+ 3a + 46 d ( A,A ) = 64 =1a2 + 48a + 736=1 ( a + 24)2+ 160 42 + 32 8080 = = 1

(a + 24)2+ 160( > 160 = 2 . 80 80 Du bng xy ra khi ch khia + 24 = 0 a = 24 . Lc 2. Mind ( A, A ) = 2 khiA (9; 24) . ( P ) ( ABC ) : x + y + z 1 = 0 abc c phng vect php tuynn = | 1 , 1 , =1 | a b c | mp (OAB ) c vect php tuynn1 = OC = (0, 0, c ) mp(OBC ) c vect php tuynn2=OA = (a, 0, 0) mp (OAC )c vect php tuynn3= OB = (0, b, 0) Gi o ,| , ln lt l gc gia cc mt phng(OAB ),(OBC ),(OCA) Do : vimp ( ABC ) . 0 1 + 0 1 + c 1 1 coso = abc = c (1) 02 + 02 + c2 1 + 1 + 11 + 1 + 1 a2 b2 c2a 2 b2 c2 a 1 + 0 1 + 0 1 1 cos|= abc = a (2) a2 + 02 + 02 1 + 1 + 11 + 1 + 1 a2 b2 c2a2 b2 c2 0 1 + b 1 + 0 1 1 cos= abc = b (3) 02 + b2 + 02 1 + 1 + 11 + 1 + 1 a2 b2 c2a 2 b2 c2 T (1), (2) v (3) suy ra:cos2o + cos2 | + cos2= 1. www.MATHVN.com Dnh cho hc sinh lp 12 n thi i hc- Cao ngB thi t lun- mn Ton 56 Vn Ph Quc- GV. Trng i hc Qung Nam D: 0982 333 443;0934 825 925 MATHVN.COM - Ton hc Vit Nam Cu VIIb iu kin: 4 < x < 1, x = 0 y > 2; y = 1 2 log( xy 2 x + y + 2) + log( x2 2x + 1) = 6 1 x 2+ y log1 x ( y + 5) log2+ y ( x + 4) = 1 2 log 1 x (1 x ) ( y + 2) + log 2+ y ( x 1)2= 6 2 + 2 log 1 x( y + 2) + 2 log 2+ y x 1 = 6 log1 x ( y + 5) log2+ y ( x + 4 ) = 1 log1 x ( y + 5) log2+ y ( x + 4) = 1 log (2 + y ) + log (1 x ) = 2 log2 (2 + y ) 2 log (2 + y ) + 1 = 0 1 x 2+ y 1 x 1 x log1 x ( y + 5) log2+ y ( x + 4) = 1 log1 x ( y + 5) log2+ y ( x + 4) = 1 log1 x (2 + y ) = 1 y = x 1 log1 x ( y + 5) log2+ y ( x + 4 ) = 1 log1 x (4 x ) log1 x ( x + 4) = 1 y = x 1 y = x 1 y = x 1 y = x 1 x = 2 log 4 x = 1 4 x = 1 x 4 x = x2 3x + 4 x2 + 2x = 0 . y = 1 1 xx + 4 x = 2 x + 4 Vy y = 1 l nghim ca h phng trnh cho. 9 A- PHN CHUNG(7,0 im) Cu I (2,0 im) . Cho hm s: y = x3 2mx2 + (m + 3) x + 4c th(Cm ) 16. Kho st s bin thin v v th(C )ca hm s khim = 1 . 2. Cho ng thngd : y = x + 4 v imE (1; 3) . Tm tt c cc gi tr ca tham s m sao cho d ct(Cm )ti ba im phn bit 4 . Cu II (2,0 im) A (0; 4) , B, Csao cho tam gic EBC c din tch bng 1.Gii phng trnh:cos 3x cos3 x sin 3x sin 3 x = 2 + 3 2 . 8 2.Gii h phng trnh: x2 + 1 + y ( y + x ) = 4 y ( x, y e ) . ( x2 + 1)( y + x 2) =y Cu III (1,0 im)Tnh tch phn: 4 I= } ln (9 x ) dx 2 ln (9 x ) +ln ( x + 3) www.MATHVN.com Dnh cho hc sinh lp 12 n thi i hc- Cao ngB thi t lun- mn Ton 57 Vn Ph Quc- GV. Trng i hc Qung Nam D: 0982 333 443;0934 825 925 MATHVN.COM - Ton hc Vit Nam Cu IV (1,0 im). Cholng tr ng tgicuABCD.A'B'C'D'cchiucaobngh. Gc giahai ng choca hai mtbn k nhau k t mt nh bng Tnh th tch khi lng tr cho. Cu V(1,0 im). Gii phng trnh: (00< < 900 ) . 3 +xx + x + 2 x x + x + 2 x2 + x x + 2 x2 + 310 ++++= x2 + x x + x + 3x2 + x x + 3x2 +x + 4x + x x + 4x x + x +x + 33 B- PHN RING(3,0 im). Th sinh ch c chn mt trong hai phn B.1. CHNG TRNH CHUN Cu VI a(2,0 im) 1. Trong mtphng Oxy,chohnh bnh hnh ABCD cdin tch bng 4.Bitto cc nhA ( 2; 0) ,B (3; 0 )v Ilgiaoim cahai ng choAC v BD, I nm trn ng thngy = x . Xc nh to cc im C, D. xyz x + 1 yz 1 2.TrongkhnggianOxyz,chohaingthng:d1 : 1 = 1 = 2 vd2 :2= 1 = 1 . Chng minhd1 , d2 chonhau. TmA e d1 , B e d2 saocho ng thng AB song song vi mt phng( P ) : x y + z = 0 v di Cu VII a (1,0 im) AB =2 . Trn cc cnh AB, BC, CD, DA ca hnh vung ABCD ln lt cho 1, 2, 3 v n im phn bit khc A, B, C, D. Tm n s tam gic c 3 nh ly tn + 6im cho l 439. B.2. CHNG TRNH NNG CAO Cu VI b(2,0 im) 1.Trongmtphng Oxy,lpphngtrnh ng trn(C) hai trc ta . quaM ( 2; 4 ) vtipxcvi 2.Trong khng gianvihtaOxyz,chobaimA (1; 0; 1) , B (2; 3; 1) , C (1; 3;1)v ng thng d : x = y 1 = z 3 . Tm ta im D thuc ng thng d sao cho th tch 112 khi t din ABCD bng 1.Vit phng trnh tham s ca ng thngAqua trc tm H ca tam gic ABC v vung gc vi mt phng (ABC). Cu VII b (1,0 im) Gii phng trnh:z 2 +z= 0 . A- PHN CHUNG Cu I 1. Hc sinh t gii BI GII 2. Phng trnh honh giao im ca (Cm ) v d: 3232 x = 0 x + 2mx+ (m + 3) x + 4 = x + 4 x+ 2mx+ ( m + 2 ) x = 0

2

x + 2mx + m + 2 = 0 (*) www.MATHVN.com Dnh cho hc sinh lp 12 n thi i hc- Cao ngB thi t lun- mn Ton 58 Vn Ph Quc- GV. Trng i hc Qung Nam D: 0982 333 443;0934 825 925 MATHVN.COM - Ton hc Vit Nam y d ct(Cm )ti ba im phn bitA (0; 4) ,B,C Phng trnh (*) c hai nghim phn bit khc 0 A' = m2 m 2 > 0m s 1v m > 2 = . m + 2 = 0 m2 Gix1 , x2 theo th t l honh ca cc imB, C . Do x1 , x2 l hai nghim phn bit x+ x = 2m ca phng trnh (*). Theo nh l Viet, ta c: 12 . x1 x2= m + 2 S EBC = 4 1 d ( E, d ).BC = 4 1 22 2.BC = 4 BC = 4 2 BC 2= 32 ()2 (( ) ( ))2 ()2 x2 x1 +x2 + 4 x1 + 4 = 32 2x2 x2 = 32 ( x x)2= 16 ( x + x)2 4x x = 16 4m2 4 (m + 2) = 16 211212 m2 m 6 = 0 m = 2

m = 3 . So vi iu kin ta u ta tm c m = 3 Vym = 3 Cu II l gi tr cn tm tho mn yu cu bi ton. 1. Ta c:cos 3x cos3 x sin 3x sin 3 x = 2 + 3 2 8 cos 3x. cos 3x + 3 cos x sin 3x. 3sin x sin 3x = 2 + 3 2 448 2 cos2 3x + 6 cos 3x cos x 6 sin 3x sin x + 2 sin 2 3x = 2 + 3 2 2 (cos2 3x + sin2 3x ) + 6 (cos 3x cos x sin 3x sin x ) = 2 + 3 2 x = t + k t cos 4 x = 2

162 ( k e ) . 2 x = t + k t 162 x2 + 1 = 0 2.+ Viy = 0ta c: ( x2 + 1)( x 2) = 0 ( H v nghim). x2 + 1 + Vi y = 0 , ta c : x2 + 1 + y ( y + x ) = 4 y + ( y + x 2) = 2 ( x2 + 1)( y + x 2) =y x2 + 1 () x2 + 1 y y + x 2 = 1 Suy ra:, y + x 2l nghim ca phng trnh:t 2 2t + 1 = 0 t = 1 . y www.MATHVN.com Dnh cho hc sinh lp 12 n thi i hc- Cao ngB thi t lun- mn Ton 59 Vn Ph Quc- GV. Trng i hc Qung Nam D: 0982 333 443;0934 825 925 MATHVN.COM - Ton hc Vit Nam \. \. } h D x2 + 1 x = 1

x = 1 = 1 y = x2 + 1 x2 + x 2 = 0

y = 2 Do : y y = 3 x y = 3 x x = 2 .

x = 2 y + x 2 = 1 Vy tp hp nghim ca h phng trnh l: Cu III y = 3 x S = {(1; 2), (2; 5)} .

y = 5 t 2 x = 6 t dx = dt ln (t + 3) ( dt ) 4 ln (t + 3) 4 ln ( x + 3) I= } 4 ln (t + 3) + 4 } = ln (9 t )2 4 2 ln (t + 3) + ln (9 t ) dt= } 2 ln ( x + 3) + dx ln (9 x ) 2I= I + I= Cu IV dx = x 2 = 2 I= 1 . A' D' Gi sB'AD = vxl cnh y ca hnh lng tr. p dng nh l Csin trong A'BDta c:B' C' BD2= A'B2 + A'D2 2 A'B.A'D cos BD2= 2 A'B 2 2 A'B2 cos 2 x2= 2 ( x2 + h2 )(1 cos ) 2 x2= 2 ( x2 + h2 ).2 sin 2 A 2 2h2 sin2 x2= 2 cos (00< < 900 ) . B C 2h3 sin2 Th tch khi lng tr cho l:V= AA'S Cu V ABCD = x2 h = 2 . cos ta = 2,b =x + 1,c = x + 1,d = x x + 1,e = x2 + 1 . Phng trnh cho tr thnh: a + b + b + c + c + d + d + e + e + a = 10 c + d + ed + e + ae + a + ba + b + cb + c + d3 |a + b + 1| + | b + c + 1| + | c + d + 1| + | d + e + 1| + | e + a + 1| = 10 + 5 c + d + e | d + e + a | e + a + b | a + b + c | b + c + d | 3 \.\.\.\.\. ( a + b + c + d + e) |1 + 1 + 1 + 1 + = 1 | = 25 c + d + ed + e + ae + a + ba + b + cb + c + d | 3 |1 + 1 + 1 + 1 + =1| c + d + ed + e + ae + a + ba + b + cb + c + d |

(c + d + e ) + ( d + e + a ) + (e + a + b ) + (a + b + c ) + (b + c + d ) = 25 (*) p dng bt ng thc Cauchy cho v trica (*) th ta c VT (*) > 25 . www.MATHVN.com Dnh cho hc sinh lp 12 n thi i hc- Cao ngB thi t lun- mn Ton 60 Vn Ph Quc- GV. Trng i hc Qung Nam D: 0982 333 443;0934 825 925 MATHVN.COM - Ton hc Vit Nam \.\. 8 C Du = xy ra khi v ch khia = b = c = d = e x = 1 . Vy x = 1 l nghim duy nht ca phng trnh cho. B- PHN RING B.1. CHNG TRNH CHUN D Cu VI a I 1.Ta c:S Mt khc IAB = 1 S 4 ABCD = 1 . AHB S= 1 IH .AB = 1 IH . (3 2 )2+ 02= 1 IH= 1 IH= 2 AIAB 222 I e d : y = x I ( a; a ) , ta c phng trnh ng thng AB l :y = 0 . Ta c :IH= 2 d ( I , AB ) = 2 a a = 2 = 2

a = 2 . + Nua = 2th suy ra :I ( 2; 2 ) ,C (2; 4 ), D (2; 4 ) + Nua = 2th suy ra :I ( 2; 2) ,C (5; 4 ) , D ( 6; 4) . 2.d1 quaO (0; 0; 0 )nhnu1 = (1;1; 2)lm vect ch phng d2quaI ( 1; 0;1) nhnu2= (2;1;1)

lm vect ch phng. Ta tnh c

u1 , u2 ( OI= 4 = 0 . Suy rad1 , d2cho nhau. A e d1 A ( a; a; 2a );B e d 2 B ( 1 2b; b;1 + b ) AB / / ( P ) AB.nP= 0 b = a . Do B (1 + 2a; a;1 a )vAB = (1+ a; 2a;1 3a ) .

a = 0 AB =2 AB2= 2 (a 1)2+ 4a2 + (1 3a )2= 2 7a2 4a = 0 4 .

a = 7 + Via = 0ta cA O( loi) + Via = 4 7 Cu VIIa ta cA | 4 ; 4 ; =8 | , B | 1 ; 4 ; =3 | . 7 7 7 | 77 7 | Nun s 2thn + 6 s 8v s tam gic c 3 nh c to thnh tn + 6im khng vt quC 3= 56 < 439 . Vyn > 3 . Mi tam gic c to thnh ng vi mt t hp chp 3 tn + 6phn t. Nhng trn cnh CD c 3 im, trn cnh DA c n im nn s tam gic c to thnh ch l C 3 C 3 C 3= (n + 6) (n + 5) ( n + 4 ) 1 n (n 1) ( n 2) . n+63n 66 ( n + 6) ( n + 5) ( n + 4) n (n 1) ( n 2) Theo gi thit ta c: n2 + 4n 140 = 0 n = 10

n = 14 6 . Chnn = 10 . = 439 + 1 = 440 www.MATHVN.com Dnh cho hc sinh lp 12 n thi i hc- Cao ngB thi t lun- mn Ton 61 Vn Ph Quc- GV. Trng i hc Qung Nam D: 0982 333 443;0934 825 925 MATHVN.COM - Ton hc Vit Nam

a = 10 O Vyn = 10l gi tr cn tm. B.2. CHNG TRNH NNG CAO Cu VI b(2,0 im) 1. GiI (a; b )l tmv R l bn knh ca ng trn(C) . y V(C) tip xc vi hai trc ta nnR =a=b . - M ( 2; 4 ) (C)qua imM ( 2; 4 )nn(C) nm gc phn t th I trn mt phng Oxy. Do a > 0; b > 0 .- I Nh vyR = a = b . Khi , phng trnh ng trn(C) c dng:- x ( x a )2+ ( y a )2= a2 . M ( 2; 4) e (C ) (2 a )2 + ( 4 a )2= a 2 a 2 12a + 20 = 0 a = 2 Kt lun: Phng trnh ng trn(C) l: ( x 2)2 + ( y 2)2= 4 ( x 10)2+ ( y 10)2= 100 2. Ta c: V ABCD = 1 S 3 AABC .h trong S AABC l din tch tam gic ABC, h l khong cch t D n mt phng( ABC ) .

AB = (1; 3; 0), AC = (0; 3; 2) AB, AC ( = (6; 2; 3)

S= 1 AB, AC ( = 7 AABC 2 2 Do :h = 3V = 3 = 6 . SAABC 7 7 2

Mt phng( ABC )qua A (1; 0; 1)nhn AB, AC ( = (6; 2; 3)lm vect php tuyn nn c phng trnh: 6 ( x 1) 2 y + 3 ( z + 1) = 0 6 x 2 y + 3z 3 = 0 x = t Dng tham s cad : y = 1 + t z = 3 2t DoD e d D (t;1 + t; 3 2t )nn h = d ( D, ( ABC )) = 6t 2 (1 + t ) + 3 (3 2t ) 3 = 2t + 46 = Suy ra:t = 1v t = 5 62 + 22 + 32 77 Vy c hai im D vD'tha mn yu cu bi tonD ( 1; 0; 5) ,D' (5; 6; 7 ) . www.MATHVN.com Dnh cho hc sinh lp 12 n thi i hc- Cao ngB thi t lun- mn Ton 62 Vn Ph Quc- GV. Trng i hc Qung Nam D: 0982 333 443;0934 825 925 MATHVN.COM - Ton hc Vit Nam \. b) ng thngAi qua trc tm H ca tam gic ABC v vung gc vi mt phng (ABC) l giao tuyn ca mt phng qua A- vung gc vi ng thng BC vi mt phng qua B-vung gc vi ng thng AC ( cng l giao tuyn vi mt phng qua C, vung gc vi ng thng AB). Mt phng quaA (1; 0; 1)nhnBC = (1; 0; 2) lm vect php tuyn nn c phng trnh: ( x 1) + 2 ( z + 1) = 0 x 2z 3 = 0 . Mt phng quaB ( 2; 3; 1) nhnAC = (0; 3; 2) lm vect php tuyn nn c phng trnh: ( x 1) + 2 ( z + 1) = 0 x 2z 3 = 0 . D thyM 0 (7;1; 2) e A | 0 Vect ch phng caAl:u = 2 21 10 | ;;= (6; 2; 3) 3220 03 | Phng trnh tham s caAl: Cu VII b (1,0 im) x = 7 + 6t y = 1 2t z = 2 + 3t tz = x + yi . Suy ra:z 2= x2 y 2 + 2xyiv z=x2 + y2. Phng trnh cho tr thnh: () x2 y2 + x2 + y2=() x2 y2 + x2 + y2 + 2xyi = 0 2xy = 0 01 . (2) Cc trng hp sau: +x =y = 0 . +x = 0 , y = 0 . Khi t (1)ta c: y 2 + y2= 0 y2 +y = 0 y= 1 y = 1. +x = 0 , y = 0 . Khi h (1) v (2) tr thnh: x2 +x= 0 x+ 1 = 0 ( V nghim). 10 A- PHN CHUNG(7,0 im) Cu I (2,0 im) . Cho hm s: y = x4 2 (m2 m + 1) x2 + m 1c th (Cm ) 17. Kho st s bin thin v v th(C )ca hm s khim = 1 . 2. Tm m th(Cm )c khong cch gia hai im cc tiu ngn nht. Cu II (2,0 im) www.MATHVN.com Dnh cho hc sinh lp 12 n thi i hc- Cao ngB thi t lun- mn Ton 63 Vn Ph Quc- GV. Trng i hc Qung Nam D: 0982 333 443;0934 825 925 MATHVN.COM - Ton hc Vit Nam 1. Tm ccnghim thc ca phng trnh: mn 1 + log 1x > 0 . 3 sin x tan 2 x +3 (sin x 3 tan 2 x ) = 3 3 tha 2.Gii h phng trnh: x2 + y2 + 2xy= 1 x + y . x + y= x2 y Cu III (1,0 im)Tnh din tch hnh phng gii hn bi cc ng: y = x 1 + sin x ,y = 0,x = 0,x = t Cu IV (1,0 im). Cho hnh chp tam gic S.ABC c y l tam gic vung B,cnhSA (ABC) . T AkAD SB S.ADE? vAE SC .BitAB=a,BC=b,SA=c.Tnhthtchcakhi chp Cu V(1,0 im). Choa, b, cl cc s dng tha mn 1 + 1 + 1 = 2011. Tm gi tr ln nht ca biu thc: abc P = 1 + 1 + 1 2a + b + ca + 2b + ca + b + 2c B- PHN RING(3,0 im). Th sinh ch c chn mt trong hai phn B.1. CHNG TRNH CHUN Cu VI a(2,0 im) 1.Trongmtphng Oxy,chobn imA (1; 0) ,B ( 2; 4 ) ,C ( 1; 4 ),D (3; 5) .Tm ta im M thuc ng thng tch bng nhau. A : 3x y 5 = 0saochohai tam gicMAB vMCD cdin 2.TrongkhnggianOxyz,chomtphng ( P ) : 2 x y + z 1 = 0vhaingthng d: x 1 = y + 2 = z 3 1 213 ,d2: x + 1 = y 1 = z 2 . Vit phng trnh ng thngAsong 232 song vi mt phng (P), vung gc vi ng thng c honh bng 3. Cu VII a (1,0 im) d1 v ct ng thngd2 ti im C Tm phn thc ca s phcz = (1 + i )n , n e . Trong n tha mn : log 4 (n 3) + log5 ( n + 6 ) = 4 . B.2. CHNG TRNH NNG CAO Cu VI b(2,0 im) 1. Trong mt phng Oxy, cho elip ( E ) : x 2 y2 += 1 v hai im A ( 5; 1) ,B ( 1;1) . Tm 165 mt ta im M nm trn (E) sao cho din tch tam gic MAB ln nht. www.MATHVN.com Dnh cho hc sinh lp 12 n thi i hc- Cao ngB thi t lun- mn Ton 64 Vn Ph Quc- GV. Trng i hc Qung Nam D: 0982 333 443;0934 825 925 MATHVN.COM - Ton hc Vit Nam x = 2 y = 3 1 2.TrongkhnggianvihtaOxyz,chomtphng ( P ) : 2 x + 2 y z + 16 = 0,( S ) : x2 + y 2 + z 2 4 x + 2 y 6 z + 5 = 0 . im M di ng trn (S), im N di ng trn (P). Tnh di ngn nht ca MN. Xc nh v tr ca M , N . Cu VII b (1,0 im) Gii h phng trnh sau: y 2 2xy + y 2x + 2 = 0 . 2 log2 ( 2x y ) + 3 log 2 ( y + 1) = 4 A- PHN CHUNG Cu I 1. Hc sinh t gii BI GII 2. Ta c:y' = 4x3 4 (m2 m + 1) m e , th(Cm)lun c ba im cc tr. x = 0 y' = 0

x = m2 m + 1 2 Khong cch gia hai im cc tiu l: d = 2 m2 m + 1 = 2 | m =1 | + 3 >3 . 2 | 4 \. Du = xy ra khim = 1 . 2 Cu II x > 0 1. Ta c:1 + log 1x > 0 logx > 1 0 < x s 3 . 3 3 sin x tan 2 x +3 (sin x 3 tan 2 x ) = 3 3 (sin x tan 2 x + 3 sin x ) ( 3 tan 2 x + 3 3 ) = 0 sin x (tan 2 x +3 ) 3 (tan 2 x +3 ) = 0 (tan 2 x +3 )(sin x 3) = 0 tan 2x = 3 2x = t + kt x = t + k t 362 (k e ) . Kt hp vi iu kin0 < x s 3 , ta chn ck = 1; 2 . Vy ta c c cc nghim tha mn yu cu bi ton l:x = t ; x = 5t . 36 2.iu kin:x + y > 0 . Ta c: x2 + y2 + 2xy = 1 ( x + y )2 2xy + 2xy 1 = 0 ( x + y 1) ( x2 + y2 + x + y ) = 0 x + yx + y Vx + y > 0nnx2 + y2 + x + y > 0 . Do :x + y 1 = 0 y = 1 x . Thay vo phng trnh th hai ta c: 1 = x2 (1 x ) x2 + x 2 = 0 x = 1 y = 0 . www.MATHVN.com Dnh cho hc sinh lp 12 n thi i hc- Cao ngB thi t lun- mn Ton 65 Vn Ph Quc- GV. Trng i hc Qung Nam D: 0982 333 443;0934 825 925 MATHVN.COM - Ton hc Vit Nam 00 2 y t Vy tp hp nghim ca h phng trnh trn l:S = {(1; 0) , ( 2; 3)} . x2 + 1 = 0 2.+ Viy = 0ta c: ( x2 + 1)( x 2) = 0 ( H v nghim). x2 + 1 + Vi y = 0 , ta c : x2 + 1 + y ( y + x ) = 4 y + ( y + x 2) = 2 ( x2 + 1)( y + x 2) =y x2 + 1 () x2 + 1 y y + x 2 = 1 Suy ra:, y + x 2l nghim ca phng trnh:t 2 2t + 1 = 0 t = 1 . y x = 1 x2 + 1 x = 1 = 1 y = x2 + 1 x2 + x 2 = 0

y = 2 Do : y y = 3 x y = 3 x x = 2 .

x = 2 y + x 2 = 1 Vy tp hp nghim ca h phng trnh l: Cu III y = 3 x S = {(1; 2), (2; 5)} .

y = 5 Nhn xt:y = x 1 + sin x > 0, x e |0; t |. Gi S l din tch hnh phng cn tm. t x t x 1 t x Ta c:S = } 1 + sin x dx = } 0 | sin x + cos =x | 2dx = 2 } | x =t | . cos 22 | 24 | \. \. p dng cng thc tch phn tng phn ta c: |t | t | t t ||t | I= x tan x x | } tan x |dx = t + 2 ln cos | = t \ Cu IV S D 4 . 00 E \ 24 .\ 24 . 0 AD,AE l cc ng cao trong tam gic SAB,SAC AABCvung ti B nnAB BC Gi thit cho : SA (ABC) SA BC BC (ABC) AD BC AD l ng cao trong tam gic SAB AD SB AD (SBC) AC AD SC Mt khc :AE SC SC (ADE) Hay SE l ng cao ca hnh chp S.ADE B www.MATHVN.com Dnh cho hc sinh lp 12 n thi i hc- Cao ngB thi t lun- mn Ton 66 Vn Ph Quc- GV. Trng i hc Qung Nam D: 0982 333 443;0934 825 925 MATHVN.COM - Ton hc Vit Nam \. \. AD = AS.AB = AS.AB = a.c SB AS2 + AB2a 2 + c2 AS.ACSA.ACc. a 2 + b2 AE === SB SA2 + AC2 p dng Pytago trong tam gic SAE c: a 2 + b2 + c2 2222 SE = AS2 AE2= c2 c(a + b) = c a 2 + b2 + c2 a 2 + b2 + c2 DE = AE2 + AD2 = c2 .b2 (a 2 + b2 + c2 ).(a 2 + c2 ) 2 2 S = 1 .AD.AE = 1 . c.b . ac 22 (a 2 + b2 + c2 ).(a 2 + c2 )a 2 + c2 Th tch: 1a.c3.b3 =. 2 (a 2 + b2 + c2 ).(a 2 + c2 ) 3 3 V =1 .SE. 1 .AD.DE = 1 . c1 . a.c.b 323 a 2 + b2 + c22(a 2 + b2 + c2 ).(a 2 + c2 ) 1a.b2 .c4 =. 6(a 2 + c2 )(a 2 + b2 + c2 ) Cu V p dng bt ng thc Cauchy, ta c:4xy s ( x + y )2 Du = xy ra khix =y . 1 x + y s 1 | 1 + 1 | . 4 xy | Ta c: 1 s =1 |1 + =1 | s =1=1 | 1 + =1 | + =1 | 1 + =1 |( = =1 | 1 + 1 + = 1| 2a + b + c 4 a + ba + c | 4 4 ab | 4 ac |( 8 a 2b2c | \.\.\.\. Tng t: 1 s =1 |1 + 1 + = 1| , 1 s =1 |1 + 1+ =1 | a + 2b + c 8 2ab 2c | a + b + 2c 8 2a 2bc | \.\. Cng v theo v ca ba bt ng thc trn ta c: P s =1 | 1 + 1 + =1 | = 2011 . B- PHN RING B.1. CHNG TRNH CHUN Cu VI a 1. Ta c:M e A M (t; 3t 5) . 4 abc | 4 www.MATHVN.com Dnh cho hc sinh lp 12 n thi i hc- Cao ngB thi t lun- mn Ton 67 Vn Ph Quc- GV. Trng i hc Qung Nam D: 0982 333 443;0934 825 925 MATHVN.COM - Ton hc Vit Nam | 1 19 22 =

t SAMAB = SAMCD 1 AB.d ( M , AB ) = 1 CD.d (M , CD ) AB.d (M , AB ) = CD.d ( M , CD ) . 22 ngthngABquaA (1; 0) ,nhnAB = (3; 4)lmvectochphngnncphng trnh: x 1 = y 0 4x + 3 y 4 = 0 . 34 ngthngCDquaC (1; 4 ) ,nhnCD = ( 4;1)lmvectochphngnncphng trnh: x + 1 = y 4 x 4 y + 17 = 0 . 41 AB = 5,CD = 17 Do :5. 4t + 3(3t 5) 4 = 17. t 4 (3t 5) + 17 13t 19=11t + 37 517

13t 19 = 11t + 37 7 M | 7 ; 2 | 3 \ 3 .. 13t 19 = 11t 37

t = 9 M ( 9; 32) 2. Ta c:C e d 2 C (3; 7; 6 ) . Gi uA , nP ,udlnltlvectoch phng caA ,vectophp tuyn ca(P),vectoch phng ca d1 . Khi :uA=

nP , ud( = ( 4; 4; 4) . 1 ng thngAquaC (3; 7; 6 ) nhnuA lm vecto ch phng nn c phng trnh: x = 3 + t Cu VIIa y = 7 + t z = 6 t (t e ) Phng trnh:log 4 (n 3) + log5 ( n + 6 ) = 4 c nghim duy nht l n =19 ( v VT ca phng trnh l mt hm s ng bin nn th ca n ct ng