Tong Hop 63 de Thi Tuyen Sinh Lop 10 Mon Toan Co Dap An

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S GIO DC V O TO

K THI TUYN SINH LP 10 THPT

H NINm hoc: 2012 2013

Mn thi: Ton

Ngy thi: 21 thng 6 nm 2012

Thi gian lm bi: 120 pht

Bi I (2,5 im)

1) Cho biu thc . Tnh gi tr ca A khi x = 36

2) Rt gn biu thc (vi )

3) Vi cc ca biu thc A v B ni trn, hy tm cc gi tr ca x nguyn gi tr ca biu thc B(A 1) l s nguyn

Bi II (2,0 im). Gii bi ton sau bng cch lp phng trnh hoc h phng trnh:

Hai ngi cng lm chung mt cng vic trong gi th xong. Nu mi ngi lm mt mnh th ngi th nht hon thnh cng vic trong t hn ngi th hai l 2 gi. Hi nu lm mt mnh th mi ngi phi lm trong bao nhiu thi gian xong cng vic?

Bi III (1,5 im)

1) Gii h phng trnh:

2) Cho phng trnh: x2 (4m 1)x + 3m2 2m = 0 (n x). Tm m phng trnh c hai nghim phn bit x1, x2 tha mn iu kin :

Bi IV (3,5 im)

Cho ng trn (O; R) c ng knh AB. Bn knh CO vung gc vi AB, M l mt im bt k trn cung nh AC (M khc A, C); BM ct AC ti H. Gi K l hnh chiu ca H trn AB.

1) Chng minh CBKH l t gic ni tip.

2) Chng minh

3) Trn an thng BM ly im E sao cho BE = AM. Chng minh tam gic ECM l tam gic vung cn ti C

4) Gi d l tip tuyn ca (O) ti im A; cho P l im nm trn d sao cho hai im P, C nm trong cng mt na mt phng b AB v . Chng minh ng thng PB i qua trung im ca on thng HK

Bi V (0,5 im). Vi x, y l cc s dng tha mn iu kin , tm gi tr nh nht ca biu thc:

GI P N

Bai I: (2,5 im)

1) Vi x = 36, ta c : A =

EMBED Equation.DSMT4 2) Vi x , x ( 16 ta c:

B = =

3) Ta c: .

nguyn, x nguyn th l c ca 2, m (2) =

Ta c bng gi tr tng ng:

1

2

x17151814

Kt hp K , nguyn th

Bai II: (2,0 im)

Gi thi gian ngi th nht hon thnh mt mnh xong cng vic l x (gi), K

Th thi gian ngi th hai lm mt mnh xong cng vic l x + 2 (gi)

Mi gi ngi th nht lm c(cv), ngi th hai lm c(cv)

V c hai ngi cng lm xong cng vic trong gi nn mi gi c hai i lm c=(cv)

Do ta c phng trnh

( 5x2 14x 24 = 0

( = 49 + 120 = 169,

=> (loi) v (TMK)

Vy ngi th nht lm xong cng vic trong 4 gi,

ngi th hai lm xong cng vic trong 4+2 = 6 gi.

Bai III: (1,5 im) 1)Gii h: , (K: ).

H .(TMK)

Vy h c nghim (x;y)=(2;1).2)+ Phng trnh cho c ( = (4m 1)2 12m2 + 8m = 4m2 + 1 > 0, (m

Vy phng trnh c 2 nghim phn bit (m

+ Theo L Vi t, ta c: .

Khi :

( (4m 1)2 2(3m2 2m) = 7 ( 10m2 4m 6 = 0 ( 5m2 2m 3 = 0

Ta thy tng cc h s: a + b + c = 0 => m = 1 hay m = .

Tr li: Vy....

Bai IV: (3,5 im)

1) Ta c ( do chn na ng trn k AB)

(do K l hnh chiu ca H trn AB)

=> nn t gic CBKH ni tip trong ng trn ng knh HB.

2) Ta c (do cng chn ca (O))

v (v cng chn .ca trn k HB)

Vy

3) V OC ( AB nn C l im chnh gia ca cung AB ( AC = BC v

Xt 2 tam gic MAC v EBC c

MA= EB(gt), AC = CB(cmt) v = v cng chn cung ca (O)

(MAC v EBC (cgc) ( CM = CE ( tam gic MCE cn ti C (1)

Ta li c (v chn cung )

. ((tnh cht tam gic MCE cn ti C)

M (Tnh cht tng ba gc trong tam gic)( (2)

T (1), (2) (tam gic MCE l tam gic vung cn ti C (pcm).

4) Gi S l giao im ca BM v ng thng (d), N l giao im ca BP vi HK.

Xt (PAM v ( OBM :

Theo gi thit ta c (v c R = OB).

Mt khc ta c (v cng chn cung ca (O))

( (PAM ( OBM

.(do OB = OM = R) (3)

V (do chn na trn(O))

( tam gic AMS vung ti M. (

v (4)

M PM = PA(cmt) nn

T (3) v (4) ( PA = PS hay P l trung im ca AS.

V HK//AS (cng vung gc AB) nn theo L Ta-lt, ta c: hay

m PA = PS(cmt) hay BP i qua trung im N ca HK. (pcm)

Bai V: (0,5 im)

Cch 1(khng s dng BT C Si) Ta c M = =

V (x 2y)2 0, du = xy ra ( x = 2y

x 2y ( , du = xy ra ( x = 2y

T ta c M 0 + 4 -=, du = xy ra ( x = 2y

Vy GTNN ca M l , t c khi x = 2y

Cch 2:

Ta c M =

V x, y > 0 , p dng bdt C si cho 2 s dng ta c ,

du = xy ra ( x = 2y

V x 2y (, du = xy ra ( x = 2y

T ta c M 1 +=, du = xy ra ( x = 2y


Cch 3:

Ta c M =

V x, y > 0 , p dng bdt C si cho 2 s dng ta c ,

du = xy ra ( x = 2y

V x 2y (, du = xy ra ( x = 2y

T ta c M 4-=, du = xy ra ( x = 2y


Cch 4:

Ta c M =

V x, y > 0 , p dng bdt Co si cho 2 s dng ta c ,

du = xy ra ( x = 2y

V x 2y (, du = xy ra ( x = 2y

T ta c M += 1+=, du = xy ra ( x = 2y


S GIO DC V O TOK THI TUYN SINH LP 10 THPT

TP.HCMNm hoc: 2012 2013

MN: TON


Bai 1: (2 im)

Gii cc phng trnh v h phng trnh sau:

a)

b)

c)

d)

Bai 2: (1,5 im)

a) V th (P) ca hm s v ng thng (D): trn cng mt h trc to .

b) Tm to cc giao im ca (P) v (D) cu trn bng php tnh.

Bai 3: (1,5 im)

Thu gn cc biu thc sau:

vi x > 0;

Bai 4: (1,5 im)

Cho phng trnh (x l n s)

a) Chng minh rng phng trnh lun lun c 2 nghim phn bit vi mi m.

b) Gi x1, x2 l cc nghim ca phng trnh. Tm m biu thc M = t gi tr nh nhtBai 5: (3,5 im)

Cho ng trn (O) c tm O v im M nm ngoi ng trn (O). ng thng MO ct (O) ti E v F (ME 0;

Cu 4:

a/ Phng trnh (1) c = m2 - 4m +8 = (m - 2)2 +4 > 0 vi mi m nn phng trnh (1) c 2 nghim phn bit vi mi m.

b/ Do , theo Viet, vi mi m, ta c: S = ; P =

M = =

. Khi m = 1 ta c nh nht

ln nht khi m = 1 nh nht khi m = 1

Vy M t gi tr nh nht l - 2 khi m = 1

Cu 5

a) V ta c do hai tam gic ng dng MAE v MBF

Nn MA.MB = ME.MF

(Phng tch ca M i vi ng trn tm O)

b) Do h thc lng trong ng trn ta c

MA.MB = MC2, mt khc h thc lng

trong tam gic vung MCO ta c

MH.MO = MC2 MA.MB = MH.MO

nn t gic AHOB ni tip trong ng trn.

c) Xt t gic MKSC ni tip trong ng

trn ng knh MS (c hai gc K v C vung).

Vy ta c : MK2 = ME.MF = MC2 nn MK = MC.

Do MF chnh l ng trung trc ca KC

nn MS vung gc vi KC ti V.

d) Do h thc lng trong ng trn ta c MA.MB = MV.MS ca ng trn tm Q.

Tng t vi ng trn tm P ta cng c MV.MS = ME.MF nn PQ vung gc vi MS v l ng trung trc ca VS (ng ni hai tm ca hai ng trn). Nn PQ cng i qua trung im ca KS (do nh l trung bnh ca tam gic SKV). Vy 3 im T, Q, P thng hng.S GIO DC V O TOK THI TUYN SINH LP 10 THPT

TP. NNG Nm hoc: 2012 2013

MN: TON


Bi 1: (2,0 im)

1) Gii phng trnh:(x + 1)(x + 2) = 0

2) Gii h phng trnh:

Bi 2: (1,0 im)

Rt gn biu thc

Bi 3: (1,5 im)

Bit rng ng cong trong hnh v bn l mt parabol y = ax2.

1) Tm h s a.

2) Gi M v N l cc giao im ca ng thng

y = x + 4 vi parabol. Tm ta ca cc im M v N.

Bi 4: (2,0 im)

Cho phng trnh x2 2x 3m2 = 0, vi m l tham s.

1) Gii phng trnh khi m = 1.

2) Tm tt c cc gi tr ca m phng trnh c hai nghim x1, x2 khc 0 v tha iu kin .

Bi 5: (3,5 im)

Cho hai ng trn (O) v (O) tip xc ngoi ti A. K tip tuyn chung ngoi BC,B ( (O),C((O). ng thng BO ct (O) ti im th hai l D.

1) Ch`ng minh rng t gic COOB l mt hnh thang vung.

2) Chng minh rng ba im A, C, D thng hng.

3) T D k tip tuyn DE vi ng trn (O) (E l tip im). Chng minh rng DB = DE.

BI GII

Bi 1:

1) (x + 1)(x + 2) = 0 ( x + 1 = 0 hay x + 2 = 0 ( x = -1 hay x = -2

2)

(

EMBED Equation.DSMT4 (

Bi 2: = =

= = 4

Bi 3:

1) Theo th ta c y(2) = 2 ( 2 = a.22 ( a =

2)Phng trnhhonh giao im ca y = v ng thng y = x + 4 l :

x + 4 = ( x2 2x 8 = 0 ( x = -2 hay x = 4

y(-2) = 2; y(4) = 8. Vy ta cc im M v N l (-2; 2) v (4; 8).

Bi 4:

1)Khi m = 1, phng trnh thnh: x2 2x 3 = 0 ( x = -1 hay x = 3 (c dng ab + c = 0)

2)Vi x1, x2 ( 0, ta c: ( ( 3(x1 + x2)(x1 x2) = 8x1x2

Ta c: a.c = -3m2 ( 0 nn ( ( 0, (m

Khi ( ( 0 ta c: x1 + x2 = v x1.x2 = ( 0

iu kin phng trnh c 2 nghim ( 0 m m ( 0 ( ( > 0 v x1.x2 < 0 ( x1 < x2

Vi a = 1 ( x1 = v x2 = ( x1 x2 =

Do , ycbt ( v m ( 0

( (hin nhin m = 0 khng l nghim)

( 4m4 3m2 1 = 0 ( m2 = 1 hay m2 = -1/4 (loi) ( m = (1

Bi 5:

1)Theo tnh cht ca tip tuyn ta c OB, OC vung gc vi BC ( t gic COOB l hnh thang vung.

2)Ta c gc ABC = gc BDC ( gc ABC + gc BCA = 900 ( gc BAC = 900

Mt khc, ta c gc BAD = 900 (ni tip na ng trn)

Vy ta c gc DAC = 1800 nn 3 im D, A, C thng hng.

3)Theo h thc lng trong tam gic vung DBC ta c DB2 = DA.DC

Mt khc, theo h thc lng trong ng trn (chng minh bng tam gic ng dng) ta c DE2 = DA.DC ( DB = DE.

S GD&T

VNH PHC

K THI TUYN SINH LP 10 THPT NM HC 2012-2013

THI MN : TON

Thi gian lm bi 120 pht (khng k thi gian giao )


Cu 1 (2,0 im). Cho biu thc :P=

1. Tm iu kin xc nh ca biu thc P.

2. Rt gn P

Cu 2 (2,0 im). Cho h phng trnh :

1. Gii h phng trnh vi a=1

2. Tm a h phng trnh c nghim duy nht.

Cu 3 (2,0 im). Mt hnh ch nht c chiu rng bng mt na chiu di. Bit rng nu gim mi chiu i 2m th din tch hnh ch nht cho gim i mt na. Tnh chiu di hnh ch nht cho.

Cu 4 (3,0 im). Cho ng trn (O;R) (im O c nh, gi tr R khng i) v im M nm bn ngoi (O). K hai tip tuyn MB, MC (B,C l cc tip im ) ca (O) v tia Mx nm gia hai tia MO v MC. Qua B k ng thng song song vi Mx, ng thng ny ct (O) ti im th hai l A. V ng knh BB ca (O). Qua O k ng thng vung gc vi BB,ng thng ny ct MC v BC ln lt ti K v E. Chng minh rng:

1. 4 im M,B,O,C cng nm trn mt ng trn.

2. on thng ME = R.

3. Khi im M di ng m OM = 2R th im K di ng trn mt ng trn c nh, ch r tm v bn knh ca ng trn .

Cu 5 (1,0 im). Cho a,b,c l cc s dng tha mn a+ b + c =4. Chng minh rng :

S GD&T VNH PHC


P N THI MN : TON


Cup n, gi im

C1.1

(0,75 im)Biu thc P xc nh

0,5

0,25

C1.2 (1,25 im)P=

0,25

0,5

0,5

C2.1 (1,0 im)Vi a = 1, h phng trnh c dng:

Vy vi a = 1, h phng trnh c nghim duy nht l:

0,25

0,25

0,25

0,25

C2.2 (1,0 im)-Nu a = 0, h c dng: => c nghim duy nht

-Nu a , h c nghim duy nht khi v ch khi:

(lun ng, v vi mi a)

Do , vi a , h lun c nghim duy nht.

Vy h phng trnh cho c nghim duy nht vi mi a.0,25

0,25

0,25

0,25

C3 (2,0 im) Gi chiu di ca hnh ch nht cho l x (m), vi x > 4.

V chiu rng bng na chiu di nn chiu rng l: (m)

=> din tch hnh ch nht cho l: (m2)

Nu gim mi chiu i 2 m th chiu di, chiu rng ca hnh ch nht ln lt l: (m)

khi , din tch hnh ch nht gim i mt na nn ta c phng trnh:

.=> (tho mn x>4);

(loi v khng tho mn x>4)

Vy chiu di ca hnh ch nht cho l (m).0,25

0,25

0,25

0,25

0,25

0,5

0,25

C4.1 (1,0 im)1) Chng minh M, B, O, C cng thuc 1 ng trn

Ta c: (v MB l tip tuyn)

(v MC l tip tuyn)=> MBO + MCO =

= 900 + 900 = 1800=> T gic MBOC ni tip

(v c tng 2 gc i =1800)

=>4 im M, B, O, C cng thuc 1 ng trn0,25

0,25

0,25

0,25

C4.2 (1,0 im)2) Chng minh ME = R:

Ta c MB//EO (v cng vung gc vi BB)

=> O1 = M1 (so le trong)

M M1 = M2 (tnh cht 2 tip tuyn ct nhau) => M2 = O1 (1)

C/m c MO//EB (v cng vung gc vi BC)

=> O1 = E1 (so le trong) (2)

T (1), (2) => M2 = E1 => MOCE ni tip

=> MEO = MCO = 900

=> MEO = MBO = BOE = 900 => MBOE l hnh ch nht

=> ME = OB = R (iu phi chng minh)0,25

0,25

0,25

0,25

C4.3 (1,0 im)3) Chng minh khi OM=2R th K di ng trn 1 ng trn c nh:

Chng minh c Tam gic MBC u => BMC = 600=> BOC = 1200

=> KOC = 600 - O1 = 600 - M1 = 600 300 = 300Trong tam gic KOC vung ti C, ta c:

M O c nh, R khng i => K di ng trn ng trn tm O, bn knh = (iu phi chng minh)0,25

0,25

0,25

0,25

C5 (1,0 im)

Do ,

0,25

0,25

0,25

0,25

Ch : -Cu 4, tha gi thit tia Mx v im A ( gy ri.

-Mi cu u c cc cch lm khc

cu 5

Cach 2: t x = => x, y , z > 0 v x4 + y4 + z4 = 4.

BT cn CM tng ng: x3 + y3 + z3 >

hay (x3 + y3 + z3 ) > 4 = x4 + y4 + z4( x3(-x) + y3(-y)+ z3(-z) > 0 (*).

Ta xt 2 trng hp:

- Nu trong 3 s x, y, z tn ti it nht mt s , gi s x th x3 .

Khi o: x3 + y3 + z3 > ( do y, z > 0).

- Nu c 3 s x, y, z u nh th BT(*) lun ung.

Vy x3 + y3 + z3 > c CM.

Cach 3: C th dng BT thc Csi kt hp phng php lm tri v nh gi cng cho kt qu (nhng hi di, phc tp).S GD V O TOK THI TUYN SINH VO 10 THPT NM HC 2012-2013

KLK

MN THI : TON

Thi gian lm bi: 120 pht,(khng k giao )

Ngy thi: 22/06/2012

Cu 1. (2,5)

1) Gii phng trnh:

a) 2x2 7x + 3 = 0.

b) 9x4 + 5x2 4 = 0.

2) Tm hm s y = ax + b, bit th hm s ca n i qua 2 im A(2;5) ; B(-2;-3).

Cu 2. (1,5)

1) Hai t i t A n B di 200km. Bit vn tc xe th nht nhanh hn vn tc xe th hai l 10km/h nn xe th nht n B sm hn xe th hai 1 gi. Tnh vn tc mi xe.

2) Rt gn biu thc:

EMBED Equation.DSMT4 vi x 0.

Cu 3. (1,5 )

Cho phng trnh: x2 2(m+2)x + m2 + 4m +3 = 0.

1) Chng minh rng : Phng trnh trn lun c hai nghim phn bit x1, x2 vi mi gi tr ca m.

2) Tm gi tr ca m biu thc A = t gi tr nh nht.

Cu 4. (3,5)

Cho tam gic ABC c ba gc nhn ni tip ng trn tm O (AB < AC). Hai tip tuyn ti B v C ct nhau ti M. AM ct ng trn (O) ti im th hai D. E l trung im on AD. EC ct ng trn (O) ti im th hai F. Chng minh rng:

1) T gic OEBM ni tip.

2) MB2 = MA.MD.

3) .

4) BF // AM

Cu 5. (1)

Cho hai s dng x, y tha mn: x + 2y = 3. Chng minh rng:

Bi gii s lc:

Cu 1. (2,5)

1) Gii phng trnh:

a) 2x2 7x + 3 = 0.

EMBED Equation.DSMT4 = (-7)2 4.2.3 = 25 > 0

= 5. Phng trnh c hai nghim phn bit:

b) 9x4 + 5x2 4 = 0. t x2 = t , k : t 0.

Ta c pt: 9t2 + 5t 4 = 0.

a b + c = 0 t1 = - 1 (khng TMK, loi)

t2 = (TMK)

t2 = x2 =

EMBED Equation.DSMT4 x =.

Vy phng trnh cho c hai nghim: x1,2 =

EMBED Equation.DSMT4 2) th hm s y = ax + b i qua hai im A(2;5) v B(-2;-3)

Vy hm s cn tm l : y = 2x + 1

Cu 2.

1) Gi vn tc xe th hai l x (km/h). k: x > 0

Vn tc xe th nht l x + 10 (km/h)

Thi gian xe th nht i qung ng t A n B l : (gi)

Thi gian xe th hai i qung ng t A n B l : (gi)

Xe th nht n B sm 1 gi so vi xe th hai nn ta c phng trnh:

Gii phng trnh ta c x1 = 40 , x2 = -50 ( loi)

x1 = 40 (TMK). Vy vn tc xe th nht l 50km/h, vn tc xe th hai l 40km/h.

2) Rt gn biu thc:

= = x, vi x 0.

Cu 3. (1,5 )

Cho phng trnh: x2 2(m+2)x + m2 + 4m +3 = 0.

1) Chng minh rng : Phng trnh trn lun c hai nghim phn bit x1, x2 vi mi gi tr ca m.

Ta c > 0 vi mi m.

Vy phng trnh cho lun c hai nghim phn bit x1, x2 vi mi gi tr ca m.

2) phng trnh cho lun c hai nghim phn bit x1, x2 vi mi gi tr ca m. Theo h thc Vi-t ta c :

A = = (x1 + x2)2 2 x1x2 = 4(m + 2)2 2(m2 + 4m +3) = 2m2 + 8m+ 10

= 2(m2 + 4m) + 10

= 2(m + 2)2 + 2 2 vi mi m.

Suy ra minA = 2 m + 2 = 0 m = - 2

Vy vi m = - 2 th A t min = 2

Cu 4.

1) Ta c EA = ED (gt) OE AD ( Quan h gia ng knh v dy)

EMBED Equation.DSMT4

EMBED Equation.DSMT4 = 900; = 900 (Tnh cht tip tuyn)

E v B cng nhn OM di mt gc vung T gic OEBM ni tip.

2) Ta c s ( gc ni tip chn cung BD)

s ( gc to bi tia tip tuyn v dy cung chn cung BD)

EMBED Equation.DSMT4 . Xt tam gic MBD v tam gic MAB c:

Gc M chung,

EMBED Equation.DSMT4 ng dng vi


MB2 = MA.MD

3) Ta c:

EMBED Equation.DSMT4 = s ( Tnh cht hai tip tuyn ct nhau);

EMBED Equation.DSMT4 s (gc ni tip)

EMBED Equation.DSMT4 .

4) T gic MFOC ni tip ( = 1800)

EMBED Equation.DSMT4 ( hai gc ni tip cng chn cung MC), mt khc (theo cu 3)


EMBED Equation.DSMT4 BF // AM.

Cu 5.

Ta c x + 2y = 3 x = 3 2y , v x dng nn 3 2y > 0

Xt hiu = 0 ( v y > 0 v 3 2y > 0)

EMBED Equation.DSMT4 du = xy ra

EMBED Equation.DSMT4 S GIO DC VO O TO HI DNG

-----------------


NM HC 2012-2013

MN THI: TON


Ngy thi: Ngy 12 thng 7 nm 2012

( thi gm: 01 trang)

Cu 1 (2,0 im):

Gii cc phng trnh sau:

a) x(x-2)=12-x.

b)

Cu 2 (2,0 im):

a) Cho h phng trnh c nghim (x;y). Tm m biu thc (xy+x-1) t gii tr ln nht.

b) Tm m ng thng y = (2m-3)x-3 ct trc honh ti im c honh bng .

Cu 3 (2,0 im):

a) Rt gn biu thc vi v .

b) Nm ngoi, hai n v sn xut nng nghip thu hoch c 600 tn thc. Nm nay, n v th nht lm vt mc 10%, n v th hai lm vt mc 20% so vi nm ngoi. Do c hai n v thu hoch c 685 tn thc. Hi nm ngoi, mi n v thu hoch c bao nhiu tn thc?

Cu 4 (3,0 im):

Cho tam gic ABC c ba gc nhn, ni tip ng trn (O). V cc ng cao BE, CF ca tam gic y. Gi H l giao im ca BE v CF. K ng knh BK ca (O) .

a) Chng minh t gic BCEF l t gic ni tip.

b) Chng minh t gic AHCK l mnh bnh hnh.

c) ng trn ng knh AC ct BE M, ng trn ng knh AB ct CF N. Chng minh AM = AN.

Cu 5 (1,0 im):

Cho a, b, c, d l cc s thc tha mn: b + d 0 v . Chng minh rng phng trnh (x2 + ax +b)(x2 + cx + d)=0 (x l n) lun c nghim.

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

HNG DN - P N

Cu 1: a ) x = - 3 v x = 4. b) x = - 2; loi x = 4.

Cu 2: a) H => x = m + 2 v y = 3 - m => A = (xy+x-1) = = 8 - ( m -1)2 Amax = 8 khi m = 1.

b) Thay x = 2/3 v y = 0 vo pt ng thng => m = 15/4

Cu 3: a) A = 1

b) x + y = 600 v 0,1x + 0,2y = 85 hay x + 2y = 850.

T tnh c y = 250 tn, x = 350 tn

Cu 4 (3,0 im):

a)

b) AH//KC ( cng vung gc vi BC)

CH // KA ( cng vung gc vi AB)

c) C AN2 = AF.AB; AM2 = AE.AC

( H thc lng trong tam gic vung)

AM = AN

Cu 5 (1,0 im)Xt 2 phng trnh:

x2 + ax + b = 0 (1) v x2 + cx + d = 0 (2)

+ Vi b+d 0 hoc >0 pt cho c nghim

+ Vi . T ac > 2(b + d) =>

=> t nht mt trong hai biu gi tr

EMBED Equation.3 => t nht mt trong hai pt (1) v (2) c nghim.Vy vi a, b, c, d l cc s thc tha mn: b + d 0 v ,

phng trnh (x2 + ax +b)(x2 + cx + d)=0 (x l n) lun c nghim.

S GIO DC VO O TO HI DNG

-----------------


NM HC 2012-2013

MN THI: TON


Ngy thi: Ngy 14 thng 7 nm 2012

( thi gm: 01 trang)

Cu 1 (2,0 im): Gii cc phng trnh sau:

a)

b) | 2x 3 | = 1.

Cu 2 (2,0 im): Cho biu thc:

A = vi a v b l cc s dng khc nhau.

a) Rt gn biu thc A .

b) Tnh gi tr ca A khi a = v b = .

Cu 3 (2,0 im):a) Tm m cc ng thng y = 2x + m v y = x 2m + 3 ct nhau ti mt im nm trn trc tung.

b) Cho qung ng t a im A ti a im B di 90 km. Lc 6 gi mt xe my i t A ti B Lc 6 gi 30 pht cng ngy, mt t cng i t A ti B vi vn tc ln hn vn tc xe my 15 km/h (Hai xe chy trn cng mt con ng cho). Hai xe ni trn u n B cng lc. Tnh vn tc mi xe.Cu 4 (3,0 im): Cho na ng trn tm O ng knh AB = 2R (R l mt di cho trc). Gi C, D l hai im trn na ng trn sao cho C thuc cung v = 1200 . Gi giao im ca hai dy AD v BC l E, giao im ca cc ng thng AC v BD l F.a) Chng minh rng bn im C, D, E, F cng nm trn mt ng trn.

b) Tnh bn knh ca ng trn i qua C, E, D, F ni trn theo R.

c) Tm gi tr ln nht ca in tch tam gic FAB theo R khi C, D thay i nhung vn tha mn gi thit bi ton

Cu 5 (1,0 im): Khng dng my tnh cm tay , tm s nguyn ln nht khng vt qu S, trong S =

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

HNG DN GII .

Cu 1.

Vy nghim ca phng trnh cho l S = {} b)

Vy nghim ca phng trnh cho l S = {1;2}

Cu 2 .

Ta c :

a) Ta c :

Vy = 0

b) Ta c :

Thay vo biu thc ta c :

Vy vi a = 7 - ; b = 7 + 4 th A = .

Cu 3 .

a) hai ng thng y = 2x + m v y = x 2m + 3 ct nhau ti mt im trn trc tung th m = -2m + 3 => 3m = 3 => m = 1.

Vy vi m = 1 th hai ng thng y = 2x + m v y = x 2m + 3 ct nhau ti mt im trn trc tung.

b) Xe my i trc t thi gian l : 6 gi 30 pht - 6 gi = 30 pht = .

Gi vn tc ca xe my l x ( km/h ) ( x > 0 )

V vn tc t ln hn vn tc xe my 15 km/h nn vn tc ca t l x + 15 (km/h)

Thi gian xe my i ht qung ng AB l :

Thi gian t i ht qung ng AB l :

Do xe my i trc t gi v hai xe u ti B cng mt lc nn ta c phng trnh :

Ta c :

( khng tha mn iu kin )

( tha mn iu kin )

Vy vn tc ca xe my l 45 ( km/h ) , vn tc ca t l 45 + 15 = 60 ( km/h ).

Cu 4.

a) Ta c : C, D thuc ng trn nn :

( gc ni tip chn na ng trn )

=> ( gc k b )

Hai im C v D cng nhn on thng FE di mt gc bng nhau bng 900 nn 4 im C,D,E,F cng thuc ng trn ng knh EF.

b) Gi I l trung im EF th ID = IC l bn knh ng trn i qua 4 im C, D, E, F ni trn.

Ta c : IC = ID ; OC = OD ( bn knh ng trn tm O )

suy ra IO l trung trc ca CD => OI l phn gic ca

=>

Do O l trung im AB v tam gic ADB vung ti D nn tam gic ODB cn ti O

=> (1)

Do ID = IF nn tam gic IFD cn ti I => (2)

Tam gic AFB c hai ng cao AD, BC ct nhau ti E nn E l trc tm tam gic => FE l ng cao th ba => FE vung gc AB ti H => (3)

T (1) , (2) , (3) suy ra => .

Xt tam gic vung IDO c .

Ta c : ID = OD.tan = R.tan600 = R.

Vy bn knh ng trn i qua 4 im C,D,E,F l R.

c) Theo phn b) : OI = .

t OH = x th => IH = .

=> FH = R + .

Ta c : 4R2 - x2 4R2 . Du bng xy ra khi x = 0.

Khi : SFAB = R2 + 2R2 v H O => O, I, F thng hng => CD // AB => => BD = AC = 2RSin150 .

Vy din tch ln nht t c ca tam gic AFB l R2 + 2R2 khi AC = BD = 2Rsin150 .

Cu 5 Xt hai s a = 2 + v b = 2 - .

Ta c : a + b = 4 v ab = 1, 0< b < 1.

(a+b)3 = 43 = 64 => a3 + b3 = 64 - 3ab(a + b) = 64 - 3.1.4 = 52

(a3+b3)(a3 + b3) = 52.52 => a6 + b6 = 2704 - 2(ab)3 = 2704 - 2 = 2702

=> a6 = S = 2702 - b6 (*).

Do 0 0 , a (1)

1. Chng minh rng :

2. Tm gi tr ca a P = a

Cu 2 (2,0 im ) : Trong mt phng to Oxy, cho Parabol (P) : y = x2 v ng thng (d) : y = 2x + 3

1. Chng minh rng (d) v (P) c hai im chung phn bit

2. Gi A v B l cc im chung ca (d) v (P) . Tnh din tch tam gic OAB ( O l gc to )

Cu 3 (2.0 im) : Cho phng trnh : x2 + 2mx + m2 2m + 4 = 0

1. Gii phng trnh khi m = 4

2. Tm m phng trnh c hai nghim phn bit

Cu 4 (3.0 im) : Cho ng trn (O) c ng knh AB c nh, M l mt im thuc (O) ( M khc A v B ) . Cc tip tuyn ca (O) ti A v M ct nhau C. ng trn (I) i qua M v tip xc vi ng thng AC ti C. CD l ng knh ca (I). Chng minh rng:

1. Ba im O, M, D thng hng

2. Tam gic COD l tam gic cn

3. ng thng i qua D v vung gc vi BC lun i qua mt im c nh khi M di ng trn ng trn (O)

Cu 5 (1.0 im) : Cho a,b,c l cc s dng khng m tho mn :

Chng minh rng :

HNG DN GII:CUNI DUNGIM

11. Chng minh rng :

(PCM)

1.0

2. Tm gi tr ca a P = a. P = a

=> .

Ta c 1 + 1 + (-2) = 0, nn phng trnh c 2 nghim

a1 = -1 < 0 (khng tho mn iu kin) - Loi

a2 = (Tho mn iu kin)

Vy a = 2 th P = a1.0

21. Chng minh rng (d) v (P) c hai im chung phn bit

Honh giao im ng thng (d) v Parabol (P) l nghim ca phng trnh

x2 = 2x + 3 => x2 2x 3 = 0 c a b + c = 0

Nn phng trnh c hai nghim phn bit

x1 = -1 v x2 =

Vi x1 = -1 => y1 = (-1)2 = 1 => A (-1; 1)

Vi x2 = 3 => y2 = 32 = 9 => B (3; 9)

Vy (d) v (P) c hai im chung phn bit A v B1.0

2. Gi A v B l cc im chung ca (d) v (P) . Tnh din tch tam gic OAB ( O l gc to )

Ta biu din cc im A v B trn mt phng to Oxy nh hnh v

Theo cng thc cng din tch ta c:

S(ABC) = S(ABCD) - S(BCO) - S(ADO) = 20 13,5 0,5 = 6 (vdt)1.0

31. Khi m = 4, ta c phng trnh

x2 + 8x + 12 = 0 c ( = 16 12 = 4 > 0

Vy phng trnh c hai nghim phn bit

x1 = - 4 + 2 = - 2 v x2 = - 4 - 2 = - 61.0

2. Tm m phng trnh c hai nghim phn bit

x2 + 2mx + m2 2m + 4 = 0

C D = m2 (m2 2m + 4) = 2m 4

phng trnh c hai nghim phn bit th D > 0

=> 2m 4 > 0 => 2(m 2) > 0 => m 2 > 0 => m > 2

Vy vi m > 2 th phng trnh c hai nghim phn bit1.0

41. Ba im O, M, D thng hng:

Ta c MC l tip tuyn ca ng trn (O) ( MC ( MO (1)

Xt ng trn (I) : Ta c ( MC ( MD (2)

T (1) v (2) => MO // MD ( MO v MD trng nhau

( O, M, D thng hng1.0

2. Tam gic COD l tam gic cn

CA l tip tuyn ca ng trn (O) ( CA (AB(3)

ng trn (I) tip xc vi AC ti C ( CA ( CD(4)

T (3) v (4) ( CD // AB => (*)

( Hai gc so le trong)

CA, CM l hai tip tuyn ct nhau ca (O) ( (**)

T (*) v (**) ( ( Tam gic COD cn ti D1.0

3. ng thng i qua D v vung gc vi BC lun i qua mt im c nh khi M di ng trn ng trn (O)

* Gi chn ng vung gc h t D ti BC l H. ( H ( (I) (Bi ton qu tch)

DH ko di ct AB ti K.

Gi N l giao im ca CO v ng trn (I)

=>

Ta c t gic NHOK ni tip

V c ( Cng b vi gc DHN) ( (5)

* Ta c : (Cng chn cung NH ca ng trn (I))

( (DHN (COB (g.g)

M

((NHO (DHC (c.g.c)

( M (5) (, ( NK ( AB ( NK // AC ( K l trung im ca OA c nh ( (PCM)1.0

5Cu 5 (1.0 im) : Cho a,b,c l cc s dng khng m tho mn :

Chng minh rng :

* C/M b : v .

Tht vy

(ng) ( PCM

p dng 2 ln , ta c:

* Ta c : , tng t Ta c: (

Ta chng minh

* p dng B trn ta c:

(

* M:

T (3) v (4) ( (2)

Kt hp (2) v (1) ta c iu phi chng minh.

Du = xy ra khi a = b = c = 11.0

S GIO DC V O TO

THNH PH CN TH


NM HC 2012-2013

Kha ngy:21/6/2012

MN: TON

Thi gian lm bi: 120 pht (khng k thi gian pht )

Cu 1: (2,0 im)

Gii h phng trnh , cc phng trnh sau y:

1.

2.

3.

4.

Cu 2: (1,5 im)

Cho biu thc: (vi )

1. Rt gn biu thc K.

2. Tm a .

Cu 3: (1,5 im)

Cho phng trnh (n s x): .

1. Chng minh phng trnh (*) lun c hai nghim phn bit vi mi m.

2. Tm gi tr ca m phng trnh (*) c hai nghim tha .

Cu 4: (1,5 im)

Mt t d nh i t A n B cch nhau 120 km trong mt thi gian quy nh. Sau khi i c 1 gi th t b chn bi xe cu ha 10 pht. Do n B ng hn xe phi tng vn tc thm 6 km/h. Tnh vn tc lc u ca t.

Cu 5: (3,5 im)

Cho ng trn , t im ngoi ng trn v hai tip tuyn v(l cc tip im). ctti E.

1. Chng minh t gic ni tip.

2. Chng minh vung gc vi v .

3. Gil trung im ca , ng thng quav vung gc ct cc tia theo th t ti v . Chng minh v cn ti .

4. Chng minh l trung im ca.

GI GII:Cu 1: (2,0 im)

Gii h phng trnh , cc phng trnh sau y:

1.

2.

3.

4.

Cu 2: (1,5 im)

Cho biu thc: (vi )

= a = 503 (TMK)

Cu 3: (1,5 im)

Cho phng trnh (n s x):.

1.

Vy (*) lun co hai nghim phn bit vi mi m.

2. Tm gi tr ca m phng trnh (*) c hai nghim tha .

Theo h thc VI-ET co :x1.x2 = - m2 + 3 ;x1+ x2 = 4; ma => x1 = - 1 ; x2 = 5

Thay x1 = - 1 ; x2 = 5 vao x1.x2 = - m2 + 3 => m =

Cu 4: (1,5 im)

Goi x (km/h) la vt d inh; x > 0 => Thi gian d inh :

Sau 1 h t i c x km => quang ng con lai 120 x ( km)

Vt luc sau: x + 6 ( km/h)

Pt => x = 48 (TMK) => KL HD C3

Tam giac BOC cn tai O => goc OBC = goc OCB

T giac OIBD co goc OID = goc OBD = 900 nn OIBD ni tip => goc ODI = goc OBI

Do o

Lai co FIOC ni tip ; nn goc IFO = goc ICO

Suy ra goc OPF = goc OFP ; vy cn ti .

HD C4

Xet t giac BPFE co IB = IE ; IP = IF ( Tam giac OPF cn co OI la ng cao=> )

Nn BPEF la Hinh binh hanh => BP // FE

Tam giac ABC co EB = EC ; BA // FE; nn EF la TB cua tam giac ABC => FA = FC

S GD T NGH AN

thi vo THPT nm hc 2012 - 2013

Mn thi: Ton

Thi gian 120 pht

Ngy thi 24/ 06/ 2012

Cu 1: 2,5 im:

Cho biu thc A =

a) Tm iu kin xc nh v t gn A.

b) Tm tt c cc gi tr ca x

c) Tm tt c cc gi tr ca x t gi tr nguyn.

Cu 2: 1,5 im:

Qung ng AB di 156 km. Mt ngi i xe my t A, mt ngi i xe p t B. Hai xe xut pht cng mt lc v sau 3 gi gp nhau. Bit rng vn tc ca ngi I xe my nhanh hn vn tc ca ngi I xe p l 28 km/h. Tnh vn tc ca mi xe?

Cu 3: 2 im:

Chjo phng trnh: x2 2(m-1)x + m2 6 =0 ( m l tham s).

a) GiI phng trnh khi m = 3

b) Tm m phng trnh c hai nghim x1, x2 tha mn

Cu 4: 4 im

Cho im M nm ngoi ng trn tm O. V tip tuyn MA, MB vi ng trn (A, B l cc tip im). V ct tuyn MCD khng I qua tm O ( C nm gia M v D), OM ct AB v (O) ln lt ti H v I. Chng minh.

a) T gic MAOB ni tip.

b) MC.MD = MA2c) OH.OM + MC.MD = MO2d) CI l tia phn gic gc MCH.

HNG DN GII

Cu 1: (2,5 im)

a, Vi x > 0 v x 4, ta c:

A = = = ... =

b, A = > ... x > 4.

c, B = . = l mt s nguyn ... l c ca 14 hay = 1, = 7, = 14.

(Gii cc pt trn v tm x)

Cu 2: (1,5 im)

Gi vn tc ca xe p l x (km/h), iu kin x > 0

Th vn tc ca xe my l x + 28 (km/h)

Trong 3 gi:

+ Xe p i c qung ng 3x (km),

+ Xe my i c qung ng 3(x + 28) (km), theo bi ra ta c phng trnh:

3x + 3(x + 28) = 156

Gii tm x = 12 (TMK)

Tr li: Vn tc ca xe p l 12 km/h v vn tc ca xe my l 12 + 28 = 40 (km/h)

Cu 3: (2,0 im)

a, Thay x = 3 vo phng trnh x2 - 2(m - 1)x + m2 - 6 = 0 v gii phng trnh:

x2 - 4x + 3 = 0 bng nhiu cch v tm c nghim x1 = 1, x2 = 3.

b, Theo h thc Vit, gi x1, x2 l hai nghim ca phng trnh

x2 - 2(m - 1)x + m2 - 6 = 0 , ta c:

v x12 + x22 = (x1 + x2)2 - 2x1.x2 = 16

Thay vo gii v tm c m = 0, m = -4

Cu 4: (4,0 im).

T vit GT-KL

A

D

C

M

I H

B

a, V MA, MB l cc tip tuyn ca ng trn (O) ti A v B nn cc gc ca t gic MAOB vung ti A v B, nn ni tip c ng trn.

b, MAC v MDA c chung v = (cng chn ), nn ng dng. T suy ra (fcm)

c, MAO v AHO ng dng v c chung gc O v (cng chn hai cung bng nhau ca ng trn ni tip t gic MAOB). Suy ra OH.OM = OA2p dng nh l Pitago vo tam gic vung MAO v cc h thc OH.OM = OA2 MC.MD = MA2 suy ra iu phi chng minh.

d, T MH.OM = MA2, MC.MD = MA2 suy ra MH.OM = MC.MD (*)

Trong MHC v MDO c (*) v chung nn ng dng.

hay (1)

Ta li c (cng chn hai cung bng nhau) AI l phn gic ca .

Theo t/c ng phn gic ca tam gic, ta c: (2)

MHA v MAO c chung v do ng dng (g.g)

(3)

T (1), (2), (3) suy ra suy ra CI l tia phn gic ca gc MCHS GIO DC V O TO

H NAM


NM HC 2012 2013

Mn: Ton


Ngy thi : 22/06/2012

Cu 1 (1,5 im)

Rt gn cc biu thc sau:

Cu 2: (2 im)

a) Gii phng trnh: x2 5x + 4 = 0

b) Gii h phng trnh:

Cu 3: (2 im)

Trong mt phng to Oxy cho Parabol (P) c phng trnh: y = x2 v ng thng (d) c phng trnh: y = 2mx 2m + 3 (m l tham s)

a) Tm to cc im thuc (P) bit tung ca chng bng 2

b) Chng minh rng (P) v (d) ct nhau ti hai im phn bit vi mi m.

Gi l cc tung giao im ca (P) v (d), tm m

Cu 4: (3,5 im)

Cho ng trn tm O, ng knh AB. Trn tip tuyn ca ng trn (O) ti A ly im M ( M khc A). T M v tip tuyn th hai MC vi (O) (C l tip im). K CH vung gc vi AB (), MB ct (O) ti im th hai l K v ct CH ti N. Chng minh rng:

a) T gic AKNH l t gic ni tip.

b) AM2 = MK.MB

c) Gc KAC bng gc OMB

d) N l trung im ca CH.

Cu 5(1 im)

Cho ba s thc a, b, c tho mn

Tm gi tr ln nht ca biu thc :

S GIO DC V OTO

QUNG TR


KHA NGY : 19/6/2012

MN : TON

Thi gian lm bi: 120 pht (khng k thi gian giao )

Cu 1:(2 im)

1.Rt gn cc biu thc (khng dng my tnh cm tay):

a) 2-

b) , vi a0,a1

2.Gii h phng trnh (khng dng my tnh cm tay):

Cu 2:(1,5 im)

Gi x1, x2 l hai nghim ca phng trnh .Khng gii phng trnh, tnh gi tr cc biu thc sau:

a, x1 + x2

b,

c,

Cu 3:(1,5 im)

Trn mt phng ta , gi (P) l th hm s

a, V (P)

b, Tm ta giao im ca (P) v ng thng d: y = -2x+3

Cu 4:(1,5 im)

Hai xe khi hnh cng mt lc i t a im A n a im B cch nhau 100km. Xe th nht chy nhanh hn xe th hai 10km/h nn n B sm hm 30 pht, Tnh vn tc mi xe.

Cu 5:(3,5 im)

Cho ng trn (O). ng thng (d) khng i qua tm (O) ct ng trn ti hai im A v B theo th t, C l im thuc (d) ngoi ng trn (O). V ng knh PQ vung gc vi dy AB ti D ( P thuc cung ln AB), Tia CP ct ng trn (O) ti im th hai l I, AB ct IQ ti K.

a) Chng minh t gic PDKI ni tip ng trn.

b) Chng minh CI.CP = CK.CD

c) Chng minh IC l phn gic ca gc ngoi nh I ca tam gic AIB.

d) Cho ba im A, B, C c nh. ng trn (O) thay i nhng vn i qua A v B. Chng minh rng IQ lun i qua mt im c nh.

S GIO DC O TO

NINH THUN

K THI TUYN SINH VO LP 10 THPT

NM HC 2012 2013

Kha ngy: 24 6 2012

Mn thi: TONThi gian lm bi: 120 pht

Bi 1: (2,0 im)a) Gii h phng trnh:

b) Xc nh cc gi tr ca m h phng trnh sau v nghim:

( m l tham s)

Bi 2: (3,0 im)

Cho hai hm s y = x2 v y = x + 2.

a) V th hai hm s cho trn cng mt h trc ta Oxy.

b) Bng php tnh hy xc nh ta cc giao im A, B ca hai th trn (im A c honh m).

c) Tnh din tch ca tam gic OAB (O l gc ta )

Bi 3: (1,0 im)Tnh gi tr ca biu thc H =

Bi 4: (3,0 im)

Cho ng trn tm O, ng knh AC = 2R. T mt im E trn on OA (E khng trng vi A v O). K dy BD vung gc vi AC. K ng knh DI ca ng trn (O).

a) Chng minh rng: AB = CI.

b) Chng minh rng: EA2 + EB2 + EC2 + ED2 = 4R2c) Tnh din tch ca a gic ABICD theo R khi OE =

Bi 5: (1,0 im)

Cho tam gic ABC v cc trung tuyn AM, BN, CP. Chng minh rng:

(AB + BC + CA) < AM + BN + CP < AB + BC + CAP N:

Bi 1: (2,0 im)a) Gii h phng trnh:

b) H phng trnh v nghim khi:

Bi 2: (3,0 im)a) V (d) v (P) trn cng mt h trc ta .

x-2-1012

(P)41014

x- 20

y = x + 2(d)02

b) Ta giao im ca (P) v (d) l nghim ca h phng trnh:

Ta cc giao im ca (d) v (P): A (-1;1) v B (2;4)

c) SOAB = .(1+4).3 - .1.1 - .2.4 = 3

Bi 3: (1,0 im)

H =

Bi 4: (3,0 im)

a) Chng minh rng: AB = CI.

Ta c: BDAC (gt)

= 900 ( gc ni tip chn na ng trn) BDBI

Do : AC // BI AB = CI

b) Chng minh rng: EA2 + EB2 + EC2 + ED2 = 4R2V BDAC nn AB = AD

Ta c: EA2 + EB2 + EC2 + ED2 = AB2 + CD2 = AD2 + CD2 = AC2 = (2R)2 = 4R2c) Tnh din tch ca a gic ABICD theo R khi OE =

SABICD = SABD + SABIC = .DE.AC + .EB.(BI + AC)

* OE = AE = v EC = + R =

* DE2 = AE.EC = . = DE = . Do : EB =

* BI = AC 2AE = 2R 2. =

Vy: SABICD = ..2R +

EMBED Equation.DSMT4 .(+ 2R) = . = (vdt)

Bi 5: (1,0 im)

Cho tam gic ABC v cc trung tuyn AM, BN, CP. Chng minh rng:

(AB + BC + CA) < AM + BN + CP < AB + BC + CA

Gi G l trng tm ca ABC, ta c: GM = AM; GN = BN; GP =CP

V AM, BN, CP cc trung tuyn, nn: M, N, P ln lt l trung im ca BC, AC, AB

Do : MN, NP, MP l cc ng trung bnh ca ABC

Nn: MN = AB; NP = BC; MP = AC

p dng bt ng thc tam gic, ta c:

* AM < MN + AN hay AM < AB + AC (1)

Tng t: BN < AB + BC (2)

CP < BC + AC (3)

T (1), (2), (3) suy ra: AM + BN + CP < AB + BC + CA

(*)

* GN + GM > MN hay BN + AM > AB(4)

Tng t: BN + CP > BC(5)

CP + AM > AC(6)

T (4), (5), (6) suy ra:

BN + AM + BN + CP + CP + AM > AB + BC+AC

(AM + BN + CP) > (AB + AC + BC)

EMBED Equation.DSMT4 (AB + BC + CA) < AM + BN + CP (**)

T (*), (**) suy ra: (AB + BC + CA) < AM + BN + CP < AB + BC + CA

S GIO DC V OTO

THA THIN HU


Kha ngy : 24/6/2012

Mn thi : TON


Bi 1:(2,0 im)

a).Cho biu thc: C = . Chng t C =

b) Gii phng trnh :

Bi 2:(2,0 im)

Cho hm s y = x2 c th (P) v ng thng (d) i qua im M (1;2) c h s gc k0.

a/ Chng minh rng vi mi gi tr k0. ng thng (d) lun ct (P) ti hai im phn bit A v B.

b/ Gi xA v xB l honh ca hai im A v B.Chng minh rng

Bi 3:(2,0 im)

a/ Mt xe la i t ga A n ga B.Sau 1 gi 40 pht, mt xe la khc i t ga A n ga B vi vn tc ln hn vn tc ca xe la th nht l 5 km/h.Hai xe la gp nhau ti mt ga cch ga B 300 km.Tm vn tc ca mi xe, bit rng qung ng st t ga A n ga B di 645 km.

b/ Gii h phng trnh :

Bi 4:(3,0 im)

Cho na ng trn (O) ng knh BC.Ly im A trn tia i ca tia CB.K tip tuyn AF vi na ng trn (O) ( F l tip im), tia AF ct tia tip tuyn Bx ca na ng trn (O) ti D ( tia tip tuyn Bx nm trong na mt phng b BC cha na ng trn (O)) .Gi H l giao im ca BF vi DO ; K l giao im th hai ca DC vi na ng trn (O).

a/ Chng minh rng : AO.AB=AF.AD.

b/ Chng minh t gic KHOC ni tip.

c/ K OM BC ( M thuc on thng AD).Chng minh

Bi 5:(1,0 im)

Cho hnh ch nht OABC, .Gi CH l ng cao ca tam gic COB, CH=20 cm.Khi hnh ch nht OABC quay mt vng quanh cnh OC c nh ta c mt hnh tr, khi tam gic OHC to thnh hnh (H).Tnh th tch ca phn hnh tr nm bn ngoi hnh (H).

(Cho )

Cu 1 (2)

a) Giai phng trinh 2x 5 =1

b) Giai bt phng trinh 3x 1 > 5

Cu 2 (2)

a) Giai h phng trinh

b) Chng minh rng

Cu 3 (2)

Cho phng trinh x2 2(m 3)x 1 = 0

a) Giai phng trinh khi m = 1

b) Tim m phng trinh co nghim x1 ; x2 ma biu thc

A = x12 x1x2 + x22 at gia tri nho nht? Tim gia tri nho nht o.

Cu 4 (3)

Cho tam giac ABC vung tai A. Ly B lam tm ve ng tron tm B ban kinh AB.Ly C lam tm ve ng tron tm C ban kinh AC, hai ng tron nay ct nhau tai im th 2 la D.Ve AM, AN ln lt la cac dy cung cua ng tron (B) va (C) sao cho AM vung goc vi AN va D nm gia M; N.

a) CMR: (ABC=(DBC

b) CMR: ABDC la t giac ni tip.

c) CMR: ba im M, D, N thng hang

d) Xac inh vi tri cua cac dy AM; AN cua ng tron (B) va (C) sao cho oan MN co dai ln nht.

Cu 5 (1) Giai H PT

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

GI GII

Cu 1 (2) a) Giai phng trinh 2x 5 = 1

b) Giai bt phng trinh 3x 1 > 5

ap an a) x = 3 ; b) x > 2

Cu 2 (2) a) Giai h phng trinh

b) Chng minh rng

ap an a) x = 2 ; y = 3

b) VT ==VP (pcm)

Cu 3 (2) Cho phng trinh x2 2(m 3)x 1 = 0

c) Giai phng trinh khi m = 1

d) Tim m phng trinh co nghim x1 ; x2 ma biu thc

A = x12 x1x2 + x22 at gia tri nho nht? Tim gia tri nho nht o.

ap an a) x1 = ; x2 =

e) Thy h s cua pt : a = 1 ; c = A 1 pt lun co 2 nghim

Theo viet ta co x1 + x2 =2(m 3) ; x1x2 = 1

Ma A=x12 x1x2 + x22 = (x1 + x2 )2 3x1x2 = 4(m 3)2 + 3 3

GTNN cua A = 3 m = 3

Cu 4 (3)

Hng dn

a) Co AB = DB; AC = DC; BC chung (ABC = (DBC (c-c-c)

b) (ABC = (DBC goc BAC =BDC = 900 ABDC la t giac ni tip

c) Co gocA1 = gocM1 ( (ABM cn tai B)

gocA4 = gocN2 ( (ACN cn tai C)

gocA1 = gocA4 ( cung phu A2;3 )

gocA1 = gocM1 =gocA4= gocN2

gocA2 = gocN1 ( cung chn cung AD cua (C) )

Lai co A1+A2 + A3 = 900 => M1 + N1 + A3 = 900

Ma (AMN vung tai A => M1 + N1 + M2 = 900

=> A3 = M2 => A3 = D1

(CDN cn tai C => N1;2 = D4

D2;3 + D1 + D4 =D2;3 + D1 + N1;2 = D2;3 + M2 + N1 + N2

= 900 + M2 + N1 + M1 ( M1 = N2) = 900 + 900 = 1800

M; D; N thng hang.

d) (AMN ng dang (ABC (g-g)

Ta co NM2 = AN2 +AM2 NM ln nht thi AN ; AM ln nht

Ma AM; AN ln nht khi AM; AN ln lt la ng kinh cua (B) va (C)

Vy khi AM; AN ln lt la ng kinh cua (B) va (C) thi NM ln nht.

Cu 5 (1): Giai H PT

Hng dn

EMBED Equation.3 (

T (2) t x +2y = a ; 2xy 1 = b (a:b 0)

Ta dc (2a-1)=(2b 1) ( ()(2= 0 ( a = b

( x = 3y + 1 thay vao (1) ta dc

2y2 y 1= 0 => y1 = 1 ; y2 = 1/2

=> x1 = 4 ; x2 = 1/2

Thy x2 + 2y2 = 1 < 0 (loai)Vy h co nghim (x; y) = (4 ; 1)

S gio dc v o to

Hng yn

( thi c 01 trang)

k thi tuyn sinh vo lp 10 thpt chuyn

Nm hc 2012 - 2013

Mn thi: Ton

(Dnh cho th sinh d thi cc lp chuyn: Ton, Tin)


Bi 1: (2 im)

a) Cho A =. Chng minh A l mt s t nhin.b) Gii h phng trnh

Bi 2: (2 im)

a) Cho Parbol (P): y = x2 v ng thng (d): y = (m +2)x m + 6. Tm m ng thng (d) ct Parabol (P) ti hai im phn bit c honh dng.

b) Gii phng trnh: 5 + x +

Bi 3: (2 im)

a) Tm tt c cc s hu t x sao cho A = x2 + x+ 6 l mt s chnh phng.

b) Cho x > 1 v y > 1. Chng minh rng :

Bi 4 (3 im)

Cho tam gic ABC nhn ni tip ng trn tm O, ng cao BE v CF. Tip tuyn ti B v C ct nhau ti S, gi BC v OS ct nhau ti M

a) Chng minh AB. MB = AE.BS

b) Hai tam gic AEM v ABS ng dng

c) Gi AM ct EF ti N, AS ct BC ti P. CMR NP vung gc vi BC

Bi 5: (1 im)

Trong mt gii bng c 12 i tham d, thi u vng trn mt lt (hai i bt k thi u vi nhau ng mt trn).a) Chng minh rng sau 4 vng u (mi i thi u ng 4 trn) lun tm c ba i bng i mt cha thi u vi nhau.b) Khng nh trn cn ng khng nu cc i thi u 5 trn?

HNG DN GII

Bi 1: (2 im)

a) Cho A =

t 2012 = a, ta c

b) t Ta c

nn

Bi 2:

a) ycbt tng ng vi PT x2 = (m +2)x m + 6 hay x2 - (m +2)x + m 6 = 0 c hai nghim dng phn bit.b) t t =

Bi 3:

a) x = 0, x = 1, x= -1 khng tha mn. Vi x khc cc gi tr ny, trc ht ta chng minh x phi l s nguyn.+) x2 + x+ 6 l mt s chnh phng nn x2 + x phi l s nguyn.

+) Gi s vi m v n c c nguyn ln nht l 1.

Ta c x2 + x = l s nguyn khi chia ht cho n2

nn chia ht cho n, v mn chia ht cho n nn m2 chia ht cho n v do m v n c c nguyn ln nht l 1, suy ra m chia ht cho n( mu thun vi m v n c c nguyn ln nht l 1). Do x phi l s nguyn.

t x2 + x+ 6 = k2

Ta c 4x2 + 4x+ 24 = 4 k2 hay (2x+1)2 + 23 = 4 k2 tng ng vi 4 k2 - (2x+1)2 = 23

=


EMBED Equation.DSMT4 .

Theo BT Csi

EMBED Equation.DSMT4 Bi 4

a) Suy ra t hai tam gic ng dng l ABE v BSM

b) T cu a) ta c (1)M MB = EM( do tam gic BEC vung ti E c M l trung im ca BC

Nn

C

Nn do

Suy ra (2)

T (1) v (2) suy ra hai tam gic AEM v ABS ng dng(pcm.)

c) D thy SM vung gc vi BC nn chng minh bi ton ta chng minh NP //SM.

+ Xt hai tam gic ANE v APB:

T cu b) ta c hai tam gic AEM v ABS ng dng nn ,

M ( do t gic BCEF ni tip)

Do hai tam gic ANE v APB ng dng nn

Li c ( hai tam gic AEM v ABS ng dng)

Suy ra nn trong tam gic AMS c NP//SM( nh l Talet o)

Do bi ton c chng minh.

Bi 5

a. Gia s kt lun cua bai toan la sai, tc la trong ba i bt ky thi co hai i a u vi nhau ri. Gia s ia gp cac i2, 3, 4, 5. Xet cac b(1; 6; i)vii {7; 8; 9;;12}, trong cac b nay phai co it nht mt cp a u vi nhau, tuy nhin1khng gp6hayinn6gp i vi mi i {7; 8; 9;;12}, v ly vi i6nh th a u hn4trn. Vy co pcm.

b. Kt lun khng ung. Chia12i thanh2nhom, mi nhom 6i. Trong mi nhom nay, cho tt ca cac i i mt a thi u vi nhau. Luc nay ro rang mi i a u5trn. Khi xet3i bt ky, phai co2i thuc cung mt nhom, do o2i nay a u vi nhau. Ta co phan vi du.

Co th giai quyt n gian hn cho cu a. nh sau:Do mi i a u4trn nn tn tai hai iA, Bcha u vi nhau. Trong cac i con lai, viAvaBchi u3trn vi ho nn tng s trn cuaA, Bvi cac i nay nhiu nht la6va do o, tn tai iCtrong s cac i con lai cha u vi caAvaB. Ta coA, B, Cla b ba i i mt cha u vi nhau.

S GIO DC V O TO

HNG YNK THI TUYN SINH VO LP 10 THPT

NM HC 2012 - 2013


PHN A: TRC NGHIM KHCH QUAN (2 im)

T cu 1 n cu 8, hy chn phng n ng v vit ch ci ng trc phng n vo bi lm

Cu 1: gi tr ca biu thc bng:

A.

B.

C.

D.

Cu 2: Biu thc c ngha khi:

A. x < 2B.

C.

D.

Cu 3: ng thng y = (2m 1)x + 3 song song vi ng thng y = 3x 2 khi:

A. m = 2B. m = - 2 C.

D.

Cu 4: H phng trnh c nghim (x;y) l:

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

Cu 5: Phng trnh x2 6x 5 = 0 c tng hai nghim l S v tch hai nghim l P th:

A. S = 6; P = -5B. S = -6; P = 5C. S = -5; P = 6D. S = 6; P = 5

Cu 6: th hm s y = -x2 i qua im:

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

Cu 7: Tam gic ABC vung ti A c AB = 4cm; AC = 3cm th di ng cao AH l:

A. cmB. cmC. cmD. cm

Cu 8: Hnh tr c bn knh y v chiu cao cng bng R th th tch l

A.

B.

C.

D.

PHN B: T LUN ( 8,0 im)

Bi 1: (1 im)

a) Tm x bit

b) Rt gn biu thc:

Bi 2: (1,5 im)

Cho ng thng (d): y = 2x + m 1

a) Khi m = 3, tm a im A(a; -4) thuc ng thng (d).

b) Tm m ng thng (d) ct cc trc ta Ox, Oy ln lt ti M v N sao cho tam gic OMN c din tch bng 1.

Bi 3: (1,5 im) Cho phng trnh x2 2(m + 1)x + 4m = 0 (1)

a) Gii phng trnh (1) vi m = 2.

b) Tm m phng trnh (1) c nghim x1, x2 tha mn (x1 + m)(x2 + m) = 3m2 + 12

Bi 4: (3 im) T im A bn ngoi ng trn (O), k cc tip tuyn Am, AN vi ng trn (M, N l cc tip im). ng thng d i qua A ct ng trn (O) ti hai im phn bit B,C (O khng thuc (d), B nm gia A v C). Gi H l trung im ca BC.

a) Chng minh cc im O, H, M, A, N cng nm trn mt ng trn,

b) Chng minh HA l tia phn gic ca .

c) Ly im E trn MN sao cho BE song song vi AM. Chng minh HE//CM.

Bi 5 (1,0 im) Cho cc s thc dng x, y , z tha mn x + y + z = 4.

Chng minh rng

HNG DN GII:

Phn trc nghim:

CuCuCuCuCuCuCuCu

BDADABBC

Phn t lun:

Bi 1:

a) Tm x bit


EMBED Equation.DSMT4 . Vy

b) Rt gn biu thc: . Vy

Bi 2:

a) Thay m = 3 vo phng trnh ng thng ta c: y = 2x + 2.

im A(a; -4) thuc ng thng (d) khi v ch khi: -4 = 2a + 2 suy ra a = -3.

b) Cho x = 0 suy ra y = m 1 suy ra: , cho y = 0 suy ra

suy ra

din tch tam gic OMN = 1 khi v ch khi: OM.ON = 2 khi v ch khi .

Khi v ch khi (m 1)2 = 4 khi v ch khi: m 1 = 2 hoc m 1 = -2 suy ra m = 3 hoc m = -1

Vy din tch tam gic OMN = 1 khi v ch khi m = 3 hoc m = -1.

Bi 3: Cho phng trnh x2 2(m + 1)x + 4m = 0 (1)

a) Gii phng trnh (1) vi m = 2.

b) Tm m phng trnh (1) c nghim x1, x2 tha mn (x1 + m)(x2 + m) = 3m2 + 12

HD:

a) Thay m = 2 vo phng trnh (1) ta c phng trnh:

x2 6x + 8 = 0 Khi v ch khi (x 2)(x 4) = 0 khi v ch khi x = 2 hoc x = 4

Vy vi m = 2 th phng trnh c 2 nghim x1 = 2 , x2 = 4.

b) Ta c vy phng trnh lun c nghim vi mi m.

p dng nh l Vi-et ta c:

(x1 + m)(x2 + m) = 3m2 + 12 khi v ch khi x1x2 + (x1 + x2) m - 2 m2 12 = 0. S khi v ch khi: 4m + m.2(m + 1) 2m2 12 = 0 khi v ch khi 6m = 12 khi v ch khi m= 2

Bi 5:

a) Theo tnh cht tip tuyn ct nhau ta c:

Do H l trung im ca BC nn ta c:

Do 3 im A, M, H, N, O thuc ng trn ng knh AO

b) Theo tnh cht hai tip tuyn ct nhau ta c: AM = AN

Do 5 im A, M, H, O, N cng thuc mt ng trn nn:

(gc ni tip chn hai cung bng nhau)

Do HA l tia phn gic ca

c) Theo gi thit AM//BE nn ( ng v) (1)

Do 5 im A, M, H, O, N cng thuc mt ng trn nn:

(gc ni tip chn cung MH) (2)

T (1) v (2) suy ra

Suy ra t gic EBNH ni tip

Suy ra

M (gc ni tip chn cung MB)

Suy ra:

Suy ra EH//MC.

Bi 5 (1,0 im) Cho cc s thc dng x, y , z tha mn x + y + z = 4.

Chng minh rng

Hng dn:

V x + y + z = 4 nn suy ra x = 4 (y + z)

Mt khc: do x dng. (*)

Thay x = 4 (y + z) vo (*) ta c:

Lun ng vi mi x, y, z dng, du bng xy ra khi v ch khi: y = z = 1, x = 2.

S GIO DC V O TO K THI TUYN SINH VO LP 10 THPT NM HC 2012

NG NAI

Kha ngy : 29 , 30 / 6 / 2012

Mn thi : TON HC

Thi gian lm bi : 120 pht

( ny c 1 trang , 5 cu )

Cu 1 : ( 1,5 im )

1 / Gii phng trnh : 7x2 8x 9 = 0 .

2 / Gii h phng trnh :

Cu 2 : ( 2,0 im )

1 / Rt gn cc biu thc :

2 / Cho x1 ; x2 l hai nghim ca phng trnh : x2 x 1 = 0 .

Tnh : .

Cu 3 : ( 1,5 im )

Trong mt phng vi h trc ta Oxy cho cc hm s :

y = 3x2 c th ( P ) ; y = 2x 3 c th l ( d ) ; y = kx + n c th l ( d1 ) vi k v n l nhng s thc .

1 / V th ( P ) .

2 / Tm k v n bit ( d1 ) i qua im T( 1 ; 2 ) v ( d1 ) // ( d ) .

Cu 4 : ( 1,5 im )

Mt tha t hnh ch nht c chu vi bng 198 m , din tch bng 2430 m2 . Tnh chiu di v chiu rng ca tha t hnh ch nht cho .

Cu 5 : ( 3,5 im )

Cho hnh vung ABCD . Ly im E thuc cnh BC , vi E khng trng B v E khng trng C . V EF vung gc vi AE , vi F thuc CD . ng thng AF ct ng thng BC ti G . V ng thng a i qua im A v vung gc vi AE , ng thng a ct ng thng DE ti im H .

1 / Chng minh .

2 / Chng minh rng t gic AEGH l t gic ni tip c ng trn .

3 / Gi b l tip tuyn ca ng trn ngoi tip tam gic AHE ti E , bit b ct ng trung trc ca on thng EG ti im K . Chng minh rng KG l tip tuyn ca ng trn ngoi tip tam gic AHE .

HNG DN GII:

Cu 1 : ( 1,5 im )

1 / Gii phng trnh : 7x2 8x 9 = 0 ( x1,2 = )

2 / Gii h phng trnh : ( x ; y ) = (1 ; 2 )

Cu 2 : ( 2,0 im )

1 / Rt gn cc biu thc :

2 / Cho x1 ; x2 l hai nghim ca phng trnh : x2 x 1 = 0 .

S = ; P =

Nn :

Cu 3 : ( 1,5 im )

1 / V th ( P ) .

2 / ( d1 ) // ( d ) nn k = 2 ; n 3 v i qua im T( 1 ; 2 ) nn x = 1 ; y = 2 . Ta c phng trnh : 2 = 1.2 + n n = 0

Cu 4 : ( 1,5 im )

Gi x ( m ) l chiu di tha t hnh ch nht ( 49,5 < x < 99 )

Chiu rng ca tha t hnh ch nht l : 99 x ( m )

Theo bi ta c phng trnh : x ( x 99 ) = 2430

Gii c : x1 = 54 ( nhn ) ; x2 = 45 ( loi )

Vy chiu di tha t hnh ch nht l 54 ( m )

Chiu rng ca tha t hnh ch nht l : 99 54 = 45 ( m )

Cu 5 : ( 3,5 im )

1 / Chng minh t gic AEFD ni tip

AEF DCE ( g g )

2 / Ta c ph vi

Ta c ph vi

M

Suy ra t gic AEFD ni tip ng trn ng knh HE

Gi I trung im ca HEI l tm ng trn ngoi tip t gic AEFD cng l ng trn ngoi tip

I nm trn ng trung trc EG IE = IG

V K nm trn ng trung trc EG KE = KG

Suy ra IEK =IGK ( c-c-c )

ti G ca ng trn ngoi tip

KG l tip tuyn ca ng trn ngoi tip

THI TUYN SINH VO LP 10 CHUYN TNH NG NAI

NM HC 2012 - 2013

Mn thi: Ton chung

Thi gian lm bi: 120 pht ( khng k thi gian giao )

( thi ny gm mt trang, c bn cu)

Cu 1: ( 2,5 im) .

1/ Gii cc phng trnh :

a/

b/

2/ Gii h phng trnh :

Cu 2 : ( 2,0 im) .

Cho parabol y = x2 (P) v ng thng y = mx (d), vi m l tham s.

1/ Tm cc gi tr ca m (P) v (d) ct nhau ti im c tung bng 9.

2/ Tm cc gi tr ca m (P) v (d) ct nhau ti 2 im, m khong cch gia hai im ny bng

Cu 3 : ( 2,0 im)

1/ Tnh :

2/ Chng minh : , bit rng .

Cu 4 : (3,5 im)

Cho tam gic ABC vung A, ng cao AH. V ng trn tm O, ng knh AH, ng trn ny ct cc cnh AB, AC theo th t ti D v E .

1/ Chng minh t gic BDEC l t gic ni tip c ng trn.

2/ Chng minh 3 im D, O, E thng hng.

3/ Cho bit AB = 3 cm, BC = 5 cm. Tnh din tch t gic BDEC.

--------HT------

THI TUYN SINH VO LP 10 CHUYN TNH NG NAI

NM HC 2012 - 2013

Mn thi: Ton ( mn chuyn)

Thi gian lm bi: 150 pht ( khng k thi gian giao )

( thi ny gm mt trang, c nm cu)

Cu 1. (1,5 im)

Cho phng trnh

EMBED Equation.DSMT4 ( vi )

Chng minh rng l mt nghim ca phng trnh cho.

Cu 2. (2,5 im)

Gii h phng trnh ( vi ).

Cu 3.(1,5 im)

Cho tam gic u MNP c cnh bng 2 cm. Ly n im thuc cc cnh hoc pha trong tam gic u MNP sao cho khong cch gia hai im tu ln hn 1 cm ( vi n l s nguyn dng). Tm n ln nht tho mn iu kin cho.

Cu 4. (1 im)

Chng minh rng trong 10 s nguyn dng lin tip khng tn ti hai s c c chung ln hn 9.

Cu 5. (3,5 im)

Cho tam gic ABC khng l tam gic cn, bit tam gic ABC ngoi tip ng trn (I). Gi D,E,F ln lt l cc tip im ca BC, CA, AB vi ng trn (I). Gi M l giao im ca ng thng EF v ng thng BC, bit AD ct ng trn (I) ti im N (N khng trng vi D), gii K l giao im ca AI v EF.

1) Chng minh rng cc im I, D, N, K cng thuc mt ng trn.

2) Chng minh MN l tip tuyn ca ng trn (I).

----------HT-----------

GII THI VO LP 10

CHUYN LNG TH VINH NG NAI

NM 2012 2013

Mn: Ton chung

-----------------

Cu 1: ( 2,5 im) .

1/ Gii cc phng trnh :

a/ (*) t

(*)( t2 t 20 = 0 ( (t1 = 5 (nhn) v t2 = - 4 ( loi)); Vi t = 5 => x2 = 5 ( x =

Vy phng trnh c hai nghim x = v x = -

b/ ( iu kin )

( x(x-3) = 0

( x = 0 ( loi) v x = 3 ( nhn).

Vy phng trnh c mt nghim x = 3.

2/ Gii h phng trnh :

T

(nhn)

Vy h phng trnh c 2 nghim (x; y):

Cu 2 : ( 2,0 im) .

1/ P.trnh honh giao im (P) v (d) :

V giao im . Vi y = 9 => m2 = 9 ( (m = 3 v m = -3)

Vy vi th (P) v (d) ct nhau ti im c tung bng 9.

2/ T cu 1 => (P) v (d) lun ct nhau ti hai im phn bit khi .

Khi giao im th nht l gc to O ( x = 0; y = 0), giao im th 2 l im A c ( x = m; y = m2).

Khong cch gia hai giao im : AO = (1)

t (1) ( (t1 = 3 ( nhn ) v t2 = - 2 ( loi))

Vi t1 = 3 ( m2 = 3 ,( ( nhn)

Vy vi th (P) ct (d) ti hai im c khong cch bng .

Cu 3 : ( 2,0 im)

1/ Tnh:

2/ Ta c:

V : (vi mi a, b ).

( theo gi thit)

( vi mi a, b )

Nn bt ng thc cui ng. Vy vi (pcm)

Cu 4 : (3,5 im)

1/ Ni H vi E .

+ ( v AH l ng knh), ( AH l ng cao)

=> (cng ph vi ) (1)

+ ( gc ni tip cng chn cung AE) (2)

T (1) v (2) => ADE = ACB =>T gic BDEC ni tip ng trn ( c gc i bng gc k b gc i)

2/ V => DE l ng knh => D, O, E thng hng (pcm).

3/ Ta c

+ vung c AH l ng cao:

=> (cm2)

(cm) ( cng l ng knh t O).

+ADE vABC c : A chung , ADE = ACB ( cu 1)

=> ADE ~ ABC (g.g) => t s din tch bng bnh phng t ng dng :

(

+ = 4,6176 (cm2)

---------HT---------

GII THI VO LP 10

CHUYN LNG TH VINH NG NAI

NM 2012 2013

Mn: Ton chuyn

-----------------

Cu 1: Phng trnh cho : ( vi ) ( (1)

Vi (

=>

Th x vo v phi ca (1) ta c:

= ( v phi bng v tri)

Vy l mt nghim ca phng trnh cho ( pcm)

Cu 2: H pt cho (

Thay x = 0, y = 0 th h khng tho . Thay x = -1 v y = -1 vo, h khng tho => (*)- Chia tng v ca hai phng trnh cho nhau : =>

Thay x = y, h pt c v phi bng nhau, v tri khc nhau (khng tho) =>) (**)

=> (3)

- Cng tng v (1) v (2) ca h ta c pt: 2(x+y)(x+1)(y+1) + 2xy = 0 (4)

( (x + y) ( x + y + xy + 1) + xy = 0 (

( ( (

- Vi x + y = 0 ( x = - y. Th vo h => -2y2 = 0 ( (y = 0 v x = 0) khng tho (*)

- Vi x + y +1 =0 ( x = -y - 1 th vo phng trnh (1) ca h ta c :

(

Vi y = - 2 => x = 1.Th vo h tho, vy c nghim 1: (x; y) = (1; - 2)

- Vi

Th x = y -6 vo pt (2) ca h :

(2) ( (

y2 - 4y - 6 = 0 (

2y +1 = 0 ( y3 =

T ba gi tr ca y trn ta tm c ba gi tr x tng ng:

Th cc gi tr (x; y) tm c vo h (tho).

Vy h phng trnh cho c 4 nghim ( x;y):

(1; -2), (

Cu 3. (Cch 1)

Tam gic u c cnh bng 2 cm th din tch bngcm2 , tam gic u c cnh bng 1 cm th din tch bng cm2 . Nu tam gic u c cnh > 1cm th din tch > cm2 Gi t l s tam gic u c cnh bng > 1cm cha c trong tam gic u c cnh 2 cm:

( vi t l s nguyn dng) => tmax = 3.

Theo nguyn l Drichen s c 1 trong t tam gic u c cnh > 1cm cha ti a 2 im tho mn khong cch gia hai im bt k lun > 1 cm.

Vy s im tho yu cu bi ton l : Vy nmax = 4

(Cch 2): Gii theo kin thc hnh hc

Nu ta chn 3 im 3 nh ca tam gic u cnh bng 2 cm v 3 ng trn ng knh 1 cm, cc ng trn ny tip xc vi nhau trung im mi cnh tam gic. => Cc im khc trong tam gic cch 3 nh > 1cm ch c th nm trong phn din tch cn li ca tam gic (ngoi phn din tch b ba hinh trn che ph), c gii hn bi 3 cung trn bn kinh 1 cm.

V 3 dy cung l 3 ng trung bnh ca tam gic c di 1 cm => khong cch gia hai im bt k nm trong phn din tch cn li ca tam gic lun 1 cm.

=> trong phn din tch ch ly c 1 im m khong cch n 3 nh ca tam gic lun > 1 cm.

Vy s im ln nht tho mn khong cch gia hai im bt k > 1cm l :

nmax = 3 + 1 = 4 im.

Cu 4. Gi a v b l hai s bt k trong 10 s nguyn dng lin tip vi a > b ( a; b nguyn dng) .

Gi n l c chung ca a v b, khi : a = n.x v b = n.y ( n, x, y l s nguyn dng).

V a > b => x > y =>

Vy trong 10 s nguyn dng lin tip khng tn ti hai s c c chung ln hn 9.

Cu 5.

1)Ni N v F, D v F.

- Xt ANF v

EMBED Equation.DSMT4 AFD c: AFN = ADF ( v AF l tt) v FAD chung =>ANFAFD (g.g) => (1)

- Xt AFI c: AFIF ( v AF tip tuyn, FI l bn knh) v FK AI ( v AF v AE tt chung v AI ni tm) => AFI vung ti F c FK l ng cao) => AK.AI = AF2 (2)

- Xt ANK v AID c:

+ IAD chung.

+ T (1) v (2) => AN.AD = AK.AI =>

=>ANKAID (c.g.c) =>NKA = IDN (3)

- T (3) => t gic DIKN ni tip t (v c gc i bng gc k b gc i)

=> cc im I,D,N,K cng thuc mt ng trn. (pcm).

2) Ta c IDDM ( DM l tip tuyn, DI l bn knh) v IKKM ( cu 1) => t gic DIKM ni tip ng trn ng knh MI. V 4 im D, I, K, N cng thuc mt ng trn ( cu 1) => hai ng trn ny cng ngoi tip DIK => hai ng trn trng nhau => N cng nm trn ng trn ng knh MI => = 900 .

V IN l bn knh ng trn (I), => MN l tip tuyn ca ng trn (I) ti tip im N. (pcm).

-----------HT----------

GI GII:

Cu 1c C = 1

Cu 2a ( 2;1) ; Cu 2b b = - 1

Cu 3a a = 1

Cu 3b A ( -1 ; 1 ) ; B (2 ; 4 )

Cu 4a1 ; nn pt lun co 2 nghim phn bit vi moi x

Cu 4 a2 => x1 + x2 = - 5 ; x1x2 = 3

Cu 4b

Goi x ( km/h) la vt xe II => vt xe I la x + 10 ( km/h ) ; x> 0

Th gian xe I i ht qg : (h)

Th gian xe II i ht qg : (h)

PT - = => x = 40

KL

Cu 5 : a

1. MH = 20 ( cm ) ; ME = 12 ( cm)

2. NPFE la h thang cn

b )

b1

b2 Tam giac ABC vung tai A co AH la g cao => AB2 = BH.BC (1)

Tam giac BHE g dang vi tam giac BDC => (2)

T (1) va (2) => AB2 = BD . BE

S GIO DC - O TOTNH NINH BNH

K THI TUYN SINH VO LP 10 THPT

NM HC 2012 2013

Mn thi: TON


thi gm 05 cu trong 01 trang

Cu 1: (2,0 im)1. Cho biu thc P = x + 5. Tnh gi tr biu thc P ti x = 1.

2. Hm s bc nht y = 2x + 1 ng bin hay nghch bin trn R? V sao?

3. Gii phng trnh x2 + 5x + 4 = 0

Cu 2: (2,5 im)1. Gii h phng trnh:

2. Cho biu thc Q = vi x > 0 v x 1.

a) Rt gn Q.b) Tnh gi tr ca Q vi x = 7 4.

Cu 3: (1,5 im)

Khong cch gia hai bn sng A v b l 30 km. Mt ca n i xui dng t bn A n bn B ri li ngc dng t bn B v bn A. Tng thi gian ca n i xui dng v ngc dng l 4 gi . Tm vn tc ca ca n khi nc yn lng, bit vn tc ca dng nc l 4 km/h.

Cu 4: (3,0 im)Cho ng trn tm O bn knh R. Mt ng thng d khng i qua O v ct ng trn ti hai im phn bit A v B. Trn d ly im M sao cho A nm gia M v B. T M k hai tip tuyn MC v MD vi ng trn (C, D l cc tip im).

1. Chng minh rng MCOD l t gic ni tip.2. Gi I l trung im ca AB. ng thng IO ct tia MD ti K. Chng minh rng KD. KM = KO. KI3. Mt ng thng i qua O v song song vi CD ct cc tia MC v MD ln lt ti E v F. Xc nh v tr ca M trn d sao cho din tch tam gic MEF t gi tr nh nht.Cu 5: (1,0 im)

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

------------------- Ht ----------------------HNG DN GII:Cu 1:

1) Thay x = 1 vo biu thc P c: P = x + 5 = 1 + 5 = 6.

2) Hm s ng bin trn R v a = 2 > 0

3)Ta thy a b + c = 1 5 + 4 = 0 nn pt c 2 nghim: x1 = 1; x2 = 4Cu 2:

1.

Vy h pt c nghim : x = 1 v y = 1.

2. Vi x > 0 v x 1, ta c:

a) Q

b) Vi

Suy ra: Cu 3:

Gi vn tc ca ca n khi nc yn lng l x(km/h) (k: )

Vn tc ca ca n khi xui dng: x + 4 (km/h)

Vn tc ca ca n khi ngc dng: x 4 (km/h)

Thi gian ca n i xui dng: (h)

Thi gian ca n i ngc dng: (h)

Tng thi gian ca n i xui dng v ngc dng l 4h nn ta c phng trnh:

+ = 4 x2 15x 16 = 0Gii phng trnh trn ta c:

Vy vn tc ca ca n khi nc yn lng l 16km/h

Cu 5: (cch 2)p dng bt ng thc Csi ta c::

du bng xy ra a = b

Tng t ta c :

du = xy ra a = b = c

S GIO DC - O TOTNH NINH BNH

K THI TUYN SINH VO LP 10 THPT CHUYN

Mn thi: TON

Ngy thi: 26 / 6 / 2012


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

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

b) Xc nh gi tr ca m phng trnh (1) c hai nghim x1, x2 sao cho nh nht. Tm nghim ca phng trnh (1) ng vi m va tm c.

Cu 2 (2,5 im).

1. Cho biu thc A=

a) Rt gn biu thc A.

b) Tm cc gi tr nguyn ca x biu thc A nhn gi tr nguyn.

2. Gii phng trnh:

Cu 3 (1,5 im). Mt ngi i xe p t A ti B, qung ng AB di 24 km. Khi i t B tr v A ngi tng vn tc thm 4 km/h so vi lc i, v vy thi gian v t hn thi gian i l 30 pht. Tnh vn tc ca xe p khi i t A ti B.

Cu 4 (3 im). Cho ABC nhn ni tip (O). Gi s M l im thuc on thng AB (MA, B); N l im thuc tia i ca tia CA sao cho khi MN ct BC ti I th I l trung im ca MN. ng trn ngoi tip AMN ct (O) ti im P khc A.

1. C MR cc t gic BMIP v CNPI ni tip c.

2. Gi s PB = PC. Chng minh rng ABC cn.

Cu 5 (1 im). Cho , tha mn x2 + y2 = 1. Tm GTLN ca :

HNG DN GII:2) Gii pt : K :

t

Ta c

T tm c nghim ca pt l x = 0

Cu 5 :

T

V thay vo

a v pt:

Dng iu kin c nghim ca pt bc hai


EMBED Equation.3 S GIO DC V O TO

K THI TUYN SINH VO 10 - THPT

TNH LO CAI

NM HC: 2012 2013

MN: TON

Thi gian: 120 pht (khng k thi gian giao )

Cu I: (2,5 im)

1. Thc hin php tnh:

2. Cho biu thc: P =

a) Tm iu kin ca a P xc nh

b) Rt gn biu thc P.

Cu II: (1,5 im)

1. Cho hai hm s bc nht y = -x + 2 v y = (m+3)x + 4. Tm cc gi tr ca m th ca hm s cho l:

a) Hai ng thng ct nhau

b) Hai ng thng song song.

2. Tm cc gi tr ca a th hm s y = ax2(a 0) i qua im M(-1; 2).

Cu III: (1,5 im)

1. Gii phng trnh x 2 7x 8 = 0

2. Cho phng trnh x2 2x + m 3 = 0 vi m l tham s. Tm cc gi tr ca m phng trnh c hai nghim x1; x2 tha mn iu kin

Cu IV: (1,5 im)

1. Gii h phng trnh

2. Tm m h phng trnh c nghim (x; y) tha mn iu kin x + y > 1.

Cu V: (3,0 im) Cho na ng trn tm O ng knh AB = 2R v tip tuyn Ax cng pha vi na ng trn i vi AB. T im M trn Ax k tip tuyn th hai MC vi na ng trn (C l tip im). AC ct OM ti E; MB ct na ng trn (O) ti D (D khc B).

a) Chng minh AMOC l t gic ni tip ng trn.

b) Chng minh AMDE l t gic ni tip ng trn.

c) Chng mnh

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

HNG DN GII:

Cu I: (2,5 im)

1. Thc hin php tnh:

2. Cho biu thc: P =

a) Tm iu kin ca a P xc nh: P xc nh khi

b) Rt gn biu thc P.

P ==

=

==

Vy vi th P =

Cu II: (1,5 im)

1. Cho hai hm s bc nht y = -x + 2 v y = (m+3)x + 4. Tm cc gi tr ca m th ca hm s cho l:

a) hm s y = (m+3)x + 4 l hm s bc nht th m + 3 0 suy ra m -3.

th ca hai hm s cho l hai ng thng ct nhau a a

-1 m+3m -4

Vy vi m -3 v m -4 th th ca hai hm s cho l hai ng thng ct nhau.

b) th ca hm s cho l Hai ng thng song song

tha mn iu kin m -3

Vy vi m = -4 th th ca hai hm s cho l hai ng thng song song.

2. Tm cc gi tr ca a th hm s y = ax2(a 0) i qua im M(-1; 2).

V th hm s y = ax2(a 0) i qua im M(-1; 2) nn ta thay x = -1 v y = 2 vo hm s ta c phng trnh 2 = a.(-1)2 suy ra a = 2 (tha mn iu kin a 0)

Vy vi a = 2 th th hm s y = ax2(a 0) i qua im M(-1; 2).

Cu III: (1,5 im)

1. Gii phng trnh x 2 7x 8 = 0 c a b + c = 1 + 7 8 = 0 suy ra x1= -1 v x2= 8

2. Cho phng trnh x2 2x + m 3 = 0 vi m l tham s. Tm cc gi tr ca m phng trnh c hai nghim x1; x2 tha mn iu kin .

phng trnh c hai nghim x1; x2 th 0 ( 1 m + 3 0 ( m 4

Theo viet ta c: x1+ x2 =2 (1) v x1. x2 = m 3 (2)

Theo u bi: = 6 (3)

Th (1) v (2) vo (3) ta c: (m - 3)(2)2 2(m-3)=6 ( 2m =12 ( m = 6 Khng tha mn iu kin m 4 vy khng c gi tr no ca m phng trnh c hai nghim x1; x2 tha mn iu kin .

Cu IV: (1,5 im)

1. Gii h phng trnh

EMBED Equation.DSMT4 2. Tm m h phng trnh c nghim (x; y) tha mn iu kin x + y > 1.

M x + y > 1 suy ra m + m + 1 > 1 2m > 0 m > 0.

Vy vi m > 0 th h phng trnh c nghim (x; y) tha mn iu kin x + y > 1.

Cu V: (3,0 im) Cho na ng trn tm O ng knh AB = 2R v tip tuyn Ax cng pha vi na ng trn i vi AB. T im M trn Ax k tip tuyn th hai MC vi na ng trn (C l tip im). AC ct OM ti E; MB ct na ng trn (O) ti D (D khc B).

a) Chng minh AMCO l t gic ni tip ng trn.

b) Chng minh AMDE l t gic ni tip ng trn.

c) Chng mnh

Gii.

a) nn t gic AMCO ni tip

b) . T gic AMDE c

D, E cng nhn AM di cng mt gc 900Nn AMDE ni tip

c) V AMDE ni tip nn

V AMCO ni tip nn

Suy ra

S GIO DC V O TO

GIA LAI

chnh thc

Ngy thi: 26/6/2012K THI TUYN SINH VO LP 10 CHUYN

Nm hc 2012 2013

Mn thi: Ton (khng chuyn)


Cu 1. (2,0 im)

Cho biu thc , vi

a. Rt gn biu thc Q

b. Tm cc gi tr nguyn ca x Q nhn gi tr nguyn.

Cu 2. (1,5 im)

Cho phng trnh , vi x l n s,

a. Gii phng trnh cho khi m ( 2

b. Gi s phng trnh cho c hai nghim phn bit v . Tm h thc lin h gia v m khng ph thuc vo m.

Cu 3. (2,0 im)

Cho h phng trnh , vi

a. Gii h cho khi m ( 3

b. Tm iu kin ca m phng trnh c nghim duy nht. Tm nghim duy nht .

Cu 4. (2,0 im)

Cho hm s c th (P). Gi d l ng thng i qua im M(0;1) v c h s gc k.

a. Vit phng trnh ca ng thng d

b. Tm iu kin ca k t d ct th (P) ti hai im phn bit.

Cu 5. (2,5 im)

Cho tam gic nhn ABC (AB < AC < BC) ni tip trong ng trn (O). Gi H l giao im ca hai ng cao BD v CE ca tam gic ABC

a. Chng minh t gic BCDE ni tip trong mt ng trn

b. Gi I l im i xng vi A qua O v J l trung im ca BC. Chng minh rng ba im H, J, I thng hng

c. Gi K, M ln lt l giao im ca AI vi ED v BD. Chng minh rng

HNG DN GII:

Cu 1.

a.

EMBED Unknown

EMBED Unknown

EMBED Unknown

EMBED UnknownVy

b.

Q nhn gi tr nguyn

khi khi 2 chia ht cho

EMBED Unknown i chiu iu kin th

Cu 2. Cho pt , vi x l n s,

a. Gii phng trnh cho khi m ( 2

Ta c phng trnh

EMBED Unknown

EMBED UnknownVy phng trinh c hai nghim v

b.

Theo Vi-et, ta c

EMBED Unknown

EMBED Unknown

Suy ra

EMBED Unknown

Cu 3. Cho h phng trnh , vi

a. Gii h cho khi m ( 3

Ta c h phng trnh

Vy h phng trnh c nghim vi

b. iu kin c nghim ca phng trnh

EMBED Unknown

EMBED Unknown

EMBED UnknownVy phng trnh c nghim khi v

Gii h phng trnh khi

EMBED Unknown

EMBED Unknown . Vy h c nghim (x; y) vi

Cu 4.

a. Vit phng trnh ca ng thng d

ng thng d vi h s gc k c dng

ng thng d i qua im M(0; 1) nn

EMBED UnknownVy

b.

Phng trnh honh giao im ca (P) v d

EMBED Unknown, c

d ct (P) ti hai im phn bit khi

EMBED Unknown

EMBED Unknown

EMBED Unknown

EMBED Unknown

Cu 5.

a. BCDE ni tip

Suy ra BCDE ni tip ng trn ng knh BC

b. H, J, I thng hng

IB ( AB; CE ( AB (CH ( AB)

Suy ra IB // CH

IC ( AC; BD ( AC (BH ( AC)

Suy ra BH // IC

Nh vy t gic BHCI l hnh bnh hnh

J trung im BC ( J trung im IH

Vy H, J, I thng hng

c.

cng b vi gc ca t gic ni tip BCDE

v (ABI vung ti B

Suy ra , hay

Suy ra (AEK vung ti K

Xt (ADM vung ti M (suy t gi thit)

DK ( AM (suy t chng minh trn)www.VNMATH.Nh vy

S GIO DC V O TOK THI TUYN SINH LP 10 THPT QUNG NINH

NM HC 2012 2013

MN: TON(Dng cho mi th sinh d thi)Ngy thi: 28/6/2012Thi gian lm bi: 120 pht (Khng k thi gian giao )

( thi ny c 01 trang)Cu I. (2,0 im)

1) Rt gn cc biu thc sau:

a) A =

b) B = vi x ( 0, x ( 1

2. Gii h phng trnh:

Cu II. (2,0 im)

Cho phng trnh (n x): x2 ax 2 = 0 (*)

1. Gii phng trnh (*) vi a = 1.

2. Chng minh rng phng trnh (*) c hai nghim phn bit vi mi gi tr ca a.

3. Gi x1, x2 l hai nghim ca phng trnh (*). Tm gi tr ca a biu thc:

N= c gi tr nh nht.

Cu III. (2,0 im)Gii bi ton bng cch lp phng trnh hoc h phng trnh.

Qung ng sng AB di 78 km. Mt chic thuyn my i t A v pha B. Sau 1 gi, mt chic ca n i t B v pha A. Thuyn v ca n gp nhau ti C cch B 36 km. Tnh thi gian ca thuyn, thi gian ca ca n i t lc khi hnh n khi gp nhau, bit vn

tc ca ca n ln hn vn tc ca thuyn l 4 km/h.

Cu IV. (3,5 im)Cho tam gic ABC vung ti A, trn cnh AC ly im D (D A, D C). ng trn (O) ng knh DC ct BC ti E (E C).

1. Chng minh t gic ABED ni tip.

2. ng thng BD ct ng trn (O) ti im th hai I. Chng minh ED l tia phn gic ca gc AEI.

3. Gi s tg ABC Tm v tr ca D trn AC EA l tip tuyn ca ng trn ng knh DC.

CuV. (0.5 im) Gii phng trnh:

HNG DN GII:Cu IV :

c. EA l tip tuyn ca .Trn, . knh CD th gc E1 = gc C1 (1)

M t gic ABED ni tip nn gc E1 = gc B1 (2)

T (1) v (2) gc C1 = gc B1 ta li c gc BAD chung nn

( (ABD ( (ACB (

EMBED Equation.3 ( AB2 = AC.AD ( AD = ( I )

Theo bi ra ta c : tan (ABC) = = nn ( II )

T (I) v (II) ( AD = .

Vy AD = th EA l tip tuyn ca T, knh CD

Cu V:

Gii phng trnh:

t ; K v, t 0

( ( ... ( ( hoc t=2

Nu t= 2 th ( x = 3 (TM)

Nu t = v th ( x = 3,5S GIO DC V O TOK THI TUYN SINH VO LP 10 THPT

KHNH HANM HC 2011 - 2012

Mn thi: TON

Ngy thi : 21/06/2011


Bi 1( 2 im)

1) n gin biu thc: A

2) Cho biu thc:

Rt gn P v chng t P 0

Bi 2( 2 im)

1) Cho phng trnh bc hai x2 + 5x + 3 = 0 c hai nghim x1; x2. Hy lp mt phng trnh bc hai c hai nghim (x12 + 1 ) v ( x22 + 1).

2) Gii h phng trnh

Bi 3( 2 im)

Qung ng t A n B di 50km.Mt ngi d nh i xe p t A n B vi vn tc khng i.Khi i c 2 gi,ngi y dng li 30 pht ngh.Mun n B ng thi gian nh,ngi phi tng vn tc thm 2 km/h trn qung ng cn li.Tnh vn tc ban u ca ngi i xe p.Bi 4( 4 im)

Cho tam gic ABC c ba gc nhn v H l trc tm.V hnh bnh hnh BHCD.ng thng i qua D v song song BC ct ng thng AH ti E.

1) Chng minh A,B,C,D,E cng thuc mt ng trn

2) Chng minh

3) Gi O l tm ng trn ngoi tip tam gic ABC v M l trung im ca BC,ng thng AM ct OH ti G.Chng minh G l trng tm ca tam gicABC.

4) Gi s OD = a.Hy tnh di ng trn ngoi tip tam gic BHC theo a

HNG DN GII:Bi 1

3) A

4)

Bi 2 x2 + 5x + 3 = 0

1) C

Nn pt lun c 2 nghim phn bit

x1+ x2 = - 5 ; x1x2 = 3

Do S = x12 + 1 + x22 + 1 = (x1+ x2)2 - 2 x1x2 + 2 = 25 6 + 2 = 21

V P = (x12 + 1) (x22 + 1) = (x1x2)2 + (x1+ x2)2 - 2 x1x2 + 1 = 9 + 20 = 29

Vy phng trnh cn lp l x2 21x + 29 = 0

2)K

Vy HPT c nghim duy nht ( x;y) = ( 2;3)

Bi 3:Gi x(km/h) l vtc d nh; x > 0; c 30 pht = (h)

Th gian d nh:

Qung ng i c sau 2h: 2x (km)

Qung ng cn li: 50 2x (km)

Vn tc i trn qung ng cn li: x + 2 ( km/h)

Th gian i qung ng cn li:

Theo bi ta c PT:

Gii ra ta c: x = 10 (tha K bi ton)

Vy Vn tc d nh: 10 km/h

Bi 4:Gii cu c)

V BHCD l HBH nn H,M,D thng hng

Tam gic AHD c OM l TBnh => AH = 2 OM

V AH // OM

2 tam gic AHG v MOG c

( )

Hay AG = 2MG

Tam gic ABC c AM l trung tuyn; G AM

Do G l trng tm ca tam gic ABC

d) ( v BHCD l HBH)

c B;D;C ni tip (O) bn knh l a

Nn tam gic BHC cng ni tip (K) c bn knh a

Do C (K) = ( VD)

S GIO DC-O TO

K THI TUYN SINH VO 10 THPT NM 2012

BNH NH

Kha ngy 29 thng 6 nm 2012

Mn thi: TON

Ngy thi: 30/6/2012

Thi gian lm bi: 120 pht (khng k thi gian giao )Bi 1: (3, 0 im)

Hc sinh khng s dng my tnh b ti

a) Gii phng trnh: 2x 5 = 0

b) Gii h phng trnh:

c) Rt gn biu thc vi

d) Tnh gi tr ca biu thc

Bi 2: (2, 0 im)

Cho parabol (P) v ng thng (d) c phng trnh ln lt l v

(m l tham s, m 0).

a) Vi m = 1 , tm ta giao im ca (d) v (P).

b) Chng minh rng vi mi m 0 ng thng (d) lun ct parabol (P) ti hai im phn bit.

Bi 3: (2, 0 im)

Qung ng t Quy Nhn n Bng Sn di 100 km. Cng mt lc, mt xe my khi hnh t Quy Nhn i Bng Sn v mt xe t khi hnh t Bng Sn i Quy Nhn. Sau khi hai xe gp nhau, xe my i 1 gi 30 pht na mi n Bng Sn. Bit vn tc hai xe khng thay i trn sut qung ng i v vn tc ca xe my km vn tc xe t l 20 km/h. Tnh vn tc mi xe.

Bi 4: (3, 0 im) Cho ng trn tm O ng knh AB = 2R. Gi C l trung im ca OA, qua C k dy MN vung gc vi OA ti C. Gi K l im ty trn cung nh BM, H l giao im ca AK v MN.

a) Chng minh t gic BCHK l t gic ni tip.

b) Chng minh AK.AH = R2

c) Trn KN ly im I sao cho KI = KM, chng minh NI = KB.

HNG DN GII:

Bi 1:a) 2x 5 = 0

b)

c)

d)

Bi 2:a) Vi v ln lt tr thnh .

Lc phng trnh honh giao im ca v l: c nn c hai nghim l .

Vi

Vi

Vy ta giao im ca v l v .

b) Phng trnh honh giao im ca v l: .

Vi th l phng trnh bc hai n x c vi mi m. Suy ra lun c hai nghim phn bit vi mi m. Hay vi mi m 0 ng thng (d) lun ct parabol (P) ti hai im phn bit.

Bi 3:i

t a im:

- Quy Nhn l A

- Hai xe gp nhau l C

- Bng Sn l B

Gi vn tc ca xe my l . K: .

Suy ra:

Vn tc ca t l .

Qung ng BC l:

Qung ng AC l :

Thi gian xe my i t A n C l:

Thi gian t my i t B n C l:

V hai xe khi hnh cng lc, nn ta c phng trnh:

Gii pt:

Phng trnh c hai nghim phn bit: (tha mn K)

(khng tha mn K)

Vy vn tc ca xe my l .

Vn tc ca t l .

Bi 4:

a) T gic BCHK l t gic ni tip.

Ta c: (gc ni tip chn na ng trn) hay

T gic BCHK c

t gic BCHK l t gic ni tip.

b)

D thy

c)

c cn ti

c MC l ng cao ng thi l ng trung tuyn (gt) cn ti

l tam gic u

l tam gic cn (KI = KM) c nn l tam gic u .

D thy cn ti B c nn l tam gic u

Gi E l giao im ca AK v MI.

D thy KB // MI (v c cp gc v tr so le trong bng nhau) mt khc nn ti E .

Ta c: mt khc (cng chn )

hay T (pcm)S GIO DC V O TO THI TUYN SINH LP 10 THPT

BC GIANG NM HC 2012 2013

Mn thi : Ton

Thi gian : 120 pht khng k thi gian giao

Ngy thi 30 thng 6 nm 2012

Cu 1. (2 im) 1.Tnh

2 .Xc nh gi tr ca a,bit th hm s y = ax - 1 i qua im M(1;5)

Cu 2: (3 im)

1.Rt gn biu thc: vi a>0,a

2.Gii h pt:

3. Chng minh rng pt: lun c nghim vi mi gi tr ca m.

Gi s x1,x2 l 2 nghim ca pt cho,tm gi tr nh nht ca biu thc

Cu 3: (1,5 im)

Mt t ti i t A n B vi vn tc 40km/h. Sau 2 gi 30 pht th mt t taxi cng xut pht i t A n B vi vn tc 60 km/h v n B cng lc vi xe t ti.Tnh di qung ng AB.

Cu 4: (3 im)

Cho ng trn (O) v mt im A sao cho OA=3R. Qua A k 2 tip tuyn AP v AQ ca ng trn (O),vi P v Q l 2 tip im.Ly M thuc ng trn (O) sao cho PM song song vi AQ.Gi N l giao im th 2 ca ng thng AM v ng trn (O).Tia PN ct ng thng AQ ti K.

1.Chng minh APOQ l t gic ni tip.

2.Chng minh KA2=KN.KP

3.K ng knh QS ca ng trn (O).Chng minh tia NS l tia phn gic ca gc.

4. Gi G l giao im ca 2 ng thng AO v PK .Tnh di on thng AG theo bn knh R.

Cu 5: (0,5im)

Cho a,b,c l 3 s thc khc khng v tho mn:

Hy tnh gi tr ca biu thc

HNG DN CHM (tham kho)

CuNi dungim

11

KL:

1

2Do th hm s y = ax-1 i qua M(1;5) nn ta c a.1-1=5a=6

KL:1

21

KL:

0,5

0,5

2

KL:

1

3 Xt Pt:

Vy pt lun c nghim vi mi m

Theo h thc Viet ta c

Theo bi

Vy minB=1 khi v ch khi m = -1

KL:

0,25

0,25

0,5

3Gi di qumg ng AB l x (km) x>0

Thi gian xe ti i t A n B l h

Thi gian xe Taxi i t A n B l :h

Do xe ti xut pht trc 2h30pht = nn ta c pt

Gi tr x = 300 c tho mn K

Vy di qung ng AB l 300 km.0,25

0,25

0,25

0,25

0,25

0,25

41Xt t gic APOQ c

(Do AP l tip tuyn ca (O) P)

(Do AQ l tip tuyn ca (O) Q)

,m hai gc ny l 2 gc i nn t gic APOQ l t gic ni tip

0,75

2Xt AKN v PAK c l gc chung

( Gc ntcng chn cung NP)

M (so le trong ca PM //AQ

AKN ~ PKA (gg) (pcm)

0,75

3K ng knh QS ca ng trn (O)

Ta c AQQS (AQ l tt ca (O) Q)

M PM//AQ (gt) nn PMQS

ng knh QS PM nn QS i qua im chnh gia ca cung PM nh

EMBED Equation.DSMT4 (hai gc nt chn 2 cung bng nhau)

Hay NS l tia phn gic ca gc PNM0,75

4Chng minh c AQO vung Q, c QGAO(theo Tnh cht 2 tip tuyn ct nhau)

Theo h thc lng trong tam gic vung ta c

Do KNQ ~KQP (gg) m nn AK=KQ

Vy APQ c cc trung tuyn AI v PK ct nhau G nn G l trng tm

0,75

5Ta c:

*TH1: nu a+ b=0

Ta c ta c

Cc trng hp cn li xt tng t

Vy

0,25

0,25

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

YN BI NM HC 2012 2013

Mn thi : TON

Thi gian : 120 pht (khng k thi gian giao )

Kha ngy 23 thng 6 nm 2012

( thi c 01 trang, gm 05 cu)

Cu 1: (2,0 im)

1. Cho hm s y = x + 3 (1)

a. Tnh gi tr ca y khi x = 1

b. V th ca hm s (1)

2. Gii phng trnh: 4xeq \l(\o\ac(2, )) 7x + 3 = 0

Cu 2: (2,0 im)

Cho biu thc M = eq \s\don1(\f(1,3)) + eq \s\don1(\f(,3+eq \l(\r(,x)))) eq \s\don1(\f(x+9,x9))

1. Tm iu kin ca x biu thc M c ngha. Rt gn biu thc M.

2. Tm cc gi tr ca x M > 1

Cu 3: (2,0 im)

Mt i th m phi khai thc 260 tn than trong mt thi hn nht nh. Trn thc t, mi ngy i u khai thc vt nh mc 3 tn, do h khai thc c 261 tn than v xong trc thi hn mt ngy.

Hi theo k hoch mi ngy i th phi khai thc bao nhiu tn than?

Cu 4: (3,0 im)

Cho na ng trn tm O, ng knh AB = 12 cm. Trn na mt phng b AB cha na ng trn (O) v cc tia tip tuyn Ax, By. M l mt im thuc na ng trn (O), M khng trng vi A v B. AM ct By ti D, BM ct Ax ti C. E l trung im ca on thng BD.

1. Chng minh: AC . BD = ABeq \l(\o\ac(2, )).

2. Chng minh: EM l tip tuyn ca na ng trn tm O.

3. Ko di EM ct Ax ti F. Xc nh v tr ca im M trn na ng trn tm O sao cho din tch t gic AFEB t gi tr nh nht? Tm gi tr nh nht .

Cu 5: (1,0 im)

Tnh gi tr ca biu thc T = xeq \l(\o\ac(2, )) + yeq \l(\o\ac(2, )) + zeq \l(\o\ac(2, )) 7 bit:

x + y + z = 2eq \l(\r(,x34)) + 4eq \l(\r(,y21)) + 6eq \l(\r(,z4)) + 45

S GIO DC V O TO

K THI TUYN SINH L 10 THPT

LM NG

MN THI : TON

Kha ngy : 26 thng 6 nm 2012 ( thi gm 01 trang)

Thi gian lm bi : 120 pht

Cu 1: (0,75) Tnh :

Cu 2: (0,75) Gii h phng trnh :

Cu 3: (0,75) Cho tam gic ABC vung ti A, ng cao AH. Bit BH = 9cm, Ch = 16cm.

Tnh di cc on thng AH, BH, AC.

Cu 4: (0,75) Cho hai ng thng (d) : y = (m-3)x + 16 (m3) v (d): y = x + m2.

Tm m (d) v (d) ct nhau ti mt im trn trc tung

Cu 5: (0,75) Cho AB l dy cung ca ng trn tm O bn knh 12cm. Bit AB = 12cm . Tnh

din tch hnh qut to bi hai bn knh OA, OB v cung nh AB.

Cu 6: (1) Cho hm s y = ax2 (a 0) c th (P).

a) Tm a bit (P) i qua im A(2;4)

b) Tm k ng thng (d) : y = 2x + k lun ct (P) ti 2 im phn bit.

Cu 7: (0,75) Hnh nn c th th tch l 320cm3, bn knh ng trn l 8cm. Tnh din tch ton

phn ca hnh nn .

Cu 8: (1) Cho ng trn (O) ng knh AB, M l trung im ca OA. Qua M v dy cung CD

vung gc vi OA.

a) Chng minh t gic ACOD l hnh thoi .

b) Tia CO ct BD ti I. Chng minh t gic DIOM ni tip.

Cu 9: (1) Hai i cng nhn cng o mt con mng . Nu h cng lm th trong 8 gi xong

vic. Nu h lm ring th i A hon thnh cng vic nhanh hn i B 12 gi. Hi nu

lm ring th mi i phi lm trong bao nhiu gi mi xong vic.

Cu 10: (0,75) Rt gn :

Cu 11: (1) Cho phng trnh : x2 2(m-2)x - 3m2 +2 = 0 (x l n, m l tham s )

Tm m phng trnh c 2 nghim x1; x2 tha : x1(2-x2) +x2(2-x1) = -2

Cu 12: (0,75) Cho na ng trn (O) ng knh AB, v cc tip tuyn Ax v By cng pha vi

na ng trn , M l im chnh gia cung AB, N l mt im thuc on OA

. ng thng vung gc vi MN ti M ct Ax v By ln lt ti C v D.

Chng minh : AC = BN

S GIO DC V O TO K THI TUYN SINH VO LP 10 NM HC 2012-2013

QUNG NGI Mn thi: Ton (khng chuyn)

Thi gian lm bi: 120 pht (khng k thi gian giao )Bi 1: (1,5 im)

1/ Thc hin php tnh:

2/ Gii h phng trnh:

3/ Gii phng trnh:

Bi 2: (2,0 im)

Cho parapol v ng thng (m l tham s).

1/ Xc nh tt c cc gi tr ca m song song vi ng thng .

2/ Chng minh rng vi mi m, lun ct ti hai im phn bit A v B.

3/ K hiu l honh ca im A v im B. Tm m sao cho .

Bi 3: (2,0 im)

Hai xe t cng i t cng Dung Qut n khu du lch Sa Hunh, xe th hai n sm hn xe th nht l 1 gi. Lc tr v xe th nht tng vn tc thm 5 km mi gi, xe th hai vn gi nguyn vn tc nhng dng li ngh mt im trn ng ht 40 pht, sau v n cng Dung Qut cng lc vi xe th nht. Tm vn tc ban u ca mi xe, bit chiu di qung ng t cng Dung Qut n khu du lch Sa Hunh l 120 km v khi i hay v hai xe u xut pht cng mt lc.

Bi 4: (3,5 im)

Cho ng trn tm O ng knh AB = 2R v C l mt im nm trn ng trn sao cho CA > CB. Gi I l trung im ca OA. V ng thng d vung gc vi AB ti I, ct tia BC ti M v ct on AC ti P; AM ct ng trn (O) ti im th hai K.

1/ Chng minh t gic BCPI ni tip c trong mt ng trn.

2/ Chng minh ba im B, P, K thng hng.

3/ Cc tip tuyn ti A v C ca ng trn (O) ct nhau ti Q. Tnh din tch ca t gic QAIM theo R khi BC = R.

Bi 5: (1,0 im)

Cho tha mn . Tm gi tr nh nht ca biu thc .

-------------- HT --------------

HNG DN GII:

Bi 1:

1/

2/

3/ Phng trnh c nn c hai nghim l: .

Bi 2:

1/ ng thng song song vi ng thng khi

2/ Phng trnh honh giao im ca v l l phng trnh bc hai c vi mi m nn lun c hai nghim phn bit vi mi m. Do lun ct ti hai im phn bit A v B vi mi m.

3/ Cch 1: K hiu l honh ca im A v im B th l nghim ca phng trnh .

Gii phng trnh .

Phng trnh c hai nghim l .

Do

Cch 2: K hiu l honh ca im A v im B th l nghim ca phng trnh . p dng h thc Viet ta c: do

Bi 3:

Gi vn tc ban u ca xe th nht l x (km/h), xe th hai l y (km/h). K: x > 0; y > 0.Thi gian xe th nht i t cng Dung Qut n khu du lch Sa Hunh l .

Thi gian xe th hai i t cng Dung Qut n khu du lch Sa Hunh l .

V xe th hai n sm hn xe th nht l 1 gi nn ta c phng trnh:

Vn tc lc v ca xe th nht l x+ 5 (km/h).

Thi gian xe th nht v t khu du lch Sa Hunh n cng Dung Qut .

Thi gian xe th hai v t khu du lch Sa Hunh n cng Dung Qut .

V xe th hai dng li ngh ht , sau v n cng Dung Qut cng lc vi xe th nht nn ta c phng trnh: .

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

Gii hpt:

.

Phng trnh c hai nghim phn bit: (tha mn K)

(khng tha mn K)

Thay vo pt (1) ta c: (tha mn K).

Vy vn tc ban u ca xe th nht l 40 km/h, xe th hai l 60 km/h.

Bi 4:(Bi gii vn tt)a) T gic BCPI ni tip (hs t cm).

b) D thy MI v AC l hai ng cao ca l trc tm

ca l ng cao th ba .

Mt khc (gc ni tip chn na ng trn) .

T (1) v (2) suy ra ba im B, P, Q thng hng.

c)

Khi BC = R d thy tam gic OBC l tam gic u suy ra

M (gc to bi tia tip tuyn v gc ni tip cng chn ) do .

D thy tam gic QAC cn ti Q (QA = QC) c nn l tam gic u .

D thy

Trong tam gic vung ta c .

Ta chng minh c t gic QAIM l hnh thang vung .

Do (vdt).

Bi 5:

Cch 1: Ta c

V do .

Mt khc (v )

Do . Du = xy ra khi .

T

Lc . Vy khi .

Cch 2: Vi ta c

Do .

Du = xy ra khi .

T

Vy khi .

Cch 3:

Vi v

Ta c

Du = xy ra khi . Vy khi .

UBND tnh bc ninh

S gio dc v o to

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