S gio dc v o to thi chuyn cp lp 10 tnh ngh an

S GIO DC V O TO THI CHUYN CP LP 10 TNH NGH AN

TNH NGH AN NM HC 2006 - 2007Bi 1(2)

Cho biu thc:

a) Tm iu kin v rt gn P

b) Tm x P>0

Bi 2(1,5)

Trong mt k thi tuyn sinh vo lp 10 hai trng THCS A v B c tt c 450 hc sinh d thi. Bit s hc sinh trng tuyn ca trng A bng 3/4 s hc sinh d thi ca trng A, s hc sinh trng tuyn ca trng B bng 9/10 s hc sinh d thi trng B. Tng s hc sinh trng tuyn ca hai trng bng 4/5 s hc sinh d thi ca hai trng. Tnh s hc sinh d thi ca mi trng.

Bi 3 (2,5)

Cho phng trnh: (1)

a) Gii phng trnh (1) khi m = 1

b) Tm m phng trnh (1) c hai nghim phn bit

c) Gi hai nghim ca phng trnh (1) l x1; x2. Hy xc nh m :

Bi 4 (4)

Cho na ng trn tm O ng knh AB = 2 R. M l mt im bt k trn na ng trn sao cho cung AM ln hn cung MB (M B). Qua M k tip tuyn d ca na ng trn ni trn. K AD; BC vung gc vi d trong D,C thuc ng thng d.

a) Chng minh M l trung im CD.

b) Chng minh AD.BC = CM2.

c) Chng minh ng trn ng knh CD tip xc vi ng thng AB.

d) K MH vung gc vi AB (H thuc AB) Hy xc nh v tr M din tch tam gic DHC bng 1/4 din tch tam gic AMB.

S GD&T NGH AN K THI TUYN SINH VO LP 10 THPT

NM HC 2007- 2008

Mn thi : Ton ; Thi gian 120 pht .

Phn I . Trc nghim ( 2im ) .

Em hy chn mt phng n tr li ng trong cc phng n ( A,B,C,D ) ca tng cu sau ri ghi phng n chn vo bi lm .

Cu 1 . th hm s y = 3x - 2 ct trc tung ti im c tung l

A. 2 B. -2 C. 3 D. .

Cu 2 . H phng trnh c nghim l

A. (2;1) B. (3;2) C. (0;1) D. (1;2) .

Cu 3 . Sin300 bng

B. C. D. .

Cu 4 . T gic MNPQ ni tip ng trn (O) . Bit gc MQP c s o l

A.1300 B.1200 C.1100 D.1000 .

Phn II . T lun ( 8im )

Cu 1 . Cho biu thc A = .

a ) Nu KX v rt gn A .

b ) Tm tt c cc gi tr ca x sao cho A < 0 .

c ) Tm tt c cc g tr ca tham s m phng trnh A c nghim

Cu 2 . Hai xe my khi hnh cng mt lc i t A n B . Xe my th nht c vn tc trung bnh ln hn vn tc trung bnh ca xe my th hai 10km/h , nn n trc xe my th hai 1 gi . Tnh vn tc trung bnh ca mi xe my bit rng qung ng AB di 120 km .

Cu 3 . Cho na ng trn tm O , ng knh AB . im H nm gia A v B ( H khng trng vi O ) . ng thng vung gc vi AB ti H , ct na ng trn trn ti im C . Gi D v E ln lt l chn cc ng vung gc k t H n AC v BC .

a ) T gic HDCE l hnh g ? V sao ?

b ) Chng minh ADEB l t gic ni tip .

c ) Gi K l tm ng trn ngoi tip t gic ADEB . Chng minh DE = 2KO .

.................................................................Ht .................................................................S GD&T NGH AN K THI TUYN SINH VO LP 10THPT

NM HC 2008 2009

Mn : Ton

Thi gian : 120 pht

I . PHN TRC NGHIM : (2,0 im)

Em hy chn mt phng n tr li ng trong cc phng n (A, B, C, D ) ca tng cu sau ri ghi phng n chn vo bi lm .

Cu 1 . (0,5 im)

th hm s y = -3x + 4 I qua im :

A. (0;4) B.(2;0) C(-5;3) D. (1;2)

Cu 2 . (0,5 im)

bng :

A. -7 B. -5C. 7

D. 5

Cu3 . (0,5 im)

Hnh trn bn knh 4cmth c din tch l :

A. 16 (cm2)B. 8 (cm2) C.4 (cm2) D.2 (cm2)

Cu4 . Tam gic ABC vung A , bit tgB = v AB = 4 . di cnh AC l :

A. 2

B. 3

C. 4

D. 6

II . PHN T LUN : (8 im)

Cu1. (3im)

Cho biu thc : P =

a. Nu iu kin xc nh v rt gn biu thc P .

b. Tm cc gi tr ca x P = .

c. Tm gi tr nh nht ca biu thc M = .

Cu 2 . (2,0 im)

Hai ngi th cng qut sn cho mt ngi nh trong 2 ngy th xong vic . Nu ngi th nht lm trong 4 ngy ri ngh v ngi th hai lm tip trong 1 ngy th xong vic . Hi mi ngi lm mt mnh th bao lu s xong vic ?

Cu 3 . (3,0 im)

Cho tam gic ABC vung A . ng trn ng knh AB ct BC ti M . Trn cung nh AM ly im E ( ) . Ko di BE ct AC ti F .

a. Chng minh , t suy ra MEFC l mt t gic ni tip .

b. Gi K l giao im ca ME v AC . Chng minh AK2 = KE.KM .

c. Khi im E v tr sao cho AE + BM = AB . chng minh rng giao im cc

ng phn gic ca v thuc on thng AB .

. Ht S GD & T NGH AN K THI TUYN SINH VO LP 10 THPT

NM HC 2009 - 2010

Mn thi : Ton . Thi gian : 120 pht

Cu 1 (3,0 im ) . Cho biu thc A = .

a) Nu iu kin xc nh v rt gn A .

b) Tnh gi tr ca biu thc A khi x = .

c) Tm tt c cc gi tr ca x A < 1 .

Cu 2 (2,5 im ) . Cho phng trnh bc hai vi tham s m : 2x2 + ( m + 3 )x + m = 0 (1).

a) Gii phng trnh khi m = 2 .

b) Tm cc gi tr ca tham s m phng trnh (1) c hai nghim x1 , x2 tho mn

=

c) Gi x1 , x2 l hai nghim ca phng trnh (1) . Tm gi tr nh nht ca biu thc

P = .

Cu3 ( 1,5 im ) . Mt tha rung hnh ch nht c chiu rng ngn hn chiu di 45m . Tnh din tch tha rung , bit rng nu chiu di gim 2 ln v chiu rng tng 3 ln th chu vi tha rung khng thay i .

Cu 4 ( 3,0 im ) . Cho ng trn ( O ; R ) , ng knh AB c nh v CD l mt ng knh thay i khng trng vi AB . Tip tuyn ca ng trn ( O ; R ) ti B ct ng thng AC v AD ln lt ti E v F .

a) Chng minh rng BE.BF = 4R2 .

b) Chng minh t gic CEFD ni tip c ng trn .

c) Gi I l tm ng trn ngoi tip t gic CEFD . Chng minh tm I lun nm trn mt ng thng c nh .

.......................................................... Ht..............................................................................

S GIO DC V O TO K THI TUYN SINH VO LP 10 THPT

NGH AN Nm hc 2010 - 2011

Mn thi : Ton

Thi gian: 120 pht Cu I (3,0 im). Cho biu thc A = .

1. Nu iu kin xc nh v rt gn biu thc A.

2. Tnh gi tr ca biu thc A khi x = 9.

3. Khi x tho mn iu kin xc nh. Hy tm gi tr nh nht cu biu thc B,

vi B = A(x-1).

Cu II (2,0 im). Cho phng trnh bc hai sau, vi tham s m:

x2 - (m + 1)x + 2m - 2 = 0 (1)

1. Gii phng trnh (1) khi m = 2.

2. Tm gi tr ca tham s m x = -2 l mt nghim ca phng trnh (1).

Cu III (1,5 im). Hai ngi cng lm chung mt cng vic th sau 4 gi 30 pht h lm xong cng vic. Nu mt mnh ngi th nht lm trong 4 gi, sau mt mnh ngi th hai lm trong 3 gi th c hai ngi lm c 75% cng vic.

Hi nu mi ngi lm mt mnh th sau bao lu s xong cng vic? (Bit rng nng sut lm vic ca mi ngi l khng thay i).

Cu IV (3,5 im). Cho na ng trn tm O ng knh AB. im H c nh thuc on thng AO (H khc A v O). ng thng i qua im H v vung gc vi AO ct na ng trn (O) ti C. Trn cung BC ly im D bt k (D khc B v C). Tip tuyn ca na ng trn (O) ti D ct ng thng HC ti E. Gi I l giao im ca AD v HC.

1. Chng minh t gic HBDI ni tip ng trn.

2. Chng minh tam gic DEI l tam gic cn.

3. Gi F l tm ng trn ngoi tip tam gic ICD. Chng minh gc ABF c

s o khng i khi D thay i trn cung BC (D khc B v C).

--------------Ht-------------

S GIO DC V O TO

NGH ANK THI TUYN SINH VO LP 1 THPT

NM HC 2011 2012

Cu 1: (3,0 im)

Cho biu thc A =

a) Nu iu kin xc nh v rt gn biu thc A

b) Tim gi tr ca x A = .

c) Tm gi tr ln nht cua biu thc P = A - 9

Cu 2: (2,0 im)

Cho phng trnh bc hai x2 2(m + 2)x + m2 + 7 = 0 (1) (m l tham s)

a) Gii phng trnh (1) khi m = 1.

b) Tm m phng trnh (1) c nghim x1, x2 tha mn Cu 3: (1,5 im)

Qung ng AB di 120 km. Hai xe my khi hnh cng mt lc i t A n B. Vn tc ca xe my th nht ln hn vn tc ca xe my th hai l 10 km/h nn xe my th nht n B trc xe my th hai 1 gi. Tnh vn tc ca mi xe ?

Cu 4: (3,5 im)

Cho im A nm ngoi ng trn (O). T A k hai tip tuyn AB, AC v ct tuyn ADE ti ng trn (B, C l hai tip im; D nm gia A v E). Gi H l giao im ca AO v BC.

a) Chng minh rng ABOC l t gic ni tip

b) Chng minh rng AH.AO = AD.AE

c) Tip tuyn ti D ca ng trn (O) ct AB, AC theo th t ti I v K. Qua im O k ng thng vung gc vi OA ct tia AB ti P v ct tia AC ti Q.

Chng minh rng IP + KQ PQ.

----- Ht ------

Cu 1: (3,0 im)

a). iu kin

Vi iu kin , ta c:

b). A = th (tha mn iu kin)

Vy th A =

c). Ta c P = A - 9 =

p dng bt ng thc C si cho hai s dng ta c:

Suy ra: . ng thc xy ra khi

Vy gi tr ln nht ca biu thc khi

Cu 2: (2,0 im)

a). Gii phng trnh (1) khi m = 1.

Khi m = 1 ta c phng trnh:

Vy phng trnh c hai nghim v

c) phng trnh (1) c nghim x1, x2 th

(*)

Theo nh l Vi t ta c:

Theo bi ra x1x2 2(x1 + x2) = 4 ta c:

EMBED Equation.DSMT4i chiu iu kin (*) ta c m = 5 l gi tr cn Tm.

Cu 3: (1,5 im)

Gi vn tc ca xe my th hai l

Vn tc ca xe my th nht l

thi gian xe my th hai i ht Q AB l: 120/x

thi gian xe my th nht i ht Q AB l :120/x+10

Theo bi ra ta c phng trnh:

EMBED Equation.DSMT4i chiu iu kin ta c x = 30.

Vy vn tc ca xe th nht l 40 (km/h) v vn tc ca xe th hai l 30 (km/h)

Cu 4:

a) V AB, AC l tip tuyn ca (O) nn

Suy ra

Vy t gic ABOC ni tip.

b) Ta c ABO vung ti B c ng cao BH, ta c :

AH.AO = AB2 (1)

Li c ABD AEB (g.g)

EMBED Equation.DSMT4 AB2 = AD.AE (2)

T (1), (2) suy ra:

AH.AO = AD.AE

c). Xt tam gic v

Ta c (V tam gic APQ cn ti A)

v

Ta c: (3)

Li c:

Suy ra (4)

T (3), (4) suy ra :

Do (g.g)

T suy ra IP.KQ = OP.OQ = hay PQ2 = 4.IP.KQ

Mt khc ta c: 4.IP.KQ (IP + KQ)2 (V )

Vy .

CHNH THC

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