K THI CHN HC SINH GII CP TNH THCS

Nm hc 2013-2014

Mn: TON

Thi gian lm bi : 150 pht

(khng k thi gian pht )

15-3-2014

Cho a thc

a. Hy phn tch a thc thnh tch cc nhn t.

b. Chng t rng nu l s nguyn th lun chia ht cho 5.

Bi 3: (4,0 im)

Cho .

a. Chng minh rng :

b. Chng minh rng :

Bi 4: (4,0 im)

Cho h phng trnh

a. Gii h phng trnh.

b. Tm mt phng trnh bc nht hai n nhn mt nghim l nghim

ca h phng trnh cho v mt nghim l (0,0).

Bi 5: (5,0 im)

Cho ng trn tm O ng knh AB . Ly mt im M trn ng trn

sao cho . Tip tuyn vi ng trn ti im A v im M ct nhau ti C,

CM ct AB ti D.

a. Chng minh rng BM song song OC.

b. Tnh din tch tam gic ACD.

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

S GIO DC V O TO

AN GIANG

CHNH THC

SBD

PHNG

Bi 1: (3,0 im)

Tnh tng:

Bi 2: (4,0 im)

S GIO DC V O TO HNG DN CHM HC SINH GII CP TNH

AN GIANG Nm hc 2013 2014

MN TON

A.P N

P N

Vy

Theo trn lun chia ht cho 5 vi

mi s nguyn x

Mt khc

nn l tch ca 5 s nguyn lin tip chia ht cho 5

vy lun chia ht cho 5

Chng minh

Xt

Bi

3a Do

Vy

Hay

Do

Ta c

Xt

Bi

3b Vy

Du bng xy ra khi

Bi

Bi 1

Bi

2a

Bi

2b

im

3,0

im

2,0

im

2,0

im

2,0

im

du bng xy ra khi

2,0

im

Phng trnh bc nht hai n c dng

Phng trnh c nghim

Phng trnh c nghim

Ta c nhiu phng trnh nh th nn c th chn

vy mt phng trnh tha bi l

C

M

D

3,0

im

- Theo bi ta c , tam gic AMB vung ti M (do

gc ni tip chn na ng trn) (*)

- Tam gic MOB cn c gc nn tam gic MOB u

- CA, CM l hai tip tuyn cng xut pht t im C nn CO l

ng phn gic ca gc , hay CO l phn gic ca gc

(**)

B A

O

Bi

4a

Bi

4b

Bi

5a

3,0

im

1,0

im

T (*) v (**) suy ra BM song song OC (gc ng v)

Nhn xt: Ba tam gic OAC, OMC v OMB l ba tam gic vung bng

nhau do c mt cnh gc vung bng nhau v mt gc nhn bng nhau

vy

Tam gic ACO vung c cnh gc vung Bi

5b

Vy din tch tam gic ACD l

B. HNG DN CHM:

+ Hc sinh lm cch khc ng vn cho im ti a.

+ im tng phn c th chia nh n 0,25 v phi c thng nht trong t chm./.

2,0

im

S GIO DC V O TO HI DNG

K THI CHN HC SINH GII TNH LP 9 NM HC 2013-2014

MN THI: TON

Thi gian lm bi: 150 pht Ngy thi 20 thng 03 nm 2014

( thi gm 01 trang)

Cu 1 (2 im).

a) Rt gn biu thc 2 3 3

2

1 1 . (1 ) (1 )

2 1

x x xA

x

vi 1 1x .

b) Cho a v b l cc s tha mn a > b > 0 v 3 2 2 36 0a a b ab b .

Tnh gi tr ca biu thc

4 4

4 4

4

4

a bB

b a

.

Cu 2 (2 im).

a) Gii phng trnh 2 2 2( 2) 4 2 4.x x x x

b) Gii h phng trnh

3

3

2

2

x x y

y y x

.

Cu 3 (2 im).

a) Tm cc s nguyn dng x, y tha mn phng trnh 2 2 32xy xy x y .

b) Cho hai s t nhin a, b tha mn 2 22 3a a b b .

Chng minh rng 2 2 1a b l s chnh phng.

Cu 4 (3 im). Cho tam gic u ABC ni tip ng trn (O, R). H l mt im di ng trn on OA (H khc A). ng thng

i qua H v vung gc vi OA ct cung nh AB ti M. Gi K l hnh chiu ca M trn OB.

a) Chng minh HKM 2AMH.

b) Cc tip tuyn ca (O, R) ti A v B ct tip tuyn ti M ca (O, R) ln lt ti D v E. OD, OE ct AB ln lt ti F v G. Chng minh OD.GF = OG.DE.

c) Tm gi tr ln nht ca chu vi tam gic MAB theo R.

Cu 5 (1 im).

Cho a, b, c l cc s thc dng tha mn 2 6 2 7ab bc ac abc . Tm gi tr nh nht ca biu thc 4 9 4

2 4

ab ac bcC

a b a c b c

.

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

H v tn thi sinh..s bo danh...

Ch k ca gim th 1..ch k ca gim th 2

S GIO DC V O TO HI DNG

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

HNG DN CHM THI CHN HC SINH GII TNH

LP 9 NM HC 2013-2014 MN THI: TON

Ngy thi 20 thng 03 nm 2014 (Hng dn chm gm c 03 trang)

Lu : Nu hc sinh lm theo cch khc m kt qu ng th gim kho vn cho im ti a.

THI CHNH THC

Cu Ni dung im

Cu

1a:

(1,0 )

2 22

1 1 . 1 1 2 1

2 1

x x x xA

x

0.25

21 1 . 1 1x x x 0.25

2

2 2 21 1 1 1 1 1 2 2 1x x x x x 0.25

22x = 2x 0.25

Cu

1b:

(1,0 )

3 2 2 3 2 26 0 ( 2 )( 3 ) 0 (*)a a b ab b a b a ab b 0.25

V a > b > 0 2 23 0a ab b nn t (*) ta c a = 2 b 0.25

Vy biu thc 4 4 4 4

4 4 4 4

4 16 4

4 64

a b b bB

b a b b

0.25

4

4

12 4

63 21

bB

b

0.25

Cu

2a:

(1,0 )

t 2 2 4 22 4 2 2t x x t x x 2

2 2 22

tx x

0.25

ta c phng trnh 2

24

4 2 8 022

ttt t t

t

0.25

Vi t = -4 ta c

2

4 2 4 2

0 02 4 4

2 2 16 2 8 0

x xx x

x x x x

2

02

2

xx

x

0.25

Vi t =2 ta c

2

4 2 4 2

0 02 4 2

2 2 4 2 2 0

x xx x

x x x x

2

03 1

3 1

xx

x

. Kt lun nghim ca phng trnh.

0.25

Cu

2b:

(1,0 )

T h ta c 3 3 2 2 2 2(2 ) (2 ) ( ) 2 0x y x y x y x y xy x y 0.25

3( ) ( ) 0x y

x y x yx y

0.25

* Vi x = y ta tm c (x ; y) = (0; 0); ( 3; 3 );( 3; 3 ) 0.25

* Vi x = - y ta tm c (x ; y) = (0; 0); (1; 1 );( 1;1 )

Vy h phng trnh c nghim

(x ; y) = (0; 0); ( 3; 3 );( 3; 3 );( 1;1 );(1; 1 )

0.25

Cu

3a:

(1,0 )

2 2 32xy xy x y 2( 1) 32x y y

Do y nguyn dng 2

321 0

( 1)

yy x

y

0.25

V 2( , 1) 1 ( 1)y y y (32)U 0.25

m 532 2 2 2( 1) 2y v 2 4( 1) 2y (Do 2( 1) 1y ) 0.25

*Nu 2 2( 1) 2 1; 8y y x

*Nu 2 4( 1) 2y 3; 6y x

Vy nghim nguyn dng ca phng trnh l:

8

1

x

y

v

6

3

x

y

0.25

Cu

3b:

(1,0 )

2 22 3a a b b 2( )(2 2 1)a b a b b (*) 0.25

Gi d l c chung ca (a - b, 2a + 2b + 1) ( *d ). Th

2

2 2

( )2 2 1

(2 2 1)

a b da b a b d

a b d

b d b d

0.25

M ( ) (2 2 )a b d a d a b d m (2 2 1) 1 1a b d d d 0.25

Do (a - b, 2a + 2b + 1) = 1. T (*) ta c a b v 2 2 1a b l s chnh phng => 2 2 1a b l s chnh phng.

0.25

Cu

4a:

(1,0 )

Qua A k tia tip tuyn Ax ca (O). Ta c

1 1

1 1A O

2 2sAM (1)

0.25

C Ax // MH (cng vung gc vi OA) 1 1A M (2) 0.25

T gic MHOK ni tip 1 1O K (cng chn MH ) (3) 0.25

T (1), (2), (3) ta c 1 1

1M K

2 hay HKM 2AMH. 0.25

Cu

4b:

(1,0 )

C t gic AOMD ni tip (4)

0.25

1

1A

2sBM ;

1 2

1O O

2sBM

1 1A O t gic AMGO ni tip (5)

0.25

T (4), (5) ta c 5 im A, D, M, G, O cng nm trn mt ng trn 0.25

x

1

1

1

1

H

K

O

A

B C

M

1

21

1

21

F

GE

D

H

O

A

B C

M

1 2 1G D D

OGF v ODE ng dng

OG GF

OD DE hay OD.GF = OG.DE.

0.25

Cu

4c:

(1,0 )

Trn on MC ly im A sao cho MA = MA AMA' u 01 2A A 60 BAA'

MAB A'AC MB A'C

0.25

MA MB MC Chu vi tam gic MAB l MA MB AB MC AB 2R AB

0.25

ng thc xy ra khi MC l ng knh ca (O) => M l im chnh gia cung AM => H l trung im on AO Vy gi tr ln nht ca chu vi tam gic MAB l 2R + AB

0.25

Gi I l giao im ca AO v BC 3 AB 3

AI R AB R 32 2

Gi tr ln nht ca chu vi tam gic MAB l 2R + AB = (2 3)R

0.25

Cu 5:

(1,0 )

T gt : 2 6 2 7ab bc ac abc v a,b,c > 0

Chia c hai v cho abc > 0 2 6 2

7c a b

t 1 1 1

, ,x y za b c

, , 0

2 6 2 7

x y z

z x y

Khi 4 9 4

2 4

ab ac bcC

a b a c b c

4 9 4

2 4x y x z y z

0.25

4 9 42 4 (2 4 )

2 4C x y x z y z x y x z y z

x y x z y z

0.25

2 22

2 3 22 4 17 17

2 4x y x z y z

x y x z y z

0.25

Khi 1

x ,y z 12

th C = 7

Vy GTNN ca C l 7 khi a =2; b =1; c = 1

0.25

21

A'

I

H

O

A

B C

M

S GIO DC V O TO PH TH

CHNH THC

K THI CHN HC SINH GII CP TNH NM HC 2013 - 2014 MN: TON - THCS

Thi gian lm bi 150 pht khng k thi gian giao

Cu1( 3,0 im) a) Gii phng trnh trn tp nguyn

0128y4x4xy5yx 22

b)Cho 214x3xxP(x) 23 .

Tm cc s t nhin x nh hn 100 m P(x) chia ht cho 11 Cu 2( 4,0 im)

a) Tnh ga tr biu thc 25a4aa

23aaP

23

3

, bit 33 302455302455a

b) Cho s thc x,y,z i 1 khc nhau tha mn 13zz1,3yy1;3xx 333

Chng minh rng 6zyx 222

Cu 3( 4,0 im)

a) Gii phng trnh 13x4x

1x13x

b) Gii h phng trnh:

03y2xyx

048yx4xy2y3x

22

22

Cu 4( 7,0 im) Cho ng trn (O;R) v dy cung BC khng i qua tm .Gi A l chnh gia

cung nh BC.Gc ni tip EAF quay quanh im A v c s o bng khng i sao cho E v F khc pha vi im A qua BC ;AE v AF ct BC ln lt ti M v N .Ly im D sao cho t gic MNED l hnh bnh hnh .

a)Chng minh t gic MNEF ni tip . b) Gi I l tm ng trn ngoi tip tam gic MDF .Chng minh rng khi gc ni

tip EAF quay quanh A th I chuyn ng trn ng thng c nh. c) Khi 060 v BC=R ,tnh theo R di nh nht ca on OI.

Cu 5( 2,0 im) Cho cc s thc dng x,y,z tha mn x+y+z=3

Chng minh rng 2 2 2 2 2 2 2 2 22 2 2

44 4 4

x y z y x z z y xxyz

yz xz yx

---Ht

H v tn th sinh..............................................s bo danh.....

Th sinh khng s dng ti liu,Cn b coi thi khng gii thch g thm

HNG DN

Cu1( 3,0 im)

a) Gii phng trnh trn tp nguyn

b) Cho 214x3xxP(x) 23 .

Hng dn

: a) 0(*12)8y5y()1(40128y4x4xy5yx 2222 yxx )

PT(*) c nghim nguyn x th / chnh phng 1616)1285(5)1(4 222/ yyyy

t tm c ;4;6;4;10;0;6;0;2; yx

Cch khc 222222 0416)22(0128y4x4xy5yx yyx

xt tng trng hp s ra nghim

b) ta c 2212)x-2)(x-(x214x3xxP(x) 223

P(x) chia ht 11 th 1112)x-2)(x-(x 2

m 1111)-x(x12)x-(x2 ta c 1)1( xx khng chia ht cho 11

suy ra 12)x-(x2 khng chia ht cho 11 nn x-2 chia ht co 11 m x

b) Gii h phng trnh:

03y2xyx

048yx4xy2y3x

22

22

Hng dn

a) HD kx 3

1x

136

132

1324

132413216

13x4131213x41)13(413x4x

1x13x

22

2

xx

xx

xxx

xxxxxx

xxxxxxx

gii ra pt c 2 nghim x=1; 72

1533x

b)

0(2)6y24xy22x

0(1)48yx4xy2y3x

03y2xyx

048yx4xy2y3x

22

22

22

22

ly pt(1) tr pt(2) ta c

22

12

0)22)(12(02)2(322

yx

yx

yxyxyxyx

thay vo phng trnh 03222 yxyx h c 4 nghim

6

10913;

3

1097;

6

10913;

3

1097;35;0;1; yx

Cu 4( 7,0 im) Hng dn

I

K

HPD

MN

F

A

BC

E

a) ENB=EFM suy ra ENM+EFM=1800

b)gi giao (O) v (I) tip tam gic MDF ti P ta c DPF=DMF =EAF=

mt khc EAF=EPF nn EPF=DPF nn E;D;P thng hng suy ra EP//BC m

EPAOBCAO gi AO ct EP ti H ;OI ct PF ti K th K l trung im FP v OI

vung gc FP nn t gic OHKP ni tip suy ra HOI=HPF= ( khng i)

suy ra I thuc tia Ox to vi tia AO mt gc bng

Q

I

H FD

MN

A

O

B C

E

c) khi BC=R ; EAF==600 th tam gic OBC u suy ra IO i qua B ta chng minh

c OI min khi F trng P khi EF//BC tam gic AMN; MDF u khi IM//AO ta

tnh BQ;QM c p dng Talet tam gic BIM c AO//IM tnh c OI

Hng dn Li gii 1

2 2 2

2 2 2

2 2 22

2 2 23

2

1 1 14 4 4

4 4 4

9( )

12 ( ) 12 ( )

93( )

12 ( ) 12 ( )

12 36

12 ( ) 12 3

x y z

P x y zyz xz yxyz xz yx

x y zx y zP

xy yz xz xy yz xz

xy yz xzxy yz xzP

xy yz xz xy yz xz

xy yz xz x y zP

xy yz xz x

2 23 y z

t 3 13

x y zxyz t

Xt hiu

23 2 2

2

364 12 ( 1)( 3) 0

12 3

tt t t t t

t

Bt ng thc c chng minh du = xy ra khi x=y=z=1 Li gii 2

2 2 2 2 2 2 2 2 22 2 24

4 4 4

x y z y x z z y xxyz

yz xz yx

(*)4)4()4()4(

222222222222

yxxyz

zxyz

xzxyz

zyyx

yzxyz

zxyxM

3

3

33 )4)(4)(4(3

312

)4)(4)(4(

6

)4)(4)(4(

6

)4(

1

)4(

1

)4(

12

)4(

1

)4(

1

)4(

12

)4()4()4(2

)4(

22

)4(

22

)4(

22

xyxzyzxyzxyxzyzxyzxyxzyzxyzN

yxxyzzxyzyyxyyzxyzzN

yxyx

zx

xzxz

zx

yzyz

zyN

Nyxxyz

yzxz

xzxyz

yzxy

yzxyz

xzxyM

Mt khc 44

4

123

4

4443)4)(4)(4(3

yzxyxzxyzyzxyxzxyzxyyzxzxyz

M 03339111

yzxzxyxyzxyz

xzyzxy

zyxzyx

33

4

33)4)(4)(4(3814

123)4)(4)(4(3

yzxzxyxyz

yzxyxzxyzxyyzxzxyz

Nn 433

3123

3

NM BT (*) c cm du = xy ra khi x=y=z=1

GVHD Nguyn Minh Sang THCS Lm Thao-Ph th

Li gii 3 :p dng BT AM GM cho bn s dng cho v tri ta c

4

4cyclic

x yzVT

yz

.

Ta cn chng minh

4 14 1

4 4cyclic cyclic

x yzxyz

yz yz yz

.

t , , 32cyclic

y za yz b zx c xy a b c x y z

.

Bt ng thc cn chng minh tr thnh

3

11

4cyclic a a

.

Ta c

24 3 2 231 4

9 4 4 16 1 2 9 04 9

aa a a a a a a

a a

Bt ny ng v 0 3a yz /2.

Do

3

1 4 3 41

4 9 3 9 3cyclic

a b c

a a

Du bng xy ra khi 1 1a b c x y z .

GV KIU NH PH -THCS TT SNG THAO - PH TH

Li gii 4 t 2 2 2 2 2 2 2 2 22 2 2

44 4 4

x y z y z x z x yA xyz

yz zx xy

Vi , , 0 x y z ta c:

2 2

94 0

4 4 4

y z x y zyz yz

Tng t ta cng c: 4 0, 4 0 zx xy .

p dng BT Bunhia-copx-ki ta c: 2 22 2 2 24 2 x x y z x x y z x y z

T suy ra:

22 2 2 22

4 4 4

x y zx y z

yz yz

Lm tng t ta c:

2 22 2 2 2 2 22 22 2;

4 4 4 4 4 4

y z x z x yy z x z x y

zx zx xy xy

Do :

2 2 22 2 2

4 4 4 4 4 4

x y z y z x z x yA B

yz zx xy

Mt khc theo BT C-si:

2 2

2 24 4 24 2 . 4 2 1

4 4 9 4 4 9 3

x y z x y zyz yz x y z

yz yz

2 22 24 4 2

4 2 . 4 2 24 4 9 4 4 9 3

y z x y z xzx zx y z x

zx zx

2 22 24 4 2

4 2 . 4 2 34 4 9 4 4 9 3

z x y z x yxy xy z x y

xy xy

Ly 1 2 3 theo v suy ra:

4 8 8 4

12 89 3 3 9

B xy yz zx x y z B xy yz zx

BT cho c gii quyt chn vn nu chng minh c:

8 4

4 43 9 xy yz zx xyz

Ta c: 38 4

4 . 481 27

x y z x y z xy yz zx xyz do 3 x y z .

S dng BT C-si cho 3 s dng ta c: 2 2 2333 ; 3 x y z xyz xy yz zx x y z

Suy ra: 33 2 2 233 3

8 4 8 4. . 3 .3 .3 4

81 27 81 27 x y z x y z xy yz zx xyz xyz x y z xyz

T BT cho c chng minh. Du ng thc xy ra khi 1 x y z .

Vit Tr, ngy 20 thng 03 nm 2014.

Bi Hi Quang Li gii 5

t 2 2 2 2 2 2 2 2 22 2 2

4 4 4

x y z y z x z x yA

yz zx xy

Khi 2 2 2

2 2 2 1 1 1( )( ) ( )4 4 4 4 4 4

x y zA x y z

xy yz zx yz zx xy

p dng bdt 1 1 1 9

a b c a b c

ta c

2 2 2 2 2 2 2 2 22 2 2

2 2 2

1 1 1 9( ) 18( ) 18( )( )( )

4 4 4 12 ( ) 24 2( ) 15

x y z x y z x y zx y z

xy yz zx xy yz zx xy yz zx x y z

(1)

Theo bdt AM-GM 2 2

3

2

(4 ) 1 (4 ) 13 . .

4 9 3 4 9 3

(4 ) 1 5 1

4 9 3 9 9 3

x x yz x x yzx

yz yz

x x yz x xyzx

yz

2 5 1

4 9 9 3

x x xyz

yz

Tng t ta c

2

2

5 1

4 9 9 3

5 1

4 9 9 3

y y xyz

zx

z z xyz

xy

Cng tng v 2 2 2 5 2

( ) 14 4 4 9 3 3 3

x y z xyz xyzx y z

yz zx xy

(2)

T (1) v (2) 2 2 2

2 2 2

18( ) 2

15 3 3

x y z xyzA

x y z

T 3x y z ta CM c 2 2 2 3x y z 2 2 2

2 2 2

18( )3

15

x y z

x y z

( d chng minh c theo

bin i tng ng) 2 11

33 3 3 3

xyz xyzA v chng minh kt thc nu ch ra c

11

43 3

xyzxyz

11 11

3 3xyz ( iu ny lun ng do

3

13

x y zxyz

)

3

13

x y zxyz

Vy ta c PCM du = xy ra khi x=y=z=1 , cc thy c ng gp thm li gii cho bi s 5 nh

S GIO DC V O TO THANH HO

THI CHNH THC

S bo danh

........................

K THI CHN HC SINH GII CP TNH Nm hc 2013 - 2014

Mn thi: TON - Lp 9 THCS Thi gian: 150 pht (khng k thi gian giao )

Ngy thi: 21/03/2014

( thi c 01 trang, gm 05 cu)

Cu I (4,0 im): Cho biu thc xy x xy xx 1 x 1

A 1 : 1xy 1 1 xy xy 1 xy 1

.

1. Rt gn biu thc A.

2. Cho 1 1 6x y . Tm gi tr ln nht ca A.

Cu II (5,0 im).

1.Cho phng trnh 04222 22 mmxmx . Tm m phng trnh

c hai nghim thc phn bit 1x , 2x tha mn mxxxx 15

112

21

2

2

2

1

.

2. Gii h phng trnh 4 4 4

1x y z

x y z xyz

.

Cu III (4,0 im).

1. Tm tt c cc cp s nguyn dng (a; b) sao cho (a + b2) chia ht cho (a2b 1).

2. Tm Nzyx ,, tha mn zyx 32 .

Cu IV (6,0 im) : Cho na ng trn tm O ng knh AB. Mt im C c nh thuc on thng AO (C khc A v C

khc O). ng thng i qua C v vung gc vi AO ct na ng trn cho ti D. Trn cung BD ly im M (M khc B

v M khc D). Tip tuyn ca na ng trn cho ti M ct ng thng CD ti E. Gi F l giao im ca AM v CD.

1. Chng minh tam gic EMF l tam gic cn.

2. Gi I l tm ng trn ngoi tip tam gic FDM. Chng minh ba im D, I, B thng hng.

3. Chng minh gc ABI c s o khng i khi M di chuyn trn cung BD. Cu V (1,0 im) : Cho x, y l cc s thc dng tho mn x + y = 1.

Tm gi tr nh nht ca biu thc 3 3

1 1Bxyx y

.

----- HT -----

Th sinh khng c s dng ti liu. Cn b coi thi khng gii thch g thm

Cu Li gii (vn tt) im

I

(4,0) 1

(2,5) iu kin: xy 1 . 0,25

x 1 1 xy xy x xy 1 xy 1 1 xyA :

xy 1 1 xy

xy 1 1 xy xy x xy 1 x 1 1 xy

xy 1 1 xy

0,50

x 1 1 xy xy x xy 1 xy 1 1 xy

xy 1 1 xy xy x xy 1 x 1 1 xy

0,50

1 x 1

x y xy xy

. 1,25

2

(1,5) Theo Csi, ta c: 1 1 1 16 2 9x y xy xy

.

0,50

Du bng xy ra 1 1

x y x = y =

1

9 .

0,50

Vy: maxA = 9, t c khi : x = y = 1

9.

0,50 II

(5,0) 1

(2,5) PT cho c hai nghim phn bit co iu kin:

0' 00422 22 mmmm (*) 0,50

Vi 0m theo Vi-et ta c:

42.

24

2

21

21

mmxx

mxx.

0,25

Ta c mxxxxxxmxxxx 15

11

2

2

15

112

2121

2

2121

2

2

2

1

(1) 0,50

mmmmm 15

1

42

1

46

122

0,50

15

1

24

1

64

1

mm

mm

. t tm

m 4

do 0m 0 t 0,50

Ta cos (1) tr thnh 412

4

15

1

2

1

6

1

t

t

t

tt ( do 0t ) 0,50

Vi 4t ta c 244

mm

m tha mn (*)

0,25 2

(2,5) Ta c:

4 4 4 4 4 44 4 4

2 2 2

x y y z z xx y z

2 2 2 2 2 2x y y z z x =

=

2 2 2 2 2 2 2 2 2 2 2 2

2 2 2

x y y z y z z x z x x yxyyz yzzx zxxy

=

= xyz (x + y + z) = xyz ( v x + y + z = 1).

0,50

0,50

0,50

S GIO DC V O TO THANH HO

HNG DN CHM THI CHNH THC

K THI CHN HC SINH GII CP TNH Nm hc 2013 - 2014

Mn thi: TON - Lp 9 THCS Thi gian: 150 pht (khng k thi gian giao )

Ngy thi: 21/03/2014

(Hng dn chm gm 04 trang)

Du bng xy ra 1

1 3

x y zx y z

x y z

Vy nghim ca h phng trnh l: 1 1 1

; ;3 3 3

x y z

0,50

III

(4,0) 1

(2,0) Gi s (a + b2) (a2b 1), tc l: a + b2 = k(a2b 1), vi k *

a + k = b(ka2 b) a + k = mb (1)

m m: m = ka2 b m + b = ka2 (2) 0,50

T (1) v (2) suy ra: (m 1)(b 1) = mb b m + 1

(m 1)(b 1) = (a + 1)(k + 1 ka) (3)

Do m > 0 (iu ny suy ra t (1) do a, k, b > 0) nn m 1 (v m ).

Do b > 0 nn b 1 0 (do b ) (m 1)(b 1) 0.

V th t (3) suy ra: (a + 1)(k + 1 ka) 0. 0,50 Li do a > 0 nn suy ra: k + 1 ka 0 k + 1 ka 1 k(a 1) (4)

V a 1 0 (do a , a > 0) v k , k > 0 nn t (4) c:

a 1k(a 1) 0

a 2k(a 1) 1

k 1

0,25

- Vi a = 1. Thay vo (3) ta c: (m 1)(b 1) = 2

m 1 2

b 1 1 b 2

b 3m 1 1

b 1 2

Vy, trng hp ny ta c: a = 1, b = 2 hoc a = 1, b = 3.

0,25

- Vi a = 2 (v k = 1). Thay vo (3) ta c: (m 1)(b 1) = 0 b 1

m 1

.

Khi b = 1, ta c: a = 2, b = 1.

Khi m = 1: T (1) suy ra a + k = b b = 3. Lc ny c: a = 2, b = 3. 0,25 Tm li, c 4 cp s (a; b) tha mn bi ton l: (1; 2), (1; 3), (2; 3), (2; 1). 0,25

2

(2,0) Ta c zyx 32 yzzyx 232

yzzyxzyxyzzyx 41234232 2 (1) 0,50

TH1. Nu 0 zyx Ta c zyx

zyxyz

4

1243

2

(2) v l

( do Nzyx ,, nn v phi ca (2) l s hu t ).

0,50

TH2. 0 zyx khi

3

01

yz

zyx (3) 0.50

Gii (3) ra ta c

3

1

4

z

y

x

hoc

1

3

4

z

y

x

th li tha mn 0,50

IV

(6,0)

1

(2.5)

2

(2.5)

3(1)

D

E

MI

H

F

C O BA

Ta c M thuc ng trn tm O ng knh AB (gi thit) nn 0AMB 90 (gc ni tip chn na ng trn)

hay 0FMB 90 . Mt khc 0FCB 90 (gi thit).Do 0FMB FCB 180 . Suy ra BCFM l t gic ni tip CBM EFM 1 (v cng b vi CFM ). Mt khc CBM EMF 2 (gc ni tip; gc to bi tip tuyn v dy cung cng

chn AM ). T (1) v (2) EFM EMF . Suy ra tam gic EMF l tam gic cn ti E.

(Co th nhn ra ngay EMF MBA MFE nn suy ra EMF cn)

0,50

0,50

0,50

0,50

0,50

G H l trung im ca DF. Suy ra IH DF v

DIF

DIH 32

.

Trong ng trn I ta c: DMF v DIF ln lt l gc ni tip v gc tm cng

chn cung DF. Suy ra 1

DMF DIF2

(4).

T (3) v (4) suy ra DMF DIH hay DMA DIH . Trong ng trn O ta c: DMA DBA (gc ni tip cng chn DA ) Suy ra DBA DIH . V IH v BC cng vung gc vi EC nn suy ra IH // BC. Do

oDBA HIB 180 oDIH HIB 180 Ba im D, I, B thng hng.

0,50

0,50

0,50

0,50

0,50

V ba im D, I, B thng hng ABI ABD 1

2sAD .

M C c nh nn D c nh 1

2sAD khng i.

Do gc ABI c s o khng i khi M thay i trn cung BD.

0,50

0,50

V(1)

Ta c: 3

1 2xy1 1 1 1Bxy 1 3xy xy xy(1 3xy)(x y) 3xy(x y)

.

Theo Csi:

2(x y) 1xy4 4

.

0.25

Gi Bo l mt gi tr ca B, khi , x, y : o1 2xy

Bxy(1 3xy)

3Bo(xy)2 (2 + Bo)xy + 1 = 0 (1)

tn ti x, y th (1) phi c nghim xy = Bo2 8Bo + 4 0

o

o

B 4 2 3

B 4 2 3

0.25

rng vi gi thit bi ton th B > 0. Do ta c: oB 4 2 3 .

Vi

oo

o

2 B 3 3 3 3B 4 2 3 xy x(1 x)

6B 6 2 3 6 2 3

2

2 3 2 31 1 1 1

3 3x , x

3 3x x 0

3 26 2 2

.

0.25

Vy, minB 4 2 3 , t c khi

2 3 2 31 1 1 1

3 3x , y

2 2

hoc

2 3 2 31 1 1 1

3 3x , y

2 2

.

0.25

Ch :

1) Nu hc sinh lm bi khng theo cch nu trong p n nhng ng th cho s im tng phn nh hng dn quy nh. 2) Vic chi tit ha (nu c) thang im trong hng dn chm phi bo m khng lm sai lch hng dn chm v phi c thng nht thc hin trong t chm. 3) im bi thi l tng im khng lm trn.