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THI VO LP 10

WWW.ToanTrungHocCoSo.ToanCapBa.Net

Phng GD-T Hi Hu

Trng THCSB Hi Minh

thi th vo lp10 thpt dng cho hs thi vo trng chuyn

(Thi gian lm bi 150)

Bi 1(1): Cho biu thc

Rt gn P.

Bi 2(1): Cho a, b, c l di 3 cnh ca mt tam gic. Chng minh rng phng trnh:

x2 + (a + b + c)x + ab + bc + ca = 0 v nghim.

Bi 3(1): Gii phng trnh sau:

Bi 4(1): Gii h phng trnh sau:

Bi 5(1): Chng minh rng:

Bi 6(1): Cho x, y, z> 0 tho mn:

Tm gi tr nh nht ca biu thc:

Bi 7(1): Trong mt phng 0xy cho ng thng (d) c phng trnh

2kx + (k - 1)y = 2 (k l tham s)

a) Tm k ng thng (d) song song ng thng y = x . Khi tnh gc to bi ng thng (d) vi 0x.

b) Tm k khong cch t gc to n ng thng (d) ln nht.

Bi 8(1): Cho gc vung x0y v 2 im A, B trn Ox (OB > OA >0), im M bt k trn cnh Oy(M ( O). ng trn (T) ng knh AB ct tia MA,MB ln lt ti im th hai:

C , E . Tia OE ct ng trn (T) ti im th hai F.

1. Chng minh 4 im: O, A, E, M nm trn 1 ng trn.

2. T gic OCFM l hnh g? Ti sao?

Bi 9(1): Cho tam gic ABC nhn c 3 ng cao: AA1, BB1, CC1 ng quy ti H.

Chng minh rng: .Du "=" xy ra khi no?

Bi 10(1): Cho 3 tia Ox, Oy, Oz khng ng phng, i mt vung gc vi nhau. Ly im A, B, C bt k trn Ox, Oy v Oz.

a) Gi H l trc tm ca tam gic ABC. Chng minh rng: OH vung gc vi mt phng ABC

b) Chng minh rng: .

p n:

BiBi giiim

Bi 1

(1 im) iu kin:

* Rt gn:

0.25

0.25

0.25

0.25

Bi 2

(1 im)Ta c: ( =(a + b + c)2 - 4(ab + bc + ca) = a2+b2+c2-2ab-2bc-2ca

* V a, b, c l 3 cnh ( ( a2 < (b + c)a

b2 < (a + c)b

c2 < (a + b)c

( a2 + b2 + c2 < 2ab + 2ac + 2bc

( ( < 0 ( phng trnh v nghim.0.25

0.25

0.25

0.25

Bi 3

(1 im)

Bi 4

(1 im)

* iu kin:

* Phng trnh

Gii h:

T (1) ( 2x2 + (y - 5)x - y2 + y + 2 = 0

0.25

0.25

0.25

0.25

0.25

* Vi: x = 2 - y, ta c h:

*Vi , ta c h:

Vy h c 2 nghim: (1;1) v

0.25

0.25

0.25

Bi 5

(1 im)t a = x + y, vi:

Ta phi chng minh: a8 > 36

Ta c:

(v: x > 1; y > 0 ( a > 1)

( a9 > 93.a ( a8 > 36 (pcm).0.25

0.25

0.25

0.25

Bi 6

(1 im)* p dng bt ng thc Bunhiacopsky cho: 1, v

Du "=" xy ra khi v ch khi x = y

Tng t:

T (1), (2), (3)

Suy ra: Pmin = 3 khi: x = y = z = .0.25

0.25

0.25

0.25

Bi 7

(1 im)1).* Vi k = 1 suy ra phng trnh (d): x = 1 khng song song:

y =

* Vi k ( 1: (d) c dng:

: (d) // y = (

EMBED Equation.3 Khi (d) to Ox mt gc nhn ( vi: tg( = ( ( = 600.

2)* Vi k = 1 th khong cch t O n (d): x = 1 l 1.

* k = 0 suy ra (d) c dng: y = -2, khi khong cch t O n (d) l 2.

* Vi k ( 0 v k ( 1. Gi A = d ( Ox, suy ra A(1/k; 0)

B = d ( Oy, suy ra B(0; 2/k-1)

Suy ra: OA =

Xt tam gic vung AOB, ta c :

Suy ra (OH)max = khi: k = 1/5.

Vy k = 1/5 th khong cch t O n (d) ln nht. 0.25

0.25

0.25

0.25

Bi 8

(1im) y

M

a) Xt t gic OAEM c: F

E

(V: gc ni tip...)

Suy ra: O, A, E, M Bcng thuc ng trn. O A x

C

b) T gic OAEM ni tip, suy ra:

*Mt khc: A, C, E, F cng thuc ng trn (T) suy ra:

Do : T gic OCFM l hnh thang.0.25

0.25

0.25

0.25

Bi 9 (1im)b)* Do tam gic ABC nhn, nn H nm trong tam gic.

* t S = S(ABC; S1 = SHBC; S2 = SHAC; S3 = SHAB. A

Ta c: C1 B1

HTng t: B A1 C

Suy ra:

Theo bt ng thc Csy:

Du "=" xy ra khi tam gic ABC u

0.25

0.25

0.25

0.25

Bi 10

(1im)a) Gi AM, CN l ng cao ca tam gic ABC.

Ta c: AB ( CN

AB ( OC (v: OC ( mt phng (ABO)

Suy ra: AB ( mp(ONC) ( AB ( OH (1).

Tng t: BC ( AM; BC ( OA, suy ra: BC ( mp (OAM) ( OH ( BC (2).

T (1) v (2) suy ra: OH ( mp(ABC)

b) t OA = a; OB = b; OC = c.

Ta c:

Mt khc: Do tam gic OAB vung, suy ra:

0.25

0.25

0.25

0.25

3

Bi 1: Cho biu thc:

a). Tm iu kin ca x v y P xc nh . Rt gn P.

b). Tm x,y nguyn tha mn phng trnh P = 2.

Bi 2: Cho parabol (P) : y = -x2 v ng thng (d) c h s gc m i qua im M(-1 ; -2) .

a). Chng minh rng vi mi gi tr ca m (d) lun ct (P) ti hai im A , B phn bit

b). Xc nh m A,B nm v hai pha ca trc tung.

Bi 3: Gii h phng trnh :

Bi 4: Cho ng trn (O) ng knh AB = 2R v C l mt im thuc ng trn . Trn na mt phng b AB c cha im C , k tia Ax tip xc vi ng trn (O), gi M l im chnh gia ca cung nh AC . Tia BC ct Ax ti Q , tia AM ct BC ti N.

a). Chng minh cc tam gic BAN v MCN cn .

b). Khi MB = MQ , tnh BC theo R.

Bi 5: Cho tha mn :

EMBED Equation.3 Hy tnh gi tr ca biu thc : M = + (x8 y8)(y9 + z9)(z10 x10) .

p n

Bi 1: a). iu kin P xc nh l :; .

*). Rt gn P:

EMBED Equation.DSMT4



EMBED Equation.DSMT4 Vy P =

b). P = 2

EMBED Equation.3 = 2

Ta c: 1 + ( ( x = 0; 1; 2; 3 ; 4

Thay vo ta ccc cp gi tr (4; 0) v (2 ; 2) tho mn

Bi 2: a). ng thng (d) c h s gc m v i qua im M(-1 ; -2) . Nn phng trnh ng thng (d) l : y = mx + m 2.

Honh giao im ca (d) v (P) l nghim ca phng trnh:

- x2 = mx + m 2

x2 + mx + m 2 = 0 (*)

V phng trnh (*) c nn phng trnh (*) lun c hai nghim phn bit , do (d) v (P) lun ct nhau ti hai im phn bit A v B.

b). A v B nm v hai pha ca trc tung phng trnh : x2 + mx + m 2 = 0 c hai nghim tri du m 2 < 0 m < 2.

Bi 3 :

EMBED Equation.3 KX :

Thay vo (1) => x = y = z = 3 .

Ta thy x = y = z = 3 tha mn h phng trnh . Vy h phng trnh c nghim duy nht x = y = z = 3.

Bi 4:

a). Xt v .

Ta c: AB l ng knh ca ng trn (O)

nn :AMB = NMB = 90o .

M l im chnh gia ca cung nh AC

nn ABM = MBN => BAM = BNM

=> cn nh B.

T gic AMCB ni tip

=> BAM = MCN ( cng b vi gc MCB).

=> MCN = MNC ( cng bng gc BAM).

=> Tam gic MCN cn nh M

b). Xt v c :

MC = MN (theo cm trn MNC cn ) ; MB = MQ ( theo gt)

BMC = MNQ ( v : MCB = MNC ; MBC = MQN ).

=> => BC = NQ .

Xt tam gic vung ABQ c AB2 = BC . BQ = BC(BN + NQ)

=> AB2 = BC .( AB + BC) = BC( BC + 2R)

=> 4R2 = BC( BC + 2R) => BC =

Bi 5:

T : =>

=>

Ta c : x8 y8 = (x + y)(x-y)(x2+y2)(x4 + y4).=

y9 + z9 = (y + z)(y8 y7z + y6z2 - .......... + z8)

z10- x10 = (z + x)(z4 z3x + z2x2 zx3 + x4)(z5 - x5)

Vy M = + (x + y) (y + z) (z + x).A =

4

Bi 1: 1) Cho ng thng d xc nh bi y = 2x + 4. ng thng d/ i xng vi ng thng d qua ng thng y = x l:

A.y = x + 2 ; B.y = x - 2 ; C.y = x - 2 ; D.y = - 2x - 4

Hy chn cu tr li ng.

2) Mt hnh tr c chiu cao gp i ng knh y ng y nc, nhng chm vo bnh mt hnh cu khi ly ra mc nc trong bnh cn li bnh. T s gia bn knh hnh tr v bn knh hnh cu l A.2 ; B. ; C. ; D. mt kt qu khc.

Ba2: 1) Gii phng trnh: 2x4 - 11 x3 + 19x2 - 11 x + 2 = 02) Cho x + y = 1 (x > 0; y > 0) Tm gi tr ln nht ca A = +

Bi 3: 1) Tm cc s nguyn a, b, c sao cho a thc : (x + a)(x - 4) - 7

Phn tch thnh tha s c : (x + b).(x + c)

2) Cho tam gic nhn xy, B, C ln lt l cc im c nh trn tia Ax, Ay sao cho AB < AC, im M di ng trong gc xAy sao cho =

Xc nh v tr im M MB + 2 MC t gi tr nh nht.

Bi 4: Cho ng trn tm O ng knh AB v CD vung gc vi nhau, ly im I bt k trn oan CD.

a) Tm im M trn tia AD, im N trn tia AC sao cho I lag trung im ca MN.

b) Chng minh tng MA + NA khng i.

c) Chng minh rng ng trn ngoi tip tam gic AMN i qua hai im c nh.

Hng dn

Bi 1: 1) Chn C. Tr li ng.

2) Chn D. Kt qu khc: p s l: 1

Bi 2 : 1)A = (n + 1)4 + n4 + 1 = (n2 + 2n + 1)2 - n2 + (n4 + n2 + 1)

= (n2 + 3n + 1)(n2 + n + 1) + (n2 + n + 1)(n2 - n + 1)

= (n2 + n + 1)(2n2 + 2n + 2) = 2(n2 + n + 1)2

Vy A chia ht cho 1 s chnh phng khc 1 vi mi s nguyn dng n.

2) Do A > 0 nn A ln nht A2 ln nht.

Xt A2 = (+ )2 = x + y + 2 = 1 + 2 (1)

Ta c: (Bt ng thc C si)

=> 1 > 2 (2)

T (1) v (2) suy ra: A2 = 1 + 2 < 1 + 2 = 2

Max A2 = 2 x = y = , max A = x = y =

Bi3 Cu 1Vi mi x ta c (x + a)(x - 4) - 7 = (x + b)(x + c)

Nn vi x = 4 th - 7 = (4 + b)(4 + c)

C 2 trng hp: 4 + b = 1 v 4 + b = 7

4 + c = - 7 4 + c = - 1

Trng hp th nht cho b = - 3, c = - 11, a = - 10

Ta c (x - 10)(x - 4) - 7 = (x - 3)(x - 11)

Trng hp th hai cho b = 3, c = - 5, a = 2

Ta c (x + 2)(x - 4) - 7 = (x + 3)(x - 5)

Cu2 (1,5im)

Gi D l im trn cnh AB sao cho:

AD = AB. Ta c D l im c nh

M = (gt) do =

Xt tam gic AMB v tam gic ADM c MB (chung)

= =

Do AMB ~ ADM => = = 2

=> MD = 2MD (0,25 im)

Xt ba im M, D, C : MD + MC > DC (khng i)

Do MB + 2MC = 2(MD + MC) > 2DC

Du "=" xy ra M thuc on thng DC

Gi tr nh nht ca MB + 2 MC l 2 DC

* Cch dng im M.

- Dng ng trn tm A bn knh AB

- Dng D trn tia Ax sao cho AD = AB

M l giao im ca DC v ng trn (A; AB)

Bi 4: a) Dng (I, IA) ct AD ti M ct tia AC ti N

Do MN = 900 nn MN l ng knh

Vy I l trung im ca MN

b) K MK // AC ta c : INC = IMK (g.c.g)

=> CN = MK = MD (v MKD vung cn)

Vy AM+AN=AM+CN+CA=AM+MD+CA

=> AM = AN = AD + AC khng i

c) Ta c IA = IB = IM = IN

Vy ng trn ngoi tip AMN i qua hai im A, B c nh.

5

Bi 1. Cho ba s x, y, z tho mn ng thi :

Tnh gi tr ca biu thc :.

Bi 2). Cho biu thc :.

Vi gi tr no ca x, y th M t gi tr nh nht ? Tm gi tr nh nht

Bi 3. Gii h phng trnh :

Bi 4. Cho ng trn tm O ng knh AB bn knh R. Tip tuyn ti im M bbt k trn ng trn (O) ct cc tip tuyn ti A v B ln lt ti C v D.

a.Chng minh : AC . BD = R2.

b.Tm v tr ca im M chu vi tam gic COD l nh nht .

Bi 5.Cho a, b l cc s thc dng. Chng minh rng :

Bi 6).Cho tam gic ABC c phn gic AD. Chng minh : AD2 = AB . AC - BD . DC.Hng dn gii

Bi 1. T gi thit ta c :

Cng tng v cc ng thc ta c :

Vy : A = -3.

Bi 2.(1,5 im) Ta c :

Do v

Bi 3. t : Ta c : u ; v l nghim ca phng trnh :

EMBED Equation.DSMT4 ;

;

Gii hai h trn ta c : Nghim ca h l :

(3 ; 2) ; (-4 ; 2) ; (3 ; -3) ; (-4 ; -3) v cc hon v.

Bi 4. a.Ta c CA = CM; DB = DM

Cc tia OC v OD l phn gic ca hai gc AOM v MOB nn OC OD

Tam gic COD vung nh O, OM l ng cao thuc cnh huyn CD nn :

MO2 = CM . MD

R2 = AC . BD

b.Cc t gic ACMO ; BDMO ni tip

(0,25)

Do : (MH1 AB)

Do MH1 OM nn

Chu vi chu vi

Du = xy ra MH1 = OM MO M l im chnh gia ca cung

Bi 5 (1,5 im) Ta c : a , b > 0

a , b > 0

Mt khc

Nhn tng v ta c :

Bi 6. (1 im) V ng trn tm O ngoi tip

Gi E l giao im ca AD v (O)

Ta c: (g.g)

Li c :

6

Cu 1: Cho hm s f(x) =

a) Tnh f(-1); f(5)

b) Tm x f(x) = 10

c) Rt gn A = khi x (

Cu 2: Gii h phng trnh

Cu 3: Cho biu thcA = vi x > 0 v x ( 1

a) Rt gn A

b) Tm gi tr ca x A = 3

Cu 4: T im P nm ngoi ng trn tm O bn knh R, k hai tip tuyn PA; PB. Gi H l chn ng vung gc h t A n ng knh BC.

a) Chng minh rng PC ct AH ti trung im E ca AH

b) Gi s PO = d. Tnh AH theo R v d.

Cu 5: Cho phng trnh 2x2 + (2m - 1)x + m - 1 = 0

Khng gii phng trnh, tm m phng trnh c hai nghim phn bit x1; x2 tha mn: 3x1 - 4x2 = 11

p n

Cu 1a)f(x) =

Suy ra f(-1) = 3; f(5) = 3

b)

c)

Vi x > 2 suy ra x - 2 > 0 suy ra

Vi x < 2 suy ra x - 2 < 0 suy ra

Cu 2

Cu 3 a)Ta c: A = =

= = = = =

b) A = 3 => = 3 => 3x + - 2 = 0 => x = 2/3

Cu 4

Do HA // PB (Cng vung gc vi BC)a) nn theo nh l Ta let p dng cho CPB ta c

;

(1)

Mt khc, do PO // AC (cng vung gc vi AB)

=>

POB = ACB (hai gc ng v)

=>( AHC ( POB

Do :

(2)

Do CB = 2OB, kt hp (1) v (2) ta suy ra AH = 2EH hay E l trung im ca AH.

b) Xt tam gic vung BAC, ng cao AH ta c AH2 = BH.CH = (2R - CH).CH

Theo (1) v do AH = 2EH ta c

AH2.4PB2 = (4R.PB - AH.CB).AH.CB

4AH.PB2 = 4R.PB.CB - AH.CB2

AH (4PB2 +CB2) = 4R.PB.CB

Cu 5 phng trnh c 2 nghim phn bit x1 ; x2 th ( > 0

(2m - 1)2 - 4. 2. (m - 1) > 0

T suy ra m ( 1,5

(1)

Mt khc, theo nh l Vit v gi thit ta c:

Gii phng trnh

ta c m = - 2 v m = 4,125

(2)

i chiu iu kin (1) v (2) ta c: Vi m = - 2 hoc m = 4,125 th phng trnh cho c hai nghim phn bit tha mn: x1 + x2 = 11

7

Cu 1: Cho P = + -

a/. Rt gn P.

b/. Chng minh: P < vi x 0 v x 1.

Cu 2: Cho phng trnh : x2 2(m - 1)x + m2 3 = 0 ( 1 ) ; m l tham s.

a/. Tm m phng trnh (1) c nghim.

b/. Tm m phng trnh (1) c hai nghim sao cho nghim ny bng ba ln nghim kia.

Cu 3: a/. Gii phng trnh : + = 2

b/. Cho a, b, c l cc s thc tha mn :

Tm gi tr ln nht v gi tr b nht ca Q = 6 a + 7 b + 2006 c.

Cu 4: Cho cn ti A vi AB > BC. im D di ng trn cnh AB, ( D khng trng vi A, B). Gi (O) l ng trn ngoi tip . Tip tuyn ca (O) ti C v D ct nhau K .

a/. Chng minh t gic ADCK ni tip.

b/. T gic ABCK l hnh g? V sao?

c/. Xc nh v tr im D sao cho t gic ABCK l hnh bnh hnh.

p n

Cu 1: iu kin: x 0 v x 1. (0,25 im)

P = + -

= + -

=

= =

b/. Vi x 0 v x 1 .Ta c: P <

EMBED Equation.DSMT4 <

3 < x + + 1 ; ( v x + + 1 > 0 )

x - 2 + 1 > 0

( - 1)2 > 0. ( ng v x 0 v x 1)

Cu 2:a/. Phng trnh (1) c nghim khi v ch khi 0.

(m - 1)2 m2 3 0

4 2m 0

m 2.

b/. Vi m 2 th (1) c 2 nghim.

Gi mt nghim ca (1) l a th nghim kia l 3a . Theo Viet ,ta c:

a= 3()2 = m2 3

m2 + 6m 15 = 0

m = 32 ( tha mn iu kin).

Cu 3:

iu kin x 0 ; 2 x2 > 0 x 0 ; < .

t y = > 0

Ta c:

T (2) c : x + y = 2xy. Thay vo (1) c : xy = 1 hoc xy = -

* Nu xy = 1 th x+ y = 2. Khi x, y l nghim ca phng trnh:

X2 2X + 1 = 0 X = 1 x = y = 1.

* Nu xy = - th x+ y = -1. Khi x, y l nghim ca phng trnh:

X2 + X - = 0 X =

V y > 0 nn: y = x =

Vy phng trnh c hai nghim: x1 = 1 ; x2 =

Cu 4: c/. Theo cu b, t gic ABCK l hnh thang. Do , t gic ABCK l hnh bnh hnh AB // CK

M s = s =

Nn

Dng tia Cy sao cho .Khi , D l giao im ca v Cy.

Vi gi thit > th > > .

D AB .

Vy im D xc nh nh trn l im cn tm.

8

Cu 1: a) Xc nh x R biu thc :A = L mt s t nhin

b. Cho biu thc: P = Bit x.y.z = 4 , tnh .

Cu 2:Cho cc im A(-2;0) ; B(0;4) ; C(1;1) ; D(-3;2)a. Chng minh 3 im A, B ,D thng hng; 3 im A, B, C khng thng hng.

b. Tnh din tch tam gic ABC.

Cu3 Gii phng trnh:

Cu 4 Cho ng trn (O;R) v mt im A sao cho OA = R. V cc tip tuyn AB, AC vi ng trn. Mt gc (xOy = 450 ct on thng AB v AC ln lt ti D v E.

Chng minh rng:

a.DE l tip tuyn ca ng trn ( O ).

b.

p n

Cu 1: a.

A =

A l s t nhin -2x l s t nhin x =

(trong k Z v k 0 )

b.iu kin xc nh: x,y,z 0, kt hp vi x.y.z = 4 ta c x, y, z > 0 v

Nhn c t v mu ca hng t th 2 vi ; thay 2 mu ca hng t th 3 bi ta c:

P =

(1)

EMBED Equation.3 v P > 0

Cu 2:a.ng thng i qua 2 im A v B c dng y = ax + b

im A(-2;0) v B(0;4) thuc ng thng AB nn b = 4; a = 2

Vy ng thng AB l y = 2x + 4.

im C(1;1) c to khng tho mn y = 2x + 4 nn C khng thuc ng thng AB A, B, C khng thng hng.

im D(-3;2) c to tho mn y = 2x + 4 nn im D thuc ng thng AB A,B,D thng hn

b.Ta c :

AB2 = (-2 0)2 + (0 4)2 =20

AC2 = (-2 1)2 + (0 1)2 =10

BC2 = (0 1)2 + (4 1)2 = 10

AB2 = AC2 + BC2 (ABC vung ti C

Vy S(ABC = 1/2AC.BC = ( n v din tch )

Cu 3:kx x1, t ta c h phng trnh:

Gii h phng trnh bng phng php th ta c: v = 2

x = 10.

Cu 4a.p dng nh l Pitago tnh c

AB = AC = R ABOC l hnh

vung (0.5)

K bn knh OM sao cho

(BOD = (MOD

(MOE = (EOC (0.5)

Chng minh (BOD = (MOD

(OMD = (OBD = 900Tng t: (OME = 900

D, M, E thng hng. Do DE l tip tuyn ca ng trn (O).

b.Xt (ADE c DE < AD +AE m DE = DB + EC

2ED < AD +AE +DB + EC hay 2DE < AB + AC = 2RDE < R

Ta c DE > AD; DE > AE ; DE = DB + EC

Cng tng v ta c: 3DE > 2R DE > R

Vy R > DE > R

9


a) Tnh f(-1); f(5)

b) Tm x f(x) = 10



Cu 3: Cho biu thc

A = vi x > 0 v x ( 1

a) Rt gn A

2) Tm gi tr ca x A = 3






p n

Cu 1

a)f(x) =

Suy ra f(-1) = 3; f(5) = 3

b)

c)



Cu 2

Cu 3a)Ta c: A =

=

=

=

= = =

b) A = 3 => = 3 => 3x + - 2 = 0 => x = 2/3

Cu 4

a) Do HA // PB (Cng vung gc vi BC)

b) nn theo nh l Ta let p dng cho tam gic CPB ta c

;

(1)


=>POB = ACB (hai gc ng v)

=>( AHC ( POB

Do :

(2)

Do CB = 2OB, kt hp (1) v (2) ta suy ra AH = 2EH hay E l trug im ca AH.






Cu 5 (1)

phng trnh c 2 nghim phn bit x1 ; x2 th ( > 0

(2m - 1)2 - 4. 2. (m - 1) > 0

T suy ra m ( 1,5

(1)


Gii phng trnh

ta c m = - 2 v m = 4,125

(2)

i chiu iu kin (1) v (2) ta c: Vi m = - 2 hoc m = 4,125 th phng trnh cho c hai nghim phn bit t

10

Cu I : Tnh gi tr ca biu thc:A = + ++ .....+

B = 35 + 335 + 3335 + ..... +

Cu II :Phn tch thnh nhn t :1) X2 -7X -18

2) (x+1) (x+2)(x+3)(x+4)

3) 1+ a5 + a10Cu III :

1) Chng minh : (ab+cd)2 (a2+c2)( b2 +d2)

2) p dng : cho x+4y = 5 . Tm GTNN ca biu thc : M= 4x2 + 4y2

Cu 4 : Cho tam gic ABC ni tip ng trn (O), I l trung im ca BC, M l mt im trn on CI ( M khc C v I ). ng thng AM ct (O) ti D, tip tuyn ca ng trn ngoi tip tam gic AIM ti M ct BD v DC ti P v Q.a) Chng minh DM.AI= MP.IB

b) Tnh t s :

Cu 5:

Cho P =

Tm iu kin biu thc c ngha, rt gn biu thc.

p n

Cu 1 :

1) A = + ++ .....+

= (+ + + .....+ ) = ()

2) B = 35 + 335 + 3335 + ..... + =

=33 +2 +333+2 +3333+2+.......+ 333....33+2

= 2.99 + ( 33+333+3333+...+333...33)

= 198 + ( 99+999+9999+.....+999...99)

198 + ( 102 -1 +103 - 1+104 - 1+ ....+10100 1) = 198 33 +

B = +165

Cu 2: 1)x2 -7x -18 = x2 -4 7x-14 = (x-2)(x+2) - 7(x+2) = (x+2)(x-9) (1)2)(x+1)(x+2)(x+3)(x+4) -3= (x+1)(x+4)(x+2)(x+3)-3

= (x2+5x +4)(x2 + 5x+6)-3= [x2+5x +4][(x2 + 5x+4)+2]-3

= (x2+5x +4)2 + 2(x2+5x +4)-3=(x2+5x +4)2 - 1+ 2(x2+5x +4)-2

= [(x2+5x +4)-1][(x2+5x +4)+1] +2[(x2+5x +4)-1]

= (x2+5x +3)(x2+5x +7)

3) a10+a5+1

= a10+a9+a8+a7+a6 + a5 +a5+a4+a3+a2+a +1

- (a9+a8+a7 )- (a6 + a5 +a4)- ( a3+a2+a )

= a8(a2 +a+1) +a5(a2 +a+1)+ a3(a2 +a+1)+ (a2 +a+1)-a7(a2 +a+1)

-a4(a2 +a+1)-a(a2 +a+1)

=(a2 +a+1)( a8-a7+ a5 -a4+a3 - a +1)

Cu 3: 4

1) Ta c : (ab+cd)2 (a2+c2)( b2 +d2)

a2b2+2abcd+c2d2 a2b2+ a2d2 +c2b2 +c2d2

0 a2d2 - 2cbcd+c2b2

0 (ad - bc)2 (pcm )

Du = xy ra khi ad=bc.

2) p dng hng ng thc trn ta c :

52 = (x+4y)2 = (x. + 4y) (x2 + y2)=>

x2 + y2 => 4x2 + 4y2 du = xy ra khi x= , y = (2)

Cu 4 : 5

Ta c : gc DMP= gc AMQ = gc AIC. Mt khc gc ADB = gc BCA=>

MPD ng dng vi ICA => => DM.IA=MP.CI hay DM.IA=MP.IB (1).

Ta c gc ADC = gc CBA,

Gc DMQ = 1800 - AMQ=1800 - gc AIM = gc BIA.

Do DMQ ng dng vi BIA =>

=> DM.IA=MQ.IB (2)

T (1) v (2) ta suy ra = 1

Cu 5

P xc nh th : x2-4x+3 0 v 1-x >0

T 1-x > 0 => x < 1

Mt khc : x2-4x+3 = (x-1)(x-3), V x < 1 nn ta c :

(x-1) < 0 v (x-3) < 0 t suy ra tch ca (x-1)(x-3) > 0

Vy vi x < 1 th biu thc c ngha.

Vi x < 1 Ta c :

P = =

11

Cu 1 : a. Rt gn biu thc . Vi a > 0.

b. Tnh gi tr ca tng.

Cu 2 : Cho pt

a. Chng minh rng pt lun lun c nghim vi .

b. Gi l hai nghim ca pt. Tm GTLN, GTNN ca bt.

Cu 3 : Cho Chng minh.

Cu 4 Cho ng trn tm o v dy AB. M l im chuyn ng trn ng trn, tM k MH ( AB (H ( AB). Gi E v F ln lt l hnh chiu vung gc ca H trn MA v MB. Qua M k ng thng vung gc vi ct dy AB ti D.

1. Chng minh rng ng thng MD lun i qua 1 im c nh khi M thay i trn ng trn.

2. Chng minh.

Hng dn

Cu 1 a. Bnh phng 2 v

(V a > 0).

c. p dng cu a.

Cu 2 a. : cm

B (2 ) p dng h thc Viet ta c:

(1) Tm k pt (1) c nghim theo n.

Cu 3 : Chuyn v quy ng ta c.

bt

ng v

Cu 4: a

- K thm ng ph.

- Chng minh MD l ng knh ca (o)

=> ........

b.

Gi E', F' ln lt l hnh chiu ca D trn MA v MB.

t HE = H1

HF = H2

Thay vo (1) ta c:

12

Cu 1: Cho biu thc D = :

a) Tm iu kin xc nh ca D v rt gn D

b) Tnh gi tr ca D vi a =

c) Tm gi tr ln nht ca D

Cu 2: Cho phng trnh x2- mx + m2 + 4m - 1 = 0 (1)

a) Gii phng trnh (1) vi m = -1

b) Tm m phng trnh (1) c 2 nghim tho mn

Cu 3: Cho tam gic ABC ng phn gic AI, bit AB = c, AC = b, Chng minh rng AI = (Cho Sin2)

Cu 4: Cho ng trn (O) ng knh AB v mt im N di ng trn mt na ng trn sao cho V vo trong ng trn hnh vung ANMP.

a) Chng minh rng ng thng NP lun i qua im c nh Q.

b) Gi I l tm ng trn ni tip tam gic NAB. Chng minh t gic ABMI ni tip.

c) Chng minh ng thng MP lun i qua mt im c nh.

Cu 5: Cho x,y,z; xy + yz + zx = 0 v x + y + z = -1

Hy tnh gi tr ca:

B =

p n

Cu 1: a) - iu kin xc nh ca D l

EMBED Equation.3 - Rt gn D

D = :

D =

b) a =

Vy D =

c) p dng bt ng thc cauchy ta c

Vy gi tr ca D l 1

Cu 2: a) m = -1 phng trnh (1)

b) phng trnh 1 c 2 nghim th (*)+ phng trnh c nghim khc 0 (*)+

Kt hp vi iu kin (*)v (**) ta c m = 0 v

Cu 3:

+

+

+

Cu 4: a) Gi Q = NP

Suy ra Q c nh

b)

T gic ABMI ni tip

c) Trn tia i ca QB ly im F sao cho QF = QB, F c nh.

Tam gic ABF c: AQ = QB = QF

ABF vung ti A

Li c T gic APQF ni tip

Ta c:

M1,P,F Thng hng

Cu 5: Bin i B = xyz =

13

Bi 1: Cho biu thc A =

a) Tm iu kin ca x A xc nh

b) Rt gn A

Bi 2 : Trn cng mt mt phng ta cho hai im A(5; 2) v B(3; -4)

a) Vit phng tnh ng thng AB

b) Xc nh im M trn trc honh tam gic MAB cn ti M

Bi 3 : Tm tt c cc s t nhin m phng trnh n x sau:

x2 - m2x + m + 1 = 0

c nghim nguyn.

Bi 4 : Cho tam gic ABC. Phn gic AD (D ( BC) v ng trn tm O qua A v D ng thi tip xc vi BC ti D. ng trn ny ct AB v AC ln lt ti E v F. Chng minh

a) EF // BC

b) Cc tam gic AED v ADC; D v ABD l cc tam gic ng dng.

c) AE.AC = .AB = AC2Bi 5 : Cho cc s dng x, y tha mn iu kin x2 + y2 ( x3 + y4. Chng minh:

x3 + y3 ( x2 + y2 ( x + y ( 2

p n

Bi 1:

a) iu kin x tha mn

(

( x > 1 v x ( 2

KL: A xc nh khi 1 < x < 2 hoc x > 2

b) Rt gn A

A =

A =

Vi 1 < x < 2 A =

Vi x > 2 A =

Kt lun

Vi 1 < x < 2 th A =

Vi x > 2 th A =

Bi 2:

a) A v B c honh v tung u khc nhau nn phng trnh ng thng AB c dng y = ax + b

A(5; 2) ( AB ( 5a + b = 2

B(3; -4) ( AB ( 3a + b = -4

Gii h ta c a = 3; b = -13

Vy phng trnh ng thng AB l y = 3x - 13

b) Gi s M (x, 0) ( xx ta c

MA =

MB =

MAB cn ( MA = MB (

( (x - 5)2 + 4 = (x - 3)2 + 16

( x = 1

Kt lun: im cn tm: M(1; 0)

Bi 3:

Phng trnh c nghim nguyn khi = m4 - 4m - 4 l s chnh phng

Ta li c: m = 0; 1 th < 0 loi

m = 2 th = 4 = 22 nhn

m ( 3 th 2m(m - 2) > 5 ( 2m2 - 4m - 5 > 0

( - (2m2 - 2m - 5) < < + 4m + 4

( m4 - 2m + 1 < < m4( (m2 - 1)2 < < (m2)2 khng chnh phng

Vy m = 2 l gi tr cn tm.

Bi 4:

a) (0,25)

(0,25)

m (0,25)

( EF // BC (2 gc so le trong bng nhau)

b) AD l phn gic gc BAC nn

ss() = s = s

do v

( DADC (g.g)

Tng t: s = (

do AFD ~ (g.g

c) Theo trn:

+ AED ~ DB

( hay AD2 = AE.AC (1)

+ ADF ~ ABD (

( AD2 = AB.AF (2)

T (1) v (2) ta c AD2 = AE.AC = AB.AF

Bi 5 (1):

Ta c (y2 - y) + 2 ( 0 ( 2y3 ( y4 + y2( (x3 + y2) + (x2 + y3) ( (x2 + y2) + (y4 + x3)

m x3 + y4 ( x2 + y3 do

x3 + y3 ( x2 + y2 (1)

+ Ta c: x(x - 1)2 ( 0: y(y + 1)(y - 1)2 ( 0

( x(x - 1)2 + y(y + 1)(y - 1)2 ( 0

( x3 - 2x2 + x + y4 - y3 - y2 + y ( 0

( (x2 + y2) + (x2 + y3) ( (x + y) + (x3 + y4)

m x2 + y3 ( x3 + y4( x2 + y2 ( x + y (2)

v (x + 1)(x - 1) ( 0.(y - 1)(y3 -1) ( 0

x3 - x2 - x + 1 + y4 - y - y3 + 1 ( 0

( (x + y) + (x2 + y3) ( 2 + (x3 + y4)

m x2 + y3 ( x3 + y4( x + y ( 2

T (1) (2) v (3) ta c:

x3 + y3 ( x2 + y2 ( x + y ( 2

14

Cu 1: x- 4(x-1) + x + 4(x-1) 1

cho A= ( 1 - )

x2- 4(x-1) x-1

a/ rt gn biu thc A.

b/ Tm gi tr nguyn ca x A c gi tr nguyn.

Cu 2: Xc nh cc gi tr ca tham s m phng trnh

x2-(m+5)x-m+6 =0

C 2 nghim x1 v x2 tho mn mt trong 2 iu kin sau:

a/ Nghim ny ln hn nghim kia mt n v.

b/ 2x1+3x2=13

Cu 3Tm gi tr ca m h phng trnh

mx-y=1

m3x+(m2-1)y =2

v nghim, v s nghim.

Cu 4: tm max v min ca biu thc: x2+3x+1

x2+1

Cu 5: T mt nh A ca hnh vung ABCD k hai tia to vi nhau mt gc 450. Mt tia ct cnh BC ti E ct ng cho BD ti P. Tia kia ct cnh CD ti F v ct ng cho BD ti Q.

a/ Chng minh rng 5 im E, P, Q, F v C cng nm trn mt ng trn.

b/ Chng minh rng: SAEF=2SAQPc/ K trung trc ca cnh CD ct AE ti M tnh s o gc MAB bit CPD=CM

hng dn Cu 1: a/ Biu thc A xc nh khi x2 v x>1

( x-1 -1)2+ ( x-1 +1)2 x-2

A= . ( )

(x-2)2 x-1

x- 1 -1 + x-1 + 1 x- 2 2 x- 1 2

= . = =

x-2 x-1 x-1 x-1

b/ A nguyn th x- 1 l c dng ca 1 v 2

* x- 1 =1 th x=0 loi

* x- 1 =2 th x=5

vy vi x = 5 th A nhn gi tr nguyn bng 1

Cu 2: Ta c x = (m+5)2-4(-m+6) = m2+14m+10 phng trnhc hai nghimphn bit khi vch khi m-7-4 3 v m-7+4 3 (*)

a/ Gi s x2>x1 ta c h x2-x1=1 (1) x1+x2=m+5 (2) x1x2 =-m+6 (3)

Gii h tac m=0 v m=-14 tho mn (*)

b/ Theo gi thit ta c: 2x1+3x2 =13(1) x1+x2 = m+5(2) x1x2 =-m+6 (3)

gii h ta c m=0 v m= 1 Tho mn (*)

Cu 3: * h v nghim th m/m3=-1/(m2-1) 1/2

3m3-m=-m3 m2(4m2- 1)=0 m=0 m=0

3m2-1-2 3m2-1 m=1/2 m=1/2

m

*Hv s nghim th: m/m3=-1/(m2-1) =1/2

3m3-m=-m3 m=0

3m2-1= -2 m=1/2

V nghim

Khng c gi tr no ca m h v s nghim.

Cu 4: Hm s xc nh vi x(v x2+10) x2+3x+1

gi y0 l 1 gi trca hmphng trnh: y0=

x2+1

(y0-1)x2-6x+y0-1 =0 c nghim

*y0=1 suy ra x = 0 y0 1; =9-(y0-1)20 (y0-1)2 9 suy ra -2 y0 4

Vy: ymin=-2 v y max=4

Cu 5: ( Hc sinh t v hnh)Gii

a/ A1 v B1 cng nhn on QE di mt gc 450

( t gic ABEQ ni tip c.

( FQE = ABE =1v.

chng minh tng t ta c FBE = 1v

( Q, P, C cng nm trn ng trn ng kinh EF.

b/ T cu a suy ra AQE vung cn.

( = (1)tng t APF cng vung cn

( = (2)t (1) v (2) ( AQP ~ AEF (c.g.c)

= ( )2 hay SAEF = 2SAQPc/ thy CPMD ni tip, MC=MD v APD=CPD

(MCD= MPD=APD=CPD=CMD

(MD=CD ( MCD u ( MPD=600

m MPD l gc ngoi ca ABM ta c APB=450 vy MAB=600-450=150 15

Bi 1: Cho biu thc M =

a. Tm iu kin ca x M c ngha v rt gn M

b. Tm x M = 5

c. Tm x Z M Z.

bi 2: a) Tm x, y nguyn dng tho mn phng trnh

3x2 +10 xy + 8y2 =96

b)tm x, y bit / x - 2005/ + /x - 2006/ +/y - 2007/+/x- 2008/ = 3

Bi 3: a. Cho cc s x, y, z dng tho mn + + = 4

Chng ming rng: + +

b. Tm gi tr nh nht ca biu thc: B = (vi x )

Bi 4: Cho hnh vung ABCD. K tia Ax, Ay sao cho = 45

Tia Ax ct CB v BD ln lt ti E v P, tia Ay ct CD v BD ln lt ti F v Q

a. Chng minh 5 im E; P; Q; F; C cng nm trn mt ng trn

b. S

EMBED Equation.3 = 2 S

K ng trung trc ca CD ct AE ti M. Tnh s o gc MAB bit =

Bi 5: (1)

Cho ba s a, b , c khc 0 tho mn: ; Hy tnh P =

p n Bi 1:M =

a.K 0,5

Rt gn M =

Bin i ta c kt qu: M = M =

c. M =

Do M nn l c ca 4 nhn cc gi tr: -4; -2; -1; 1; 2; 4

do

EMBED Equation.3 Bi 2 a. 3x2 + 10xy + 8y2 = 96

3x2 + 4xy + 6xy + 8y2 = 96

(3x2 + 6xy) + (4xy + 8y2) = 96

3x(x + 2y) + 4y(x +2y) = 96

(x + 2y)(3x + 4y) = 96

Do x, y nguyn dng nn x + 2y; 3x + 4y nguyen dng v 3x + 4y > x + 2y

m 96 = 25. 3 c cc c l: 1; 2; 3; 4; 6; 8; 12; 24; 32; 48; 96 c biu din thnh tch 2 tha s khng nh hn 3 l: 96 = 3.32 = 4.24 = 6. 16 = 8. 12

Li c x + 2y v 3x + 4y c tch l 96 (L s chn) c tng 4x + 6y l s chn do

H PT ny v nghim

Hoc

EMBED Equation.3 Hoc H PT v nghim

Vy cp s x, y nguyn dng cn tm l (x, y) = (4, 1)

b. ta c /A/ = /-A/

Nn /x - 2005/ + / x - 2006/ = / x - 2005/ + / 2008 - x/ (1)

m /x - 2005/ + / x - 2006/ + / y - 2007/ + / x - 2008/ = 3 (2)

Kt hp (1 v (2) ta c / x - 2006/ + / y - 2007/ (3)

(3) sy ra khi v ch khi

Bi 3

a. Trc ht ta chng minh bt ng thc ph

b. Vi mi a, b thuc R: x, y > 0 ta c

(a2y + b2x)(x + y)

a2y2 + a2xy + b2 x2 + b2xy a2xy + 2abxy + b2xy

a2y2 + b2x2 2abxy

a2y2 2abxy + b2x2 0

(ay - bx)2 0 (**) bt ng thc (**) ng vi mi a, b, v x,y > 0

Du (=) xy ra khi ay = bx hay

p dung bt ng thc (*) hai ln ta c

Tng t

Cng tng v cc bt ng thc trn ta c:

V

Ta c:

V (x - 2006)2 0 vi mi x

x2 > 0 vi mi x khc 0

Bi 4a. ni tip; = 900 ( gc AQE = 900 ( gcEQF = 900Tng t gc FDP = gc FAP = 450( T gic FDAP ni tip gc D = 900 ( gc APF = 900 ( gc EPF = 900 . 0,25

Cc im Q, P,C lun nhn di 1gc900 nn 5 im E, P, Q, F, C cng nm trn 1 ng trn ng knh EF 0,25

b. Ta c gc APQ + gc QPE = 1800 (2 gc k b) gc APQ = gc AFE

Gc AFE + gc EPQ = 1800

(Tam gic APQ ng dng vi tam gic AEF (g.g)

(

c. gc CPD = gc CMD ( t gic MPCD ni tip ( gc MCD = gc CPD (cng chn cung MD)

Li c gc MPD = gc CPD (do BD l trung trc ca AC)

gc MCD = gc MDC (do M thuc trung trc ca DC)

( gc CPD = gcMDC = gc CMD = gcMCD ( tam gic MDC u ( gc CMD = 600( tam gic DMA cn ti D (v AD = DC = DM)

V gc ADM =gcADC gcMDC = 900 600 = 300( gc MAD = gc AMD (1800 - 300) : 2 = 750( gcMAB = 900 750 = 150Bi 5t x = 1/a; y =1/b; z = 1/c ( x + y + z = 0 (v 1/a = 1/b + 1/c = 0)

( x = -(y + z)

( x3 + y3 + z3 3 xyz = -(y + z)3 + y3 3xyz

(-( y3 + 3y2 z +3 y2z2 + z3) + y3 + z3 3xyz = - 3yz(y + z + x) = - 3yz .0 = 0

T x3 + y3 + z3 3xyz = 0 ( x3 + y3 + z3 = 3xyz

( 1/ a3 + 1/ b3 + 1/ c3 3 1/ a3 .1/ b3 .1/ c3 = 3/abc

Do P = ab/c2 + bc/a2 + ac/b2 = abc (1/a3 + 1/b3+ 1/c3) = abc.3/abc = 3

nu 1/a + 1/b + 1/c =o th P = ab/c2 + bc/a2 + ac/b2 = 3

16

Bi 1Cho biu thc A = +

a. Rt gn biu thc A

b. Tm nhng gi tr nguyn ca x sao cho biu thc A cng c gi tr nguyn.

Bi 2: (2 im)

Cho cc ng thng:

y = x-2 (d1)

y = 2x 4 (d2)

y = mx + (m+2) (d3)

a. Tm im c nh m ng thng (d3 ) lun i qua vi mi gi tr ca m.

b. Tm m ba ng thng (d1); (d2); (d3) ng quy .

Bi 3: Cho phng trnh x2 - 2(m-1)x + m - 3 = 0 (1)

a. Chng minh phng trnh lun c 2 nghim phn bit.

b. Tm mt h thc lin h gia hai nghim ca phng trnh (1) m khng ph thuc vo m.

c. Tm gi tr nh nht ca P = x21 + x22 (vi x1, x2 l nghim ca phng trnh (1))

Bi 4: Cho ng trn (o) vi dy BC c nh v mt im A thay i v tr trn cung ln BC sao cho AC>AB v AC > BC . Gi D l im chnh gia ca cung nh BC. Cc tip tuyn ca (O) ti D v C ct nhau ti E. Gi P, Q ln lt l giao im ca cc cp ng thng AB vi CD; AD v CE.

a. Chng minh rng DE// BC

b. Chng minh t gic PACQ ni tip

c. Gi giao im ca cc dy AD v BC l F

Chng minh h thc: = +

Bi 5: Cho cc s dng a, b, c Chng minh rng:

p n

Bi 1: - iu kin : x 0

a. Rt gn:

- Vi x x =

Bi 2:

a. (d1) : y = mx + (m +2)

m (x+1)+ (2-y) = 0

hm s lun qua im c nh vi mi m

=.>

Vy N(-1; 2) l im c nh m (d3) i qua

b. Gi M l giao im (d1) v (d2) . Ta M l nghim ca h

=>

Vy M (2; 0) .

Nu (d3) i qua M(2,0) th M(2,0) l nghim (d3)

Ta c : 0 = 2m + (m+2) => m= -

Vy m = - th (d1); (d2); (d3) ng quy

Bi 3: a. = m2 3m + 4 = (m - )2 + >0 m.

Vy phng trnh c 2 nghim phn bit

b. Theo Vit: =>

x1+ x2 2x1x2 4 = 0 khng ph thuc vo m

a. P = x12 + x12 = (x1 + x2)2 - 2x1x2 = 4(m - 1)2 2 (m-3)

= (2m - )2 +

VyPmin = vi m =

Bi 4: V hnh ng vit gi thit kt lun

a. SCDE = S DC = S BD =

=> DE// BC (2 gc v tr so le)

b. APC = s (AC - DC) = AQC

=> APQC ni tip (v APC = AQC

cng nhn oan AC)

c.T gic APQC ni tip

CPQ = CAQ (cng chn cung CQ)

CAQ = CDE (cng chn cung DC)

Suy ra CPQ = CDE => DE// PQ

Ta c: = (v DE//PQ) (1)

= (v DE// BC) (2)

Cng (1) v (2) :

=> (3)

ED = EC (t/c tip tuyn) t (1) suy ra PQ = CQ

Thay vo (3) :

Bi 5:Ta c: < < (1)

< a=-1;a=-4

T ta c

* (1)

*(2)

Gii h (1) ta c x=3, y=2

Gii h (2) ta c x=0, y=4

Vy h phng trnh c nghim l x=3, y=2 hoc x=0; y=4

b) Ta c x3-4x2-2x-15=(x-5)(x2+x+3)

m x2+x+3=(x+1/2)2+11/4>0 vi mi x

Vy bt phng trnh tng ng vi x-5>0 =>x>5

Cu 3: Phng trnh: ( 2m-1)x2-2mx+1=0

Xt 2m-1=0=> m=1/2 pt tr thnh x+1=0=> x=1

Xt 2m-1(0=> m( 1/2 khi ta c

= m2-2m+1= (m-1)2(0 mi m=> pt c nghim vi mi m

ta thy nghim x=1 khng thuc (-1,0)

vi m( 1/2 pt cn c nghim x==

pt c nghim trong khong (-1,0)=> -1m E,F thuc ng trn ng knh BK

hay 4 im E,F,B,K thuc ng trn ng knh BK.

b. BCF= BAF

M BAF= BAE=450=> BCF= 450Ta c BKF= BEF

M BEF= BEA=450(EA l ng cho ca hnh vung ABED)=> BKF=450V BKC= BCK= 450=> tam gic BCK vung cn ti B

5

Bi 1: Cho biu thc: P =

a,Rt gn Pb,Tm x nguyn P c gi tr nguyn.

Bi 2: Cho phng trnh: x2-( 2m + 1)x + m2 + m - 6= 0 (*)a.Tm m phng trnh (*) c 2 nghim m.

b.Tm m phng trnh (*) c 2 nghim x1; x2 tho mn =50

Bi 3: Cho phng trnh: ax2 + bx + c = 0 c hai nghim dng phn bit x1, x2Chng minh:

a,Phng trnh ct2 + bt + a =0 cng c hai nghim dng phn bit t1 v t2.

b,Chng minh: x1 + x2 + t1 + t2 4

Bi 4: Cho tam gic c cc gc nhn ABC ni tip ng trn tm O . H l trc tm ca tam gic. D l mt im trn cung BC khng cha im A.a, Xc nh v tr ca im D t gic BHCD l hnh bnh hnh.

b, Gi P v Q ln lt l cc im i xng ca im D qua cc ng thng AB v AC . Chng minh rng 3 im P; H; Q thng hng.

c, Tm v tr ca im D PQ c di ln nht.

Bi 5: Cho hai s dng x; y tho mn: x + y 1Tm gi tr nh nht ca: A =

p n

Bi 1: (2 im). K: x

a, Rt gn: P = P =

b. P =

P nguyn th

Vy vi x= th P c gi tr nguyn.

Bi 2: phng trnh c hai nghim m th:

b. Gii phng trnh:

Bi 3: a. V x1 l nghim ca phng trnh: ax2 + bx + c = 0 nn ax12 + bx1 + c =0. .

V x1> 0 => c. Chng t l mt nghim dng ca phng trnh: ct2 + bt + a = 0; t1 = V x2 l nghim ca phng trnh:

ax2 + bx + c = 0 => ax22 + bx2 + c =0

v x2> 0 nn c. iu ny chng t l mt nghim dng ca phng trnh ct2 + bt + a = 0 ; t2 =

Vy nu phng trnh: ax2 + bx + c =0 c hai nghim dng phn bit x1; x2 th phng trnh : ct2 + bt + a =0 cng c hai nghim dng phn bit t1 ; t2 . t1 = ; t2 =

b. Do x1; x1; t1; t2 u l nhng nghim dng nn

t1+ x1 = + x1 2 t2 + x2 = + x2 2

Do x1 + x2 + t1 + t2 4

Bi 4a. Gi s tm c im D trn cung BC sao cho t gic BHCD l hnh bnh hnh . Khi : BD//HC; CD//HB v H l trc tm tam gic ABC nn

CH v BH => BD v CD.

Do : ABD = 900 v ACD = 900 .

Vy AD l ng knh ca ng trn tm O

Ngc li nu D l u ng knh AD

ca ng trn tm O th

t gic BHCD l hnh bnh hnh.

b) V P i xng vi D qua AB nn APB = ADB

nhng ADB =ACB nhng ADB = ACB

Do : APB = ACB Mt khc:

AHB + ACB = 1800 => APB + AHB = 1800

T gic APBH ni tip c ng trn nn PAB = PHB

M PAB = DAB do : PHB = DAB

Chng minh tng t ta c: CHQ = DAC

Vy PHQ = PHB + BHC + CHQ = BAC + BHC = 1800Ba im P; H; Q thng hng

c). Ta thy APQ l tam gic cn nh A C AP = AQ = AD v PAQ = 2BAC khng i nn cnh y PQ

t gi tr ln nht ( AP v AQ l ln nht hay ( AD l ln nht

( D l u ng knh k t A ca ng trn tm O

6

Bi 1: Cho biu thc:

a). Tm iu kin ca x v y P xc nh . Rt gn P.

b). Tm x,y nguyn tha mn phng trnh P = 2.

Bi 2: Cho parabol (P) : y = -x2 v ng thng (d) c h s gc m i qua im M(-1 ; -2) .

a). Chng minh rng vi mi gi tr ca m (d) lun ct (P) ti hai im A , B phn bit

b). Xc nh m A,B nm v hai pha ca trc tung.

Bi 3: Gii h phng trnh :

Bi 4: Cho ng trn (O) ng knh AB = 2R v C l mt im thuc ng trn . Trn na mt phng b AB c cha im C , k tia Ax tip xc vi ng trn (O), gi M l im chnh gia ca cung nh AC . Tia BC ct Ax ti Q , tia AM ct BC ti N.

a). Chng minh cc tam gic BAN v MCN cn .

b). Khi MB = MQ , tnh BC theo R.

Bi 5: Cho tha mn :

EMBED Equation.3 Hy tnh gi tr ca biu thc : M = + (x8 y8)(y9 + z9)(z10 x10) .

p n

Bi 1: a). iu kin P xc nh l :; .

*). Rt gn P:




EMBED Equation.DSMT4 Vy P =

b). P = 2

EMBED Equation.3 = 2

Ta c: 1 + ( ( x = 0; 1; 2; 3 ; 4

Thay vo ta ccc cp gi tr (4; 0) v (2 ; 2) tho mn

Bi 2: a). ng thng (d) c h s gc m v i qua im M(-1 ; -2) . Nn phng trnh ng thng (d) l : y = mx + m 2.

Honh giao im ca (d) v (P) l nghim ca phng trnh:

- x2 = mx + m 2

x2 + mx + m 2 = 0 (*)

V phng trnh (*) c nn phng trnh (*) lun c hai nghim phn bit , do (d) v (P) lun ct nhau ti hai im phn bit A v B.

b). A v B nm v hai pha ca trc tung phng trnh : x2 + mx + m 2 = 0 c hai nghim tri du m 2 < 0 m < 2.

Bi 3 :

EMBED Equation.3 KX :

Thay vo (1) => x = y = z = 3 .

Ta thy x = y = z = 3 tha mn h phng trnh . Vy h phng trnh c nghim duy nht x = y = z = 3.

Bi 4:

a). Xt v .

Ta c: AB l ng knh ca ng trn (O)

nn :AMB = NMB = 90o .

M l im chnh gia ca cung nh AC

nn ABM = MBN => BAM = BNM

=> cn nh B.

T gic AMCB ni tip

=> BAM = MCN ( cng b vi gc MCB).

=> MCN = MNC ( cng bng gc BAM).

=> Tam gic MCN cn nh M

b). Xt v c :

MC = MN (theo cm trn MNC cn ) ; MB = MQ ( theo gt)

BMC = MNQ ( v : MCB = MNC ; MBC = MQN ).

=> => BC = NQ .

Xt tam gic vung ABQ c AB2 = BC . BQ = BC(BN + NQ)

=> AB2 = BC .( AB + BC) = BC( BC + 2R)

=> 4R2 = BC( BC + 2R) => BC =

Bi 5:

T : =>

=>

Ta c : x8 y8 = (x + y)(x-y)(x2+y2)(x4 + y4).=

y9 + z9 = (y + z)(y8 y7z + y6z2 - .......... + z8)

z10- x10 = (z + x)(z4 z3x + z2x2 zx3 + x4)(z5 - x5)

Vy M = + (x + y) (y + z) (z + x).A =

7

Bi 1: 1) Cho ng thng d xc nh bi y = 2x + 4. ng thng d/ i xng vi ng thng d qua ng thng y = x l:

A.y = x + 2 ; B.y = x - 2 ; C.y = x - 2 ; D.y = - 2x - 4

Hy chn cu tr li ng.

2) Mt hnh tr c chiu cao gp i ng knh y ng y nc, nhng chm vo bnh mt hnh cu khi ly ra mc nc trong bnh cn li bnh. T s gia bn knh hnh tr v bn knh hnh cu l A.2 ; B. ; C. ; D. mt kt qu khc.

Ba2: 1) Gii phng trnh: 2x4 - 11 x3 + 19x2 - 11 x + 2 = 02) Cho x + y = 1 (x > 0; y > 0) Tm gi tr ln nht ca A = +

Bi 3: 1) Tm cc s nguyn a, b, c sao cho a thc : (x + a)(x - 4) - 7

Phn tch thnh tha s c : (x + b).(x + c)

2) Cho tam gic nhn xy, B, C ln lt l cc im c nh trn tia Ax, Ay sao cho AB < AC, im M di ng trong gc xAy sao cho =

Xc nh v tr im M MB + 2 MC t gi tr nh nht.

Bi 4: Cho ng trn tm O ng knh AB v CD vung gc vi nhau, ly im I bt k trn oan CD.

a) Tm im M trn tia AD, im N trn tia AC sao cho I lag trung im ca MN.

b) Chng minh tng MA + NA khng i.

c) Chng minh rng ng trn ngoi tip tam gic AMN i qua hai im c nh.

Hng dn Bi 1: 1) Chn C. Tr li ng.

2) Chn D. Kt qu khc: p s l: 1

Bi 2 : 1)A = (n + 1)4 + n4 + 1 = (n2 + 2n + 1)2 - n2 + (n4 + n2 + 1)

= (n2 + 3n + 1)(n2 + n + 1) + (n2 + n + 1)(n2 - n + 1)

= (n2 + n + 1)(2n2 + 2n + 2) = 2(n2 + n + 1)2

Vy A chia ht cho 1 s chnh phng khc 1 vi mi s nguyn dng n.

2) Do A > 0 nn A ln nht A2 ln nht.

Xt A2 = (+ )2 = x + y + 2 = 1 + 2 (1)

Ta c: (Bt ng thc C si)

=> 1 > 2 (2)

T (1) v (2) suy ra: A2 = 1 + 2 < 1 + 2 = 2

Max A2 = 2 x = y = , max A = x = y =

Bi3 Cu 1Vi mi x ta c (x + a)(x - 4) - 7 = (x + b)(x + c)

Nn vi x = 4 th - 7 = (4 + b)(4 + c)

C 2 trng hp: 4 + b = 1 v 4 + b = 7

4 + c = - 7 4 + c = - 1

Trng hp th nht cho b = - 3, c = - 11, a = - 10

Ta c (x - 10)(x - 4) - 7 = (x - 3)(x - 11)

Trng hp th hai cho b = 3, c = - 5, a = 2

Ta c (x + 2)(x - 4) - 7 = (x + 3)(x - 5)

Cu2 (1,5im)

Gi D l im trn cnh AB sao cho:

AD = AB. Ta c D l im c nh

M = (gt) do =

Xt tam gic AMB v tam gic ADM c MB (chung)

= =

Do AMB ~ ADM => = = 2

=> MD = 2MD (0,25 im)

Xt ba im M, D, C : MD + MC > DC (khng i)

Do MB + 2MC = 2(MD + MC) > 2DC

Du "=" xy ra M thuc on thng DC

Gi tr nh nht ca MB + 2 MC l 2 DC

* Cch dng im M.

- Dng ng trn tm A bn knh AB

- Dng D trn tia Ax sao cho AD = AB

M l giao im ca DC v ng trn (A; AB)

Bi 4: a) Dng (I, IA) ct AD ti M ct tia AC ti N

Do MN = 900 nn MN l ng knh

Vy I l trung im ca MN

b) K MK // AC ta c : INC = IMK (g.c.g)

=> CN = MK = MD (v MKD vung cn)

Vy AM+AN=AM+CN+CA=AM+MD+CA

=> AM = AN = AD + AC khng i

c) Ta c IA = IB = IM = IN

Vy ng trn ngoi tip AMN i qua hai im A, B c nh.

8

Bi 1. Cho ba s x, y, z tho mn ng thi :

Tnh gi tr ca biu thc :.

Bi 2). Cho biu thc :.

Vi gi tr no ca x, y th M t gi tr nh nht ? Tm gi tr nh nht

Bi 3. Gii h phng trnh :

Bi 4. Cho ng trn tm O ng knh AB bn knh R. Tip tuyn ti im M bbt k trn ng trn (O) ct cc tip tuyn ti A v B ln lt ti C v D.

a.Chng minh : AC . BD = R2.

b.Tm v tr ca im M chu vi tam gic COD l nh nht .

Bi 5.Cho a, b l cc s thc dng. Chng minh rng :

Bi 6).Cho tam gic ABC c phn gic AD. Chng minh : AD2 = AB . AC - BD . DC.

Hng dn gii

Bi 1. T gi thit ta c :

Cng tng v cc ng thc ta c :

Vy : A = -3.

Bi 2.(1,5 im) Ta c :

Do v

Bi 3. t : Ta c : u ; v l nghim ca phng trnh :

EMBED Equation.DSMT4 ;

;

Gii hai h trn ta c : Nghim ca h l :

(3 ; 2) ; (-4 ; 2) ; (3 ; -3) ; (-4 ; -3) v cc hon v.

Bi 4. a.Ta c CA = CM; DB = DM

Cc tia OC v OD l phn gic ca hai gc AOM v MOB nn OC OD

Tam gic COD vung nh O, OM l ng cao thuc cnh huyn CD nn :

MO2 = CM . MD

R2 = AC . BD

b.Cc t gic ACMO ; BDMO ni tip

(0,25)

Do : (MH1 AB)

Do MH1 OM nn

Chu vi chu vi

Du = xy ra MH1 = OM MO M l im chnh gia ca cung

Bi 5 (1,5 im) Ta c : a , b > 0

a , b > 0

Mt khc

Nhn tng v ta c :

Bi 6. (1 im) V ng trn tm O ngoi tip

Gi E l giao im ca AD v (O)

Ta c: (g.g)

Li c :

9


a) Tnh f(-1); f(5)

b) Tm x f(x) = 10



Cu 3: Cho biu thcA = vi x > 0 v x ( 1

a) Rt gn A

b) Tm gi tr ca x A = 3






p n

Cu 1a)f(x) =

Suy ra f(-1) = 3; f(5) = 3

b)

c)



Cu 2

Cu 3 a)Ta c: A = =

= = = = =

b) A = 3 => = 3 => 3x + - 2 = 0 => x = 2/3

Cu 4

Do HA // PB (Cng vung gc vi BC)b) nn theo nh l Ta let p dng cho CPB ta c

;

(1)


=>

POB = ACB (hai gc ng v)

=>( AHC ( POB

Do :

(2)

Do CB = 2OB, kt hp (1) v (2) ta suy ra AH = 2EH hay E l trung im ca AH.






Cu 5 phng trnh c 2 nghim phn bit x1 ; x2 th ( > 0

(2m - 1)2 - 4. 2. (m - 1) > 0

T suy ra m ( 1,5

(1)


Gii phng trnh

ta c m = - 2 v m = 4,125

(2)

i chiu iu kin (1) v (2) ta c: Vi m = - 2 hoc m = 4,125 th phng trnh cho c hai nghim phn bit tha mn: x1 + x2 = 11

10

Cu 1: Cho P = + -

a/. Rt gn P.

b/. Chng minh: P < vi x 0 v x 1.

Cu 2: Cho phng trnh : x2 2(m - 1)x + m2 3 = 0 ( 1 ) ; m l tham s.

a/. Tm m phng trnh (1) c nghim.

b/. Tm m phng trnh (1) c hai nghim sao cho nghim ny bng ba ln nghim kia.

Cu 3: a/. Gii phng trnh : + = 2

b/. Cho a, b, c l cc s thc tha mn :

Tm gi tr ln nht v gi tr b nht ca Q = 6 a + 7 b + 2006 c.

Cu 4: Cho cn ti A vi AB > BC. im D di ng trn cnh AB, ( D khng trng vi A, B). Gi (O) l ng trn ngoi tip . Tip tuyn ca (O) ti C v D ct nhau K .

a/. Chng minh t gic ADCK ni tip.

b/. T gic ABCK l hnh g? V sao?

c/. Xc nh v tr im D sao cho t gic ABCK l hnh bnh hnh.

p n

Cu 1: iu kin: x 0 v x 1. (0,25 im)

P = + -

= + -

=

= =

b/. Vi x 0 v x 1 .Ta c: P <

EMBED Equation.DSMT4 <

3 < x + + 1 ; ( v x + + 1 > 0 )

x - 2 + 1 > 0

( - 1)2 > 0. ( ng v x 0 v x 1)

Cu 2:a/. Phng trnh (1) c nghim khi v ch khi 0.

(m - 1)2 m2 3 0

4 2m 0

m 2.

b/. Vi m 2 th (1) c 2 nghim.

Gi mt nghim ca (1) l a th nghim kia l 3a . Theo Viet ,ta c:

a= 3()2 = m2 3

m2 + 6m 15 = 0

m = 32 ( tha mn iu kin).

Cu 3:

iu kin x 0 ; 2 x2 > 0 x 0 ; < .

t y = > 0

Ta c:

T (2) c : x + y = 2xy. Thay vo (1) c : xy = 1 hoc xy = -

* Nu xy = 1 th x+ y = 2. Khi x, y l nghim ca phng trnh:

X2 2X + 1 = 0 X = 1 x = y = 1.

* Nu xy = - th x+ y = -1. Khi x, y l nghim ca phng trnh:

X2 + X - = 0 X =

V y > 0 nn: y = x =

Vy phng trnh c hai nghim: x1 = 1 ; x2 =

Cu 4: c/. Theo cu b, t gic ABCK l hnh thang. Do , t gic ABCK l hnh bnh hnh AB // CK

M s = s =

Nn

Dng tia Cy sao cho .Khi , D l giao im ca v Cy.

Vi gi thit > th > > .

D AB .

Vy im D xc nh nh trn l im cn tm.

11

Cu 1: a) Xc nh x R biu thc :A = L mt s t nhin

b. Cho biu thc: P = Bit x.y.z = 4 , tnh .

Cu 2:Cho cc im A(-2;0) ; B(0;4) ; C(1;1) ; D(-3;2)d. Chng minh 3 im A, B ,D thng hng; 3 im A, B, C khng thng hng.

e. Tnh din tch tam gic ABC.

Cu3 Gii phng trnh:

Cu 4 Cho ng trn (O;R) v mt im A sao cho OA = R. V cc tip tuyn AB, AC vi ng trn. Mt gc (xOy = 450 ct on thng AB v AC ln lt ti D v E.

Chng minh rng:

a.DE l tip tuyn ca ng trn ( O ).

b.

p n

Cu 1: a.

A =

A l s t nhin -2x l s t nhin x =

(trong k Z v k 0 )

b.iu kin xc nh: x,y,z 0, kt hp vi x.y.z = 4 ta c x, y, z > 0 v

Nhn c t v mu ca hng t th 2 vi ; thay 2 mu ca hng t th 3 bi ta c:

P =

(1)

EMBED Equation.3 v P > 0

Cu 2:a.ng thng i qua 2 im A v B c dng y = ax + b

im A(-2;0) v B(0;4) thuc ng thng AB nn b = 4; a = 2

Vy ng thng AB l y = 2x + 4.

im C(1;1) c to khng tho mn y = 2x + 4 nn C khng thuc ng thng AB A, B, C khng thng hng.

im D(-3;2) c to tho mn y = 2x + 4 nn im D thuc ng thng AB A,B,D thng hn

b.Ta c :

AB2 = (-2 0)2 + (0 4)2 =20

AC2 = (-2 1)2 + (0 1)2 =10

BC2 = (0 1)2 + (4 1)2 = 10

AB2 = AC2 + BC2 (ABC vung ti C

Vy S(ABC = 1/2AC.BC = ( n v din tch )

Cu 3:kx x1, t ta c h phng trnh:

Gii h phng trnh bng phng php th ta c: v = 2

x = 10.

Cu 4a.p dng nh l Pitago tnh c

AB = AC = R ABOC l hnh

vung (0.5)

K bn knh OM sao cho

(BOD = (MOD

(MOE = (EOC (0.5)

Chng minh (BOD = (MOD

(OMD = (OBD = 900Tng t: (OME = 900

D, M, E thng hng. Do DE l tip tuyn ca ng trn (O).

b.Xt (ADE c DE < AD +AE m DE = DB + EC

2ED < AD +AE +DB + EC hay 2DE < AB + AC = 2RDE < R

Ta c DE > AD; DE > AE ; DE = DB + EC

Cng tng v ta c: 3DE > 2R DE > R

Vy R > DE > R

12


a) Tnh f(-1); f(5)

b) Tm x f(x) = 10



Cu 3: Cho biu thc

A = vi x > 0 v x ( 1

a) Rt gn A

2) Tm gi tr ca x A = 3






p n

Cu 1

a)f(x) =

Suy ra f(-1) = 3; f(5) = 3

b)

c)



Cu 2

Cu 3a)Ta c: A =

=

=

=

= = =

b) A = 3 => = 3 => 3x + - 2 = 0 => x = 2/3

Cu 4

c) Do HA // PB (Cng vung gc vi BC)

d) nn theo nh l Ta let p dng cho tam gic CPB ta c

;

(1)


=>POB = ACB (hai gc ng v)

=>( AHC ( POB

Do :

(2)

Do CB = 2OB, kt hp (1) v (2) ta suy ra AH = 2EH hay E l trug im ca AH.






Cu 5 (1)

phng trnh c 2 nghim phn bit x1 ; x2 th ( > 0

(2m - 1)2 - 4. 2. (m - 1) > 0

T suy ra m ( 1,5

(1)


Gii phng trnh

ta c m = - 2 v m = 4,125

(2)

i chiu iu kin (1) v (2) ta c: Vi m = - 2 hoc m = 4,125 th phng trnh cho c hai nghim phn bit t

13

Cu I : Tnh gi tr ca biu thc:A = + ++ .....+

B = 35 + 335 + 3335 + ..... +

Cu II :Phn tch thnh nhn t :4) X2 -7X -18

5) (x+1) (x+2)(x+3)(x+4)

6) 1+ a5 + a10Cu III :

3) Chng minh : (ab+cd)2 (a2+c2)( b2 +d2)

4) p dng : cho x+4y = 5 . Tm GTNN ca biu thc : M= 4x2 + 4y2

Cu 4 : Cho tam gic ABC ni tip ng trn (O), I l trung im ca BC, M l mt im trn on CI ( M khc C v I ). ng thng AM ct (O) ti D, tip tuyn ca ng trn ngoi tip tam gic AIM ti M ct BD v DC ti P v Q.c) Chng minh DM.AI= MP.IB

d) Tnh t s :

Cu 5:

Cho P =

Tm iu kin biu thc c ngha, rt gn biu thc.

p n

Cu 1 :

1) A = + ++ .....+

= (+ + + .....+ ) = ()

2) B = 35 + 335 + 3335 + ..... + =

=33 +2 +333+2 +3333+2+.......+ 333....33+2

= 2.99 + ( 33+333+3333+...+333...33)

= 198 + ( 99+999+9999+.....+999...99)

198 + ( 102 -1 +103 - 1+104 - 1+ ....+10100 1) = 198 33 +

B = +165

Cu 2: 1)x2 -7x -18 = x2 -4 7x-14 = (x-2)(x+2) - 7(x+2) = (x+2)(x-9) (1)2)(x+1)(x+2)(x+3)(x+4) -3= (x+1)(x+4)(x+2)(x+3)-3

= (x2+5x +4)(x2 + 5x+6)-3= [x2+5x +4][(x2 + 5x+4)+2]-3

= (x2+5x +4)2 + 2(x2+5x +4)-3=(x2+5x +4)2 - 1+ 2(x2+5x +4)-2

= [(x2+5x +4)-1][(x2+5x +4)+1] +2[(x2+5x +4)-1]

= (x2+5x +3)(x2+5x +7)

3) a10+a5+1

= a10+a9+a8+a7+a6 + a5 +a5+a4+a3+a2+a +1

- (a9+a8+a7 )- (a6 + a5 +a4)- ( a3+a2+a )

= a8(a2 +a+1) +a5(a2 +a+1)+ a3(a2 +a+1)+ (a2 +a+1)-a7(a2 +a+1)

-a4(a2 +a+1)-a(a2 +a+1)

=(a2 +a+1)( a8-a7+ a5 -a4+a3 - a +1)

Cu 3: 4

1) Ta c : (ab+cd)2 (a2+c2)( b2 +d2)

a2b2+2abcd+c2d2 a2b2+ a2d2 +c2b2 +c2d2

0 a2d2 - 2cbcd+c2b2

0 (ad - bc)2 (pcm )

Du = xy ra khi ad=bc.

2) p dng hng ng thc trn ta c :

52 = (x+4y)2 = (x. + 4y) (x2 + y2)=>

x2 + y2 => 4x2 + 4y2 du = xy ra khi x= , y = (2)

Cu 4 : 5

Ta c : gc DMP= gc AMQ = gc AIC. Mt khc gc ADB = gc BCA=>

MPD ng dng vi ICA => => DM.IA=MP.CI hay DM.IA=MP.IB (1).

Ta c gc ADC = gc CBA,

Gc DMQ = 1800 - AMQ=1800 - gc AIM = gc BIA.

Do DMQ ng dng vi BIA =>

=> DM.IA=MQ.IB (2)

T (1) v (2) ta suy ra = 1

Cu 5

P xc nh th : x2-4x+3 0 v 1-x >0

T 1-x > 0 => x < 1

Mt khc : x2-4x+3 = (x-1)(x-3), V x < 1 nn ta c :

(x-1) < 0 v (x-3) < 0 t suy ra tch ca (x-1)(x-3) > 0

Vy vi x < 1 th biu thc c ngha.

Vi x < 1 Ta c :

P = =

14

Cu 1 : a. Rt gn biu thc . Vi a > 0.

b. Tnh gi tr ca tng.

Cu 2 : Cho pt

a. Chng minh rng pt lun lun c nghim vi .

b. Gi l hai nghim ca pt. Tm GTLN, GTNN ca bt.

Cu 3 : Cho Chng minh.

Cu 4 Cho ng trn tm o v dy AB. M l im chuyn ng trn ng trn, tM k MH ( AB (H ( AB). Gi E v F ln lt l hnh chiu vung gc ca H trn MA v MB. Qua M k ng thng vung gc vi ct dy AB ti D.

1. Chng minh rng ng thng MD lun i qua 1 im c nh khi M thay i trn ng trn.

2. Chng minh.

Hng dn

Cu 1 a. Bnh phng 2 v

(V a > 0).

f. p dng cu a.

Cu 2 a. : cm

B (2 ) p dng h thc Viet ta c:

(1) Tm k pt (1) c nghim theo n.

Cu 3 : Chuyn v quy ng ta c.

bt

ng v

Cu 4: a

- K thm ng ph.

- Chng minh MD l ng knh ca (o)

=> ........

b.

Gi E', F' ln lt l hnh chiu ca D trn MA v MB.

t HE = H1

HF = H2

Thay vo (1) ta c:

15

Cu 1: Cho biu thc D = :

a) Tm iu kin xc nh ca D v rt gn D

b) Tnh gi tr ca D vi a =

c) Tm gi tr ln nht ca D

Cu 2: Cho phng trnh x2- mx + m2 + 4m - 1 = 0 (1)

a) Gii phng trnh (1) vi m = -1

b) Tm m phng trnh (1) c 2 nghim tho mn

Cu 3: Cho tam gic ABC ng phn gic AI, bit AB = c, AC = b, Chng minh rng AI = (Cho Sin2)

Cu 4: Cho ng trn (O) ng knh AB v mt im N di ng trn mt na ng trn sao cho V vo trong ng trn hnh vung ANMP.

a) Chng minh rng ng thng NP lun i qua im c nh Q.

b) Gi I l tm ng trn ni tip tam gic NAB. Chng minh t gic ABMI ni tip.

c) Chng minh ng thng MP lun i qua mt im c nh.

Cu 5: Cho x,y,z; xy + yz + zx = 0 v x + y + z = -1

Hy tnh gi tr ca:

B =

p n

Cu 1: a) - iu kin xc nh ca D l

EMBED Equation.3 - Rt gn D

D = :

D =

b) a =

Vy D =

c) p dng bt ng thc cauchy ta c

Vy gi tr ca D l 1

Cu 2: a) m = -1 phng trnh (1)

b) phng trnh 1 c 2 nghim th (*)+ phng trnh c nghim khc 0 (*)+

Kt hp vi iu kin (*)v (**) ta c m = 0 v

Cu 3:

+

+

+

Cu 4: a) Gi Q = NP

Suy ra Q c nh

b)

T gic ABMI ni tip

c) Trn tia i ca QB ly im F sao cho QF = QB, F c nh.

Tam gic ABF c: AQ = QB = QF

ABF vung ti A

Li c T gic APQF ni tip

Ta c:

M1,P,F Thng hng

Cu 5: Bin i B = xyz =

16

Bi 1: Cho biu thc A =

a) Tm iu kin ca x A xc nh

b) Rt gn A

Bi 2 : Trn cng mt mt phng ta cho hai im A(5; 2) v B(3; -4)

a) Vit phng tnh ng thng AB

b) Xc nh im M trn trc honh tam gic MAB cn ti M

Bi 3 : Tm tt c cc s t nhin m phng trnh n x sau:

x2 - m2x + m + 1 = 0

c nghim nguyn.

Bi 4 : Cho tam gic ABC. Phn gic AD (D ( BC) v ng trn tm O qua A v D ng thi tip xc vi BC ti D. ng trn ny ct AB v AC ln lt ti E v F. Chng minh

a) EF // BC

b) Cc tam gic AED v ADC; D v ABD l cc tam gic ng dng.

c) AE.AC = .AB = AC2Bi 5 : Cho cc s dng x, y tha mn iu kin x2 + y2 ( x3 + y4. Chng minh:

x3 + y3 ( x2 + y2 ( x + y ( 2

p n

Bi 1:

a) iu kin x tha mn

(

( x > 1 v x ( 2

KL: A xc nh khi 1 < x < 2 hoc x > 2

b) Rt gn A

A =

A =

Vi 1 < x < 2 A =

Vi x > 2 A =

Kt lun

Vi 1 < x < 2 th A =

Vi x > 2 th A =

Bi 2:

a) A v B c honh v tung u khc nhau nn phng trnh ng thng AB c dng y = ax + b

A(5; 2) ( AB ( 5a + b = 2

B(3; -4) ( AB ( 3a + b = -4

Gii h ta c a = 3; b = -13

Vy phng trnh ng thng AB l y = 3x - 13

b) Gi s M (x, 0) ( xx ta c

MA =

MB =

MAB cn ( MA = MB (

( (x - 5)2 + 4 = (x - 3)2 + 16

( x = 1

Kt lun: im cn tm: M(1; 0)

Bi 3:

Phng trnh c nghim nguyn khi = m4 - 4m - 4 l s chnh phng

Ta li c: m = 0; 1 th < 0 loi

m = 2 th = 4 = 22 nhn

m ( 3 th 2m(m - 2) > 5 ( 2m2 - 4m - 5 > 0

( - (2m2 - 2m - 5) < < + 4m + 4

( m4 - 2m + 1 < < m4( (m2 - 1)2 < < (m2)2 khng chnh phng

Vy m = 2 l gi tr cn tm.

Bi 4:

a) (0,25)

(0,25)

m (0,25)

( EF // BC (2 gc so le trong bng nhau)

b) AD l phn gic gc BAC nn

ss() = s = s

do v

( DADC (g.g)

Tng t: s = (

do AFD ~ (g.g

c) Theo trn:

+ AED ~ DB

( hay AD2 = AE.AC (1)

+ ADF ~ ABD (

( AD2 = AB.AF (2)

T (1) v (2) ta c AD2 = AE.AC = AB.AF

Bi 5 (1):

Ta c (y2 - y) + 2 ( 0 ( 2y3 ( y4 + y2( (x3 + y2) + (x2 + y3) ( (x2 + y2) + (y4 + x3)

m x3 + y4 ( x2 + y3 do

x3 + y3 ( x2 + y2 (1)

+ Ta c: x(x - 1)2 ( 0: y(y + 1)(y - 1)2 ( 0

( x(x - 1)2 + y(y + 1)(y - 1)2 ( 0

( x3 - 2x2 + x + y4 - y3 - y2 + y ( 0

( (x2 + y2) + (x2 + y3) ( (x + y) + (x3 + y4)

m x2 + y3 ( x3 + y4( x2 + y2 ( x + y (2)

v (x + 1)(x - 1) ( 0.(y - 1)(y3 -1) ( 0

x3 - x2 - x + 1 + y4 - y - y3 + 1 ( 0

( (x + y) + (x2 + y3) ( 2 + (x3 + y4)

m x2 + y3 ( x3 + y4( x + y ( 2

T (1) (2) v (3) ta c:

x3 + y3 ( x2 + y2 ( x + y ( 2

14

Cu 1: x- 4(x-1) + x + 4(x-1) 1

cho A= ( 1 - )

x2- 4(x-1) x-1

a/ rt gn biu thc A.

b/ Tm gi tr nguyn ca x A c gi tr nguyn.

Cu 2: Xc nh cc gi tr ca tham s m phng trnh

x2-(m+5)x-m+6 =0

C 2 nghim x1 v x2 tho mn mt trong 2 iu kin sau:

a/ Nghim ny ln hn nghim kia mt n v.

b/ 2x1+3x2=13

Cu 3Tm gi tr ca m h phng trnh

mx-y=1

m3x+(m2-1)y =2

v nghim, v s nghim.

Cu 4: tm max v min ca biu thc: x2+3x+1

x2+1

Cu 5: T mt nh A ca hnh vung ABCD k hai tia to vi nhau mt gc 450. Mt tia ct cnh BC ti E ct ng cho BD ti P. Tia kia ct cnh CD ti F v ct ng cho BD ti Q.

a/ Chng minh rng 5 im E, P, Q, F v C cng nm trn mt ng trn.

b/ Chng minh rng: SAEF=2SAQPc/ K trung trc ca cnh CD ct AE ti M tnh s o gc MAB bit CPD=CM

hng dn Cu 1: a/ Biu thc A xc nh khi x2 v x>1

( x-1 -1)2+ ( x-1 +1)2 x-2

A= . ( )

(x-2)2 x-1

x- 1 -1 + x-1 + 1 x- 2 2 x- 1 2

= . = =

x-2 x-1 x-1 x-1

b/ A nguyn th x- 1 l c dng ca 1 v 2

* x- 1 =1 th x=0 loi

* x- 1 =2 th x=5

vy vi x = 5 th A nhn gi tr nguyn bng 1

Cu 2: Ta c x = (m+5)2-4(-m+6) = m2+14m+10 phng trnhc hai nghimphn bit khi vch khi m-7-4 3 v m-7+4 3 (*)

a/ Gi s x2>x1 ta c h x2-x1=1 (1) x1+x2=m+5 (2) x1x2 =-m+6 (3)

Gii h tac m=0 v m=-14 tho mn (*)

b/ Theo gi thit ta c: 2x1+3x2 =13(1) x1+x2 = m+5(2) x1x2 =-m+6 (3)

gii h ta c m=0 v m= 1 Tho mn (*)

Cu 3: * h v nghim th m/m3=-1/(m2-1) 1/2

3m3-m=-m3 m2(4m2- 1)=0 m=0 m=0

3m2-1-2 3m2-1 m=1/2 m=1/2

m

*Hv s nghim th: m/m3=-1/(m2-1) =1/2

3m3-m=-m3 m=0

3m2-1= -2 m=1/2

V nghim

Khng c gi tr no ca m h v s nghim.

Cu 4: Hm s xc nh vi x(v x2+10) x2+3x+1

gi y0 l 1 gi trca hmphng trnh: y0=

x2+1

(y0-1)x2-6x+y0-1 =0 c nghim

*y0=1 suy ra x = 0 y0 1; =9-(y0-1)20 (y0-1)2 9 suy ra -2 y0 4

Vy: ymin=-2 v y max=4

Cu 5: ( Hc sinh t v hnh)Gii

a/ A1 v B1 cng nhn on QE di mt gc 450

( t gic ABEQ ni tip c.

( FQE = ABE =1v.

chng minh tng t ta c FBE = 1v

( Q, P, C cng nm trn ng trn ng kinh EF.

b/ T cu a suy ra AQE vung cn.

( = (1)tng t APF cng vung cn

( = (2)t (1) v (2) ( AQP ~ AEF (c.g.c)

= ( )2 hay SAEF = 2SAQPc/ thy CPMD ni tip, MC=MD v APD=CPD

(MCD= MPD=APD=CPD=CMD

(MD=CD ( MCD u ( MPD=600

m MPD l gc ngoi ca ABM ta c APB=450 vy MAB=600-450=150 17

Bi 1: Cho biu thc M =

d. Tm iu kin ca x M c ngha v rt gn M

e. Tm x M = 5

f. Tm x Z M Z.

bi 2: a) Tm x, y nguyn dng tho mn phng trnh

3x2 +10 xy + 8y2 =96

b)tm x, y bit / x - 2005/ + /x - 2006/ +/y - 2007/+/x- 2008/ = 3

Bi 3: a. Cho cc s x, y, z dng tho mn + + = 4

Chng ming rng: + +

b. Tm gi tr nh nht ca biu thc: B = (vi x )

Bi 4: Cho hnh vung ABCD. K tia Ax, Ay sao cho = 45

Tia Ax ct CB v BD ln lt ti E v P, tia Ay ct CD v BD ln lt ti F v Q

c. Chng minh 5 im E; P; Q; F; C cng nm trn mt ng trn

d. S

EMBED Equation.3 = 2 S

K ng trung trc ca CD ct AE ti M. Tnh s o gc MAB bit =

Bi 5: (1)

Cho ba s a, b , c khc 0 tho mn: ; Hy tnh P =

p n Bi 1:M =

a.K 0,5

Rt gn M =

Bin i ta c kt qu: M = M =

c. M =

Do M nn l c ca 4 nhn cc gi tr: -4; -2; -1; 1; 2; 4

do

EMBED Equation.3 Bi 2 a. 3x2 + 10xy + 8y2 = 96

3x2 + 4xy + 6xy + 8y2 = 96

(3x2 + 6xy) + (4xy + 8y2) = 96

3x(x + 2y) + 4y(x +2y) = 96

(x + 2y)(3x + 4y) = 96

Do x, y nguyn dng nn x + 2y; 3x + 4y nguyen dng v 3x + 4y > x + 2y

m 96 = 25. 3 c cc c l: 1; 2; 3; 4; 6; 8; 12; 24; 32; 48; 96 c biu din thnh tch 2 tha s khng nh hn 3 l: 96 = 3.32 = 4.24 = 6. 16 = 8. 12

Li c x + 2y v 3x + 4y c tch l 96 (L s chn) c tng 4x + 6y l s chn do

H PT ny v nghim

Hoc

EMBED Equation.3 Hoc H PT v nghim

Vy cp s x, y nguyn dng cn tm l (x, y) = (4, 1)

b. ta c /A/ = /-A/

Nn /x - 2005/ + / x - 2006/ = / x - 2005/ + / 2008 - x/ (1)

m /x - 2005/ + / x - 2006/ + / y - 2007/ + / x - 2008/ = 3 (2)

Kt hp (1 v (2) ta c / x - 2006/ + / y - 2007/ (3)

(3) sy ra khi v ch khi

Bi 3

d. Trc ht ta chng minh bt ng thc ph

e. Vi mi a, b thuc R: x, y > 0 ta c

(a2y + b2x)(x + y)

a2y2 + a2xy + b2 x2 + b2xy a2xy + 2abxy + b2xy

a2y2 + b2x2 2abxy

a2y2 2abxy + b2x2 0

(ay - bx)2 0 (**) bt ng thc (**) ng vi mi a, b, v x,y > 0

Du (=) xy ra khi ay = bx hay

p dung bt ng thc (*) hai ln ta c

Tng t

Cng tng v cc bt ng thc trn ta c:

V

Ta c:

V (x - 2006)2 0 vi mi x

x2 > 0 vi mi x khc 0

Bi 4a. ni tip; = 900 ( gc AQE = 900 ( gcEQF = 900Tng t gc FDP = gc FAP = 450( T gic FDAP ni tip gc D = 900 ( gc APF = 900 ( gc EPF = 900 . 0,25

Cc im Q, P,C lun nhn di 1gc900 nn 5 im E, P, Q, F, C cng nm trn 1 ng trn ng knh EF 0,25

b. Ta c gc APQ + gc QPE = 1800 (2 gc k b) gc APQ = gc AFE

Gc AFE + gc EPQ = 1800

(Tam gic APQ ng dng vi tam gic AEF (g.g)

(

f. gc CPD = gc CMD ( t gic MPCD ni tip ( gc MCD = gc CPD (cng chn cung MD)

Li c gc MPD = gc CPD (do BD l trung trc ca AC)

gc MCD = gc MDC (do M thuc trung trc ca DC)

( gc CPD = gcMDC = gc CMD = gcMCD ( tam gic MDC u ( gc CMD = 600( tam gic DMA cn ti D (v AD = DC = DM)

V gc ADM =gcADC gcMDC = 900 600 = 300( gc MAD = gc AMD (1800 - 300) : 2 = 750( gcMAB = 900 750 = 150Bi 5t x = 1/a; y =1/b; z = 1/c ( x + y + z = 0 (v 1/a = 1/b + 1/c = 0)

( x = -(y + z)

( x3 + y3 + z3 3 xyz = -(y + z)3 + y3 3xyz

(-( y3 + 3y2 z +3 y2z2 + z3) + y3 + z3 3xyz = - 3yz(y + z + x) = - 3yz .0 = 0

T x3 + y3 + z3 3xyz = 0 ( x3 + y3 + z3 = 3xyz

( 1/ a3 + 1/ b3 + 1/ c3 3 1/ a3 .1/ b3 .1/ c3 = 3/abc

Do P = ab/c2 + bc/a2 + ac/b2 = abc (1/a3 + 1/b3+ 1/c3) = abc.3/abc = 3

nu 1/a + 1/b + 1/c =o th P = ab/c2 + bc/a2 + ac/b2 = 3

19

Bi 1Cho biu thc A = +

a. Rt gn biu thc A

b. Tm nhng gi tr nguyn ca x sao cho biu thc A cng c gi tr nguyn.

Bi 2: (2 im)

Cho cc ng thng:

y = x-2 (d1)

y = 2x 4 (d2)

y = mx + (m+2) (d3)

a. Tm im c nh m ng thng (d3 ) lun i qua vi mi gi tr ca m.

b. Tm m ba ng thng (d1); (d2); (d3) ng quy .

Bi 3: Cho phng trnh x2 - 2(m-1)x + m - 3 = 0 (1)

a. Chng minh phng trnh lun c 2 nghim phn bit.

b. Tm mt h thc lin h gia hai nghim ca phng trnh (1) m khng ph thuc vo m.

c. Tm gi tr nh nht ca P = x21 + x22 (vi x1, x2 l nghim ca phng trnh (1))

Bi 4: Cho ng trn (o) vi dy BC c nh v mt im A thay i v tr trn cung ln BC sao cho AC>AB v AC > BC . Gi D l im chnh gia ca cung nh BC. Cc tip tuyn ca (O) ti D v C ct nhau ti E. Gi P, Q ln lt l giao im ca cc cp ng thng AB vi CD; AD v CE.

a. Chng minh rng DE// BC

b. Chng minh t gic PACQ ni tip

c. Gi giao im ca cc dy AD v BC l F

Chng minh h thc: = +

Bi 5: Cho cc s dng a, b, c Chng minh rng:

p n

Bi 1: - iu kin : x 0

a. Rt gn:

- Vi x x =

Bi 2:

a. (d1) : y = mx + (m +2)

m (x+1)+ (2-y) = 0

hm s lun qua im c nh vi mi m

=.>

Vy N(-1; 2) l im c nh m (d3) i qua

b. Gi M l giao im (d1) v (d2) . Ta M l nghim ca h

=>

Vy M (2; 0) .

Nu (d3) i qua M(2,0) th M(2,0) l nghim (d3)

Ta c : 0 = 2m + (m+2) => m= -

Vy m = - th (d1); (d2); (d3) ng quy

Bi 3: a. = m2 3m + 4 = (m - )2 + >0 m.

Vy phng trnh c 2 nghim phn bit

b. Theo Vit: =>

x1+ x2 2x1x2 4 = 0 khng ph thuc vo m

b. P = x12 + x12 = (x1 + x2)2 - 2x1x2 = 4(m - 1)2 2 (m-3)

= (2m - )2 +

VyPmin = vi m =

Bi 4: V hnh ng vit gi thit kt lun

a. SCDE = S DC = S BD =

=> DE// BC (2 gc v tr so le)

b. APC = s (AC - DC) = AQC

=> APQC ni tip (v APC = AQC

cng nhn oan AC)

c.T gic APQC ni tip

CPQ = CAQ (cng chn cung CQ)

CAQ = CDE (cng chn cung DC)

Suy ra CPQ = CDE => DE// PQ

Ta c: = (v DE//PQ) (1)

= (v DE// BC) (2)

Cng (1) v (2) :

=> (3)

ED = EC (t/c tip tuyn) t (1) suy ra PQ = CQ

Thay vo (3) :

Bi 5:Ta c: < < (1)

< 0 .

2) Tm gi tr ca m biu thc P = t gi tr nh nht .

Cu 3 ( 2 im )

Gii phng trnh :

a)

b)

Cu 4 ( 3 im )

Cho hai ng trn (O1) v (O2) c bn knh bng R ct nhau ti A v B , qua A v ct tuyn ct hai ng trn (O1) v (O2) th t ti E v F , ng thng EC , DF ct nhau ti P .

1) Chng minh rng : BE = BF .

2) Mt ct tuyn qua A v vung gc vi AB ct (O1) v (O2) ln lt ti C,D . Chng minh t gic BEPF , BCPD ni tip v BP vung gc vi EF .

3) Tnh din tch phn giao nhau ca hai ng trn khi AB = R .

s 3

Cu 1 ( 3 im )

1) Gii bt phng trnh :

2) Tm gi tr nguyn ln nht ca x tho mn .

Cu 2 ( 2 im )

Cho phng trnh : 2x2 ( m+ 1 )x +m 1 = 0

a) Gii phng trnh khi m = 1 .

b) Tm cc gi tr ca m hiu hai nghim bng tch ca chng .

Cu3 ( 2 im )

Cho hm s : y = ( 2m + 1 )x m + 3

(1)

a) Tm m bit th hm s (1) i qua im A ( -2 ; 3 ) .

b) Tm im c nh m th hm s lun i qua vi mi gi tr ca m .

Cu 4 ( 3 im )

Cho gc vung xOy , trn Ox , Oy ln lt ly hai im A v B sao cho OA = OB . M l mt im bt k trn AB .

Dng ng trn tm O1 i qua M v tip xc vi Ox ti A , ng trn tm O2 i qua M v tip xc vi Oy ti B , (O1) ct (O2) ti im th hai N .

1) Chng minh t gic OANB l t gic ni tip v ON l phn gic ca gc ANB .

2) Chng minh M nm trn mt cung trn c nh khi M thay i .

3) Xc nh v tr ca M khong cch O1O2 l ngn nht .

s 4 .

Cu 1 ( 3 im )

Cho biu thc :

a) Rt gn biu thc .

b) Tnh gi tr ca khi

Cu 2 ( 2 im )

Gii phng trnh :

Cu 3 ( 2 im )

Cho hm s : y = -

a) Tm x bit f(x) = - 8 ; - ; 0 ; 2 .

b) Vit phng trnh ng thng i qua hai im A v B nm trn th c honh ln lt l -2 v 1 .

Cu 4 ( 3 im )

Cho hnh vung ABCD , trn cnh BC ly 1 im M . ng trn ng knh AM ct ng trn ng knh BC ti N v ct cnh AD ti E .

1) Chng minh E, N , C thng hng .

2) Gi F l giao im ca BN v DC . Chng minh

3) Chng minh rng MF vung gc vi AC .

s 5

Cu 1 ( 3 im )

Cho h phng trnh :

a) Gii h phng trnh khi m = 1 .

b) Gii v bin lun h phng trnh theo tham s m .

c) Tm m x y = 2 .

Cu 2 ( 3 im )

1) Gii h phng trnh :

2) Cho phng trnh bc hai : ax2 + bx + c = 0 . Gi hai nghim ca phng trnh l x1 , x2 . Lp phng trnh bc hai c hai nghim l 2x1+ 3x2 v 3x1 + 2x2 .

Cu 3 ( 2 im )

Cho tam gic cn ABC ( AB = AC ) ni tip ng trn tm O . M l mt im chuyn ng trn ng trn . T B h ng thng vung gc vi AM ct CM D .

Chng minh tam gic BMD cn

Cu 4 ( 2 im )

1) Tnh :

2) Gii bt phng trnh :

( x 1 ) ( 2x + 3 ) > 2x( x + 3 ) .

s 6

Cu 1 ( 2 im )

Gii h phng trnh :

Cu 2 ( 3 im )

Cho biu thc :

a) Rt gn biu thc A .

b) Coi A l hm s ca bin x v thi hm s A .

Cu 3 ( 2 im )

Tm iu kin ca tham s m hai phng trnh sau c nghim chung .

x2 + (3m + 2 )x 4 = 0 v x2 + (2m + 3 )x +2 =0 .

Cu 4 ( 3 im )

Cho ng trn tm O v ng thng d ct (O) ti hai im A,B . T mt im M trn d v hai tip tuyn ME , MF ( E , F l tip im ) .

1) Chng minh gc EMO = gc OFE v ng trn i qua 3 im M, E, F i qua 2 im c nh khi m thay i trn d .

2) Xc nh v tr ca M trn d t gic OEMF l hnh vung .

s 7

Cu 1 ( 2 im )

Cho phng trnh (m2 + m + 1 )x2 - ( m2 + 8m + 3 )x 1 = 0

a) Chng minh x1x2 < 0 .

b) Gi hai nghim ca phng trnh l x1, x2 . Tm gi tr ln nht , nh nht ca biu thc :

S = x1 + x2 .

Cu 2 ( 2 im )

Cho phng trnh : 3x2 + 7x + 4 = 0 . Gi hai nghim ca phng trnh l x1 , x2 khng gii phng trnh lp phng trnh bc hai m c hai nghim l : v .

Cu 3 ( 3 im )

1) Cho x2 + y2 = 4 . Tm gi tr ln nht , nh nht ca x + y .

2) Gii h phng trnh :

3) Gii phng trnh : x4 10x3 2(m 11 )x2 + 2 ( 5m +6)x +2m = 0

Cu 4 ( 3 im )

Cho tam gic nhn ABC ni tip ng trn tm O . ng phn gic trong ca gc A , B ct ng trn tm O ti D v E , gi giao im hai ng phn gic l I , ng thng DE ct CA, CB ln lt ti M , N .

1) Chng minh tam gic AIE v tam gic BID l tam gic cn .

2) Chng minh t gic AEMI l t gic ni tip v MI // BC .

3) T gic CMIN l hnh g ?

s 8

Cu1 ( 2 im )

Tm m phng trnh ( x2 + x + m) ( x2 + mx + 1 ) = 0 c 4 nghim phn bit .

Cu 2 ( 3 im )

Cho h phng trnh :

a) Gii h khi m = 3

b) Tm m phng trnh c nghim x > 1 , y > 0 .

Cu 3 ( 1 im )

Cho x , y l hai s dng tho mn x5+y5 = x3 + y3 . Chng minh x2 + y2 1 + xy

Cu 4 ( 3 im )

1) Cho t gic ABCD ni tip ng trn (O) . Chng minh

AB.CD + BC.AD = AC.BD

2) Cho tam gic nhn ABC ni tip trong ng trn (O) ng knh AD . ng cao ca tam gic k t nh A ct cnh BC ti K v ct ng trn (O) ti E .

a) Chng minh : DE//BC .

b) Chng minh : AB.AC = AK.AD .

c) Gi H l trc tm ca tam gic ABC . Chng minh t gic BHCD l hnh bnh hnh .

s 9

Cu 1 ( 2 im )

Trc cn thc mu cc biu thc sau :

;

;

Cu 2 ( 3 im )

Cho phng trnh : x2 ( m+2)x + m2 1 = 0

(1)

a) Gi x1, x2 l hai nghim ca phng trnh .Tm m tho mn x1 x2 = 2 .

b) Tm gi tr nguyn nh nht ca m phng trnh c hai nghim khc nhau .

Cu 3 ( 2 im )

Cho

Lp mt phng trnh bc hai c cc h s bng s v c cc nghim l x1 =

Cu 4 ( 3 im )

Cho hai ng trn (O1) v (O2) ct nhau ti A v B . Mt ng thng i qua A ct ng trn (O1) , (O2) ln lt ti C,D , gi I , J l trung im ca AC v AD .

1) Chng minh t gic O1IJO2 l hnh thang vung .

2) Gi M l giao dim ca CO1 v DO2 . Chng minh O1 , O2 , M , B nm trn mt ng trn

3) E l trung im ca IJ , ng thng CD quay quanh A . Tm tp hp im E.

4) Xc nh v tr ca dy CD dy CD c di ln nht .

s 10

Cu 1 ( 3 im )

1)V th ca hm s : y =

2)Vit phng trnh ng thng i qua im (2; -2) v (1 ; -4 )

3) Tm giao im ca ng thng va tm c vi th trn .

Cu 2 ( 3 im )

a) Gii phng trnh :

b)Tnh gi tr ca biu thc

vi

Cu 3 ( 3 im )

Cho tam gic ABC , gc B v gc C nhn . Cc ng trn ng knh AB , AC ct nhau ti D . Mt ng thng qua A ct ng trn ng knh AB , AC ln lt ti E v F .

1) Chng minh B , C , D thng hng .

2) Chng minh B, C , E , F nm trn mt ng trn .

3) Xc nh v tr ca ng thng qua A EF c di ln nht .

Cu 4 ( 1 im )

Cho F(x) =

a) Tm cc gi tr ca x F(x) xc nh .

b) Tm x F(x) t gi tr ln nht .

s 11

Cu 1 ( 3 im )

1) V th hm s

2) Vit phng trnh ng thng i qua hai im ( 2 ; -2 ) v ( 1 ; - 4 )

3) Tm giao im ca ng thng va tm c vi th trn .

Cu 2 ( 3 im )

1) Gii phng trnh :

2) Gii phng trnh :

Cu 3 ( 3 im )

Cho hnh bnh hnh ABCD , ng phn gic ca gc BAD ct DC v BC theo th t ti M v N . Gi O l tm ng trn ngoi tip tam gic MNC .

1) Chng minh cc tam gic DAM , ABN , MCN , l cc tam gic cn .

2) Chng minh B , C , D , O nm trn mt ng trn .

Cu 4 ( 1 im )

Cho x + y = 3 v y . Chng minh x2 + y2

s 12

Cu 1 ( 3 im )

1) Gii phng trnh :

2) Xc nh a tng bnh phng hai nghim ca phng trnh x2 +ax +a 2 = 0 l b nht .

Cu 2 ( 2 im )

Trong mt phng to cho im A ( 3 ; 0) v ng thng x 2y = - 2 .

a) V th ca ng thng . Gi giao im ca ng thng vi trc tung v trc honh l B v E .

b) Vit phng trnh ng thng qua A v vung gc vi ng thng x 2y = -2 .

c) Tm to giao im C ca hai ng thng . Chng minh rng EO. EA = EB . EC v tnh din tch ca t gic OACB .

Cu 3 ( 2 im )

Gi s x1 v x2 l hai nghim ca phng trnh :

x2 (m+1)x +m2 2m +2 = 0

(1)

a) Tm cc gi tr ca m phng trnh c nghim kp , hai nghim phn bit .

b) Tm m t gi tr b nht , ln nht .

Cu 4 ( 3 im )

Cho tam gic ABC ni tip ng trn tm O . K ng cao AH , gi trung im ca AB , BC theo th t l M , N v E , F theo th t l hnh chiu vung gc ca ca B , C trn ng knh AD .

a) Chng minh rng MN vung gc vi HE .

b) Chng minh N l tm ng trn ngoi tip tam gic HEF .

s 13

Cu 1 ( 2 im )

So snh hai s :

Cu 2 ( 2 im )

Cho h phng trnh :

Gi nghim ca h l ( x , y ) , tm gi tr ca a x2 + y2 t gi tr nh nht .

Cu 3 ( 2 im )

Gi h phng trnh :

Cu 4 ( 3 im )

1) Cho t gic li ABCD cc cp cnh i AB , CD ct nhau ti P v BC , AD ct nhau ti Q . Chng minh rng ng trn ngoi tip cc tam gic ABQ , BCP , DCQ , ADP ct nhau ti mt im .

3) Cho t gic ABCD l t gic ni tip . Chng minh

Cu 4 ( 1 im )

Cho hai s dng x , y c tng bng 1 . Tm gi tr nh nht ca :

s 14

Cu 1 ( 2 im )

Tnh gi tr ca biu thc :

Cu 2 ( 3 im )

1) Gii v bin lun phng trnh :

(m2 + m +1)x2 3m = ( m +2)x +3

2) Cho phng trnh x2 x 1 = 0 c hai nghim l x1 , x2 . Hy lp phng trnh bc hai c hai nghim l :

Cu 3 ( 2 im )

Tm cc gi tr nguyn ca x biu thc : l nguyn .

Cu 4 ( 3 im )

Cho ng trn tm O v ct tuyn CAB ( C ngoi ng trn ) . T im chnh gia ca cung ln AB k ng knh MN ct AB ti I , CM ct ng trn ti E , EN ct ng thng AB ti F .

1) Chng minh t gic MEFI l t gic ni tip .

2) Chng minh gc CAE bng gc MEB .

3) Chng minh : CE . CM = CF . CI = CA . CB

s 15

Cu 1 ( 2 im )

Gii h phng trnh :

Cu 2 ( 2 im )

Cho hm s : v y = - x 1

a) V th hai hm s trn cng mt h trc to .

b) Vit phng trnh cc ng thng song song vi ng thng y = - x 1 v ct th hm s ti im c tung l 4 .

Cu 2 ( 2 im )

Cho phng trnh : x2 4x + q = 0

a) Vi gi tr no ca q th phng trnh c nghim .

b) Tm q tng bnh phng cc nghim ca phng trnh l 16 .

Cu 3 ( 2 im )

1) Tm s nguyn nh nht x tho mn phng trnh :

2) Gii phng trnh :

Cu 4 ( 2 im )

Cho tam gic vung ABC ( gc A = 1 v ) c AC < AB , AH l ng cao k t nh A . Cc tip tuyn ti A v B vi ng trn tm O ngoi tip tam gic ABC ct nhau ti M . on MO ct cnh AB E , MC ct ng cao AH ti F . Ko di CA cho ct ng thng BM D . ng thng BF ct ng thng AM N .

a) Chng minh OM//CD v M l trung im ca on thng BD .

b) Chng minh EF // BC .

c) Chng minh HA l tia phn gic ca gc MHN .

s 16

Cu 1 : ( 2 im )

T

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