Đề và hướng dẫn giải Toán Chuyên PTNK 2008 - 2012

THI VO LP 10

GV: NGUYN TNG V www.truonglang.wordpress.com

THI VO LP 10 CHUYN TON TRNG PH THNG NNG KHIU NM HC 2008 2009 MN TON CHUYN

Bi 1:1) Cho phng trnh:

a) Chng minh rng phng trnh khng th c hai nghim u m.

b) Gi 1 2,x x l hai nghim ca phng trnh. Chng minh rng: ( )( )2 21 1 2 2

2 21 2

2 2 2 2x x x xx x

+ ++

khng ph thuc vo m.

2) Gii h phng trnh:

Bi 2: Cho tam gic ABC khng phi tam gic cn. ng trn (I ni tip tam gic v tip xc vi cc cnh BD, AC v AB ln lt ti D, E, F. EF ct BC v ID ln lt ti K v J.

a) Chng minh tam gic DIJ v AID ng dng.

b) Chng minh IK vung gc vi AD.

Bi 3: Cho gc xAy vung ti A. B thuc Ax v C thuc Ay. Hnh vung MNPQ vi M thuc AB, N thuc AC, P v Q thuc BC.

a) Tnh cnh hnh vung MNPQ theo BC = a v AH = h vi AH l ng cao h t A ca tam gic ABC.

b) Cho khng i. Tnh gi tr ln nht ca din tch hnh vung MNPQ

Bi 4: Gi s bch kim l s nguyn dng c tng cc bnh phng cc ch s bng chnh s .

a) Chng minh rng khng c s bch kim c 3 ch s.

THI VO LP 10


b) Tm tt c cc s nguyn dng bch kim n.

Bi 5: Trong mt gii bng c 6 i tham gia thi u vng trn mt lt. i thng c 3 im, ha c 1 im v thua th 0 im. Sau khi kt thc s im ca cc i ln lt l

( )1 2 3 4 5 6 1 2 3 4 5 6, , , , ,D D D D D D D D D D D D . Bit D1 thua ng mt trn v 1 2 3 4 5 6D D D D D D= + = + + . Tnh 1 6,D D

Hng dn gii Bi 1: 1) Phng trnh: ( )2 2 2 0 1x mx m + =

a) Gi s phng trnh c hai nghim 1 2,x x u m. Khi ta c:

( )2 21 2

1 2

4 2 2 0 8 8 00

0 01

2 2 00

m m m mm

S x x mm

mP x x

= + >= > (Mu thun)

Vy phng trnh khng c hai nghip u m. b) Gi 1 2,x x l hai nghim ca phng trnh. Khi theo nh l Viet ta c:

1 2

1 2 2 2S x x mP x x m= + = = =

Ta c: ( ) ( )22 2 2 21 2 1 2 1 22 2 2 2 4 4x x x x x x m m m m+ = + = = + V

( )( )( ) ( ) ( ) ( )( ) ( ) ( ) ( )

2 2 2 2 2 2 2 21 1 2 2 1 2 1 2 1 2 1 2

1 2 1 22 2 2

1 2 1 2 1 2 1 2 1 2 1 2

2 2

2 2 2

2

2 2 2 2 2 2 2 2

4 4 4 4

2 2 4 4 4

2 2 2 2 2 2 4 4 4 4 2 2 4

4 8 4 4 4 2 8 8 4 8 8 42 8 8

x x x x x x x x x x x x

x x x x

x x x x x x x x x x x x

m m m m m m m

m m m m m m m mm m

+ + = + + + +

= + + + + + += + + + += + + + + + += +

Suy ra: ( )( )2 2 21 1 2 2

2 2 21 2

2 2 2 2 2 8 8 24 4

x x x x m mx x m m

+ + += =+ + khng ph thuc vo gi tr ca m. 2)

THI VO LP 10


( )( )( )

2 2

2 2

2 2

1

2

3

x y z

y x z

z x y

= + = + = +

Ta c 2 2 2 2 2 20, 0, 0x y z y x z z x y= + = + = + Ly (1) tr (2) v theo v ta c:

( )( ) ( )( )( )

2 2 1 0

01 0

x y y x y x y x x y x y

x yx y l

= = + + + = = + + =

Ly (2) tr (3) v theo v ta c:

( )( )

( )( )( )

2 2

1 0

1

y z z y y z z y z y

y z y z

y zy z l

= = + + + =

= + =

T ta c x y z= = . T (1) suy ra ( )2 2 2 20

2 2 0 1 2 0 12

xx y z x x x x x

x

== + = = = =

Vi x = 0 th y = z = 0.

Vi 1 12 2

x y z= = =

Vy h phng trnh c hai nghim (x, y, z) l ( )0,0,0 v 1 1 1, ,2 2 2

Bi 2:

J

K

F

E

D

I

A

B C

THI VO LP 10


a) Ta c AE = AF (tnh cht hai tip tuyn ct nhau) v IE = IF (bn knh ng trn (I)) Suy ra AI l ng trung trc ca EF, suy ra AI EF ti J. Tam gic AFI vung ti c FJ l ng cao nn ta c : IJ. IA = IF2

M IF = ID, suy ra IJ. IA = ID2 IJ IDID IA

= Xt tam gic IDJ v tam gic IDA c : + Gc AID chung

+ IJ IDID IA

= (cmt) Suy ( )~ . .IDJ IAD c g c b) Tam gic ~IDJ IAD suy ra n nIJD IDA= . T gic IJKD c n n 90 90 180o o oIJK IDK+ = + = nn l t gic ni tip, suy ra n nIJD IKD= T ta c n nIDA IKD= Gi P l giao im ca AD v IK. Ta c n n n n n n90 90o oPDK PKD PDK PDI IDK DPK AD IK+ = + = = = Bi 3:

a) t x l di cnh hnh vung. Gi E l giao im ca AH v MN. Ta c MEHQ l hnh ch nht, suy ra EH = MQ = x, v AE = AH EH = h x.

D

A

B C

M N

PQ

THI VO LP 10


Ta c MN //BC nn MN ANBC AC

=

V NE // CH nn ta c AN AEAC AH

=

Do ta c: MN AEBC AH

= hay x h x ahxa h a h

= = +

Vy di cnh hnh vung MNPQ bng aha h+

b) Gi M l trung im BC, ta c 12

AM BC= suy ra 1 1hay2 2

AH AM BC h a = (1) Ta c 22 . . .ABCS AB AC AH BC a h k= = = (2) T (1) v (2) ta c: 2 22 2a ah k = Ta c ( )( )

22 42

2 2 2 2MNPQahah kS MP

a h a h aha h = = = = + + + +

Ta c 2

2 2 2 234 4aa h h a+ = + + . p dng bt ng thc Cauchy hai s ta c

:2 2

2 22 .4 4a ah h ah+ =

Suy ra:

2 2 2 2 2 2

2 2 2 2 2

3 3 5.24 4 2

5 92 22 2

a h ah a k k k

a h ah k k k

+ + +

+ + + =

Do 4

2

2

29 92

MNPQkS kk

= .Du bng xy ta khi v ch khi a = 2h 2a h M H= tam gic

ABC vung cn ti A khi v ch khi AB = AC = k.

Vy gi tr ln nht ca din tch hnh vung MNPQ bng 229

k khi tam gic ABC vung cn

ti A Bi 4: a) Xt s t nhin c 3 ch s abc trong 1 9, 0 , 9a b c . Ta chng minh abc khng th l s bch kim. Ta c ( )2100 10 100 10abc a b c a a a b c= + + = + + + Ta c 210b b

THI VO LP 10


Ta c 100 90a > v 1a suy ra ( ) 2100 90a a c > Do 2 2 2abc a b c> + + . Vy abc khng th l s bch kim. Do khng c s bch kim c 3 ch s. b) Ta chng mi s t nhin c nhiu hn 3 ch s cng khng phi l s bch kim. t 1 2... ka a a l s t nhin c k ch s vi 4k trong 11 9, 0 9 2,ia a i k = Ta c 1 21 2 1 2 1... 10 10 ... 10k kk k ka a a a a a a = + + + + Vi i = 2, 3, , k-1 th 210k i i ia a (1) V ( )1 2 1 2 2 21 1 1 1 1 110 10 990k k ka a a a a a a = + > + > + (2) ( V k > 3) T (1) v (2) ta c 1 2 2 2 21 2 1 2 1 1 2... 10 10 ... 10 ...k kk k k ka a a a a a a a a a = + + + + > + + + Vy khng c s bch kim c nhiu hn 3 ch s. Hn na t cu a) suy ra khng c s bch kim c nhiu hn hoc bng 3 ch s. Vy s bch kim nu c ch c th l s c mt hoc hai ch s. TH1: Nu s bch kim c mt ch s.

Ta c ( )2 01

a la a

a== =

. Vy s bch kim c mt ch s l 1.

TH2: S bch kim c hai ch s ab . Ta c 2 2 2 210 10ab a b a b b b a a= + = + = (3) M ( )2 1b b b b = l tch ca hai s nguyn lin tip nn chia ht cho 2. Suy ra 210a a chia ht cho 2, suy ra a phi l s chn. M 0 9 2,4,6,8a a< = Th a trc tip vo ta khng tm c s t nhin b no. Do khng c s bch kim c hai ch s. Vy ch c mt s bch kim duy nht l 1. Cu 5: Gi a l s trn c kt qu thng thua v b l s trn c kt qu ha.

Khi ta c: 6.5 152

a b+ = = v 1 2 3 4 5 6 3 2A D D D D D D a b= + + + + + = + l tng s im ca cc i t c. Ta c ( ) ( )2 3 2 3 30 3 2 45a b a b a b a b+ + + + V 1 2 3 4 5 6 1 13 2 3 10D D D D D D A a b D D= + = + + = + = Hn na do D1 thua mt trn nn 1 12D . Suy ra 110 12D . Gi m l s trn thng ca D1, n l s trn ha ca D1. Khi ta c m + n = 4 v 3m + n = D1

THI VO LP 10


+ Nu D1 = 10 => 3a+ 2b = 30 => a = 0, b = 15 v l v D1 thua mt trn nn c t nht mt trn thng thua.

+ Nu D1 = 11 ta c 3m + n = 11, m + n = 4. 1 7,2 2

n m = = ( v l v m, n l s t nhin) Vy D1 = 12. Khi ta c 3a + 2b = 3 . 12 = 36 => a = 6 v b = 9. Ta c D1 thng 4 trn thua mt trn nn s trn c D1 vi 5 i cn li l 5 trn c kt qu thng thua. Do trong 5 i cn li u vi nhau c ng mt trn thng thua, cn li l kt qu ha. (*) Ta c D2 + D3 = 12 v 2 3 3 6D D D Ta c D4 + D5 + D6 = 12 v 6 5 4 6 63 12 4D D D D D Ta c D6 u vi 4 i D2, D3, D4, D5 c nhiu lm l mt trn thua. Nn ta c 6 3D . Nu D6 = 3, suy ra D4 + D5 = 9 4 5D . Suy ra D4 phi c t nht mt trn thng (v khng th ha vi D1) v D4 khi u vi nhm (D2, D6) c t nht 3 trn ha nn suy ra

4 3 46 6 6D D D = => D5 = 3. Suy ra D5 c mt trn thua na (v l vi (*)) Vy D6 = 4. Mt vi nhn xt: + Bi 1 th chc ai cng lm c v y l dng bi tp c bn. V th nu lm c cc bi khc m khng lm c bi ny th cng nguy him. + Bi 2, 3 thuc mn hnh hc. Hai bi ny ch c cu b ca bi 3 th nhiu hc sinh d b nhm lc p dng Cauchy ngay m khng ti du bng xy ra. Cn li nhng cu khc hi vng l lm ng ht v cng thuc dng c bn. + Bi 4: Bi ny cng l mt bi hay, khng kh nhng nh gc c nhiu hc sinh v thy khng quen. Nhng nu , mt phng trnh nghim nguyn c nhiu n th phng php c bn c th ngh ti l bt ng thc. Khi gii ra cng h nghi v sao ch c s 1 l s bch kim (nh ngha s bch kim lm g ch c mt s). +Bi 5: Nm nay Euro + thy Trn Nam Dng rt khoi bong nn chc chn s l mt cu v bng ri. Bi ny ri rm, trong lc thi kh ngh ra v tm l. C l nhiu bn chun b nhiu v dng bi tp ny nhng vn gp kh khn. B bi ny m lm ht my bi khc th vn cn hi vng.

NGUYN TNG V

I HC QUC GIA TP HCM K THI TUYN SINH LP 10 NM HC 2009 - 2010 TRNG PH THNG Mn thi: TON CHUYN

NNG KHIU Thi gian lm bi: 150 pht, khng k thi gian giao _______________________________________________________________________________ Cu 1.

a) Cho , , ,a b c d l cc s thc tha mn iu kin , . 03

a c a c a cb d b d

.

Chng minh rng: 2 2b d . b) Gii h phng trnh:

2 2

2 2

1 33 72 34 7

x x yxy x yy x yxy x y

Cu 2. a) Gii bt phng trnh: 2 1 8 9x x b) Cho , ,a b c l cc s thuc 1;2 tha mn iu kin 2 2 2 6a b c . Chng minh rng: 0a b c

Cu 3.

a) Chng minh rng khng tn ti s t nhin a sao cho 2 20092010a a

b) Chng minh rng khng tn ti s t nhin a sao cho 2 3 20102009a a a

Cu 4. Cho ng trn O tm O , ng knh 2AB R . C l mt im thay i trn ng trn O sao cho tam gic ABC khng cn ti C . Gi H l chn ng cao ca tam gic ABC h t C . H ,HE HF vung gc vi ,AC BC tng ng. Cc ng thng EF v AB ct nhau ti K .

a) Tnh theo R din tch tam gic CEF v di cc on ,KA KB trong trng hp 060BAC .

b) H ,EP FQ vung gc vi AB . Chng minh rng ng trn ng knh PQ tip xc vi ng thng EF .

c) Gi D l giao im ca O v ng trn ng knh CH , D C . Chng minh rng 2.KA KB KH v giao im M ca cc ng thng CD v EF lun thuc mt ng thng c nh. Cu 5. Trn mt ng trn, ngi ta xp cc s 1,2,3,...,10 (mi s xut hin ng mt ln).

a) Chng minh khng tn ti mt cch xp m tng hai s k nhau u ln hn 10. b) Tn ti hay khng mt cch xp m tng hai s k nhau u ln hn hoc bng 10?

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

Hng dn gii

Bi 1. a)

Cch 1: Ta c 3 33 3

a c a c a cb d b d b d

(Do 3 0b d )

T ta c 3 33 3

a c a c a c a c a cb d b d

. M 2 2v 0a c ac b d b db d

Cch 2: Trng hp 1: 2 2b d b d (ccm)

Trng hp 2: b d , kt hp vi iu kin a cb d suy ra a c

Khi a c a cb d b d

(tnh cht dy t s bng nhau)

Suy ra 0

33a ca c a cb d b db d b d

Vi 0a c m 0ac suy ra 0, 0a c suy ra b d (mu thun)

Vi 2 23b d b d b d b d Vy trong hai trng hp ta u c 2 2b d Nhn xt: My em p ng ngay dy t s bng nhau l thiu trng hp ri, s b tr im. b)

2 2

2 2

2 2

1 33 7 1 2 32 3 3 4 74 7

x x yxy x y x y x yy x y xy xy x yxy x y

iu kin 2 2

34

7

xyxyx y

Trng hp 1: 73 42

xy xy xy , khi 1 2 3x y x y

Ta c h 3

72

x yVN

xy

Trng hp 2: 3 4xy xy

Khi ta c 2 23 1 2 3

7 3 4 2 7x y x y x y

x y xy xy xy

Suy ra 2 2

3 0

7 2 7

x y

x y xy

Vi 3 0x y ta c 1

1 2 02

xx y

y

Vi 22 27 2 7 0x y xy x y x y

Khi ta c 2 21 11 2

2 23 4x yx xx yx x

Th li ta thy 1;2 v 1; 1 l nghim ca h phng trnh Vy phng trnh c hai nghim ;x y l 1;2 v 1; 1 Bi 2 a) Ta c

2

2 1 08 9 0

2 1 8 92 1 0

2 1 8 9

xI

xx x

xII

x x

Gii (I): Ta c

19 12

9 8 28

xI x

x

Gii (II): Ta c

2 2

11 122 2

4 1 2 04 4 1 8 9 4 4 8 0

11 2221 2

xx xII

x xx x x x x

xx

x

Vy tp nghim ca bt phng trnh l 9 ;28

S

b) V 2 21;2 1 2 0 2 0 2a a a a a a a Du = xy ra khi v ch khi 1a hoc 2a

Chng minh tng t ta cng c 2 22, 2b b c c

Do 2 2 2 6a b c a b c suy ra 0a b c (v 2 2 2 6a b c )

Du = xy ra khi v ch khi , ,a b c l hon v ca 1; 1;2 Cu 3

a) Gi s tn ti s t nhin a tha 2 20092010a a

Ta c 2 1a a a a l tch hai s t nhin lin tip. Ta c , 1 1a a v 1 1a a .

Do , 1a a phi c dng 2009 2009, 1a p a q trong p q , . 2010, , 1p q p q

iu ny khng th xy ra v 20092009 2009, 1 1 1 1p q q p q p p Vy khng tn ti s t nhin a tha mn bi.

b) Gi s tn ti s t nhin a tha bi. Tc l 3 2 20102009a a a

R rng 0a , khi ta c 33 3 2 3 23 3 1 1a a a a a a a a

Mt khc 32010 6072009 2009

Suy ra 3 33 6702009 1a a . (V l v 33, 1a a l lp phng ca hai s t nhin lin tip. ) Vy khng tn ti s t nhin a tha mn bi. Cu 4

J

T

D

I

P QK,M

F

E

H OA B

C

a) Tnh theo R din tch tam gic CEF v di cc on ,KA KB trong trng hp

060BAC .

Ta c 090ACB (gc ni tip chn na ng trn O )

Tam gic ABC vung ti C nn ta c 0.cos 2 .cos60AC AB CAB R R V 0.sin 2 .sin 60 3CB AB CAB R R

Ta c 0 3.sin .sin 602

RCH AC ACB R

Tam gic CHE vung ti H c HE l ng cao nn 2

22

32 3.

4

RCHCE CA CH CE RCA R

Tng t ta cng c 2 3

4CH RCFCB

Do 21 1 3 3 3 3. . .

2 2 4 4 32CEFR R RS CE CF

V 060BAC nn A nm gia K v B D thy CEHF l hnh ch nht v 030KEA CEF CHF CBA , m 0 0 060 30 30AKE AEK CAB AKE CAB AEK Vy tam gic KAE cn ti A suy ra KA AE

M 3 14 4RAE AC CE R R nn 1

4KA R

V 1 924 4

KB KA AB R R R

b) Chng minh EF tip xc vi ng trn ng knh PQ

Cu b, c ta xt trng hp AC < BC, trng hp AC BC lm tng t Gi I l giao im ca EF v CH . V AEHF l hnh ch nht nn I l trung im EF. T gic EPQF l hnh thang vung (v ,EP FQ PQ ) Ta c //IH EP v I l trung im EF nn H l trung im ca PQ. Khi ng trn ng knh PQ l ng trn tm H bn knh HP. Gi T l hnh chiu ca H trn EF

Ta c PEH EAH (cng ph EHA ) v TEH IHE , IHE EAH (cng ph vi EHA . ) Suy ra PEH TEH , suy ra PEH TEH HT HP Ta c HT EF T EF v HT HP nn EF tip xc vi ng trn ng knh PQ

c) Chng minh 2.KA KB KH v M thuc mt ng c nh

Ta c KEA CEF CHF CBK , suy ra .KAE KFB g g ,

Do . .KA KE KA KB KE KFKF KB

(1)

Mt khc ta c KHE HCE HFK , suy ra .KHE KFH g g

Do 2.KH KE KE KF KHKF KH

(2)

T (1) v (2) th 2.KA KB KH Gi J l giao im ca OC v EF,

Ta c OCF OBC (tam gic OBC cn ti O) V JFE ICF (do tam gic ICF cn ti I) Do

0

0

90

90

OCF JFE OBC ICF

CJF OC EF

Tam gic CKO c CH v KJ l hai ng cao, ct nhau ti I nn I l trc tm ca tam gic CKO, do OI CK (3) Mt khc hai ng trn (O) v ng trn tm I ng knh CH ct nhau ti C v D, nn OI l ng trung trc ca CD, suy ra OI CD (4) T (3) v (4) ta c , ,C K D thng hng. Vy K cng l giao im ca CD v EF, do M K v M lun thuc ng thng AB c nh Nhn xt: y l mt bi hnh hc rt quen thuc, khng kh. hn nm ngoi nhiu. Bi 5.

a) Gi s tn ti mt cch sp xp tha bi l

a5a6

a7

a8

a9

a10

a4

a3

a2a1

Khng mt tnh tng qut ta gi s 1 1a . Khi ta c

1 2 2 2

1 10 10 10

10 9 1010 9 10

a a a aa a a a

(v l v mi s xut hin ng mt ln)

Vy khng tn ti cch sp xp tha mn bi. b) Tn ti cch sp xp nh trn. V d:

3

74

6

5

9

8

2

101

Nguyn Tng V Trng Ph Thng Nng Khiu 1

HNG DN GII V NHN XT

Hng dn gii khng phi l p n chnh thc, ch mang tnh cht tham kho.

Cu 1.

a) T a + b + c = 0, suy ra c = - (a + b). T ta c

33 3 3 3 30 3 3a b c a b a b ab a b abc . Vy abc = 0, suy ra mt trong 3 s a, b, c bng khng. @

b) Nhn xt trong h phng trnh trn vai tr ca x, y, z l nh nhau. Cch 1: t 1, 1, 1x a y b z c . Thay vo phng trnh (1) ta c a + b + c = 0 Thay vo (2) vi a + b + c = 0 ta c ab + ac + bc = - 4. (4) Thay vo (3) ta c 3 3 3 0a b c . p dng cu a ta suy ra mt trong 3 s a, b, c bng 0. Khng mt tnh tng qut, gi s a = 0, Khi b = - c, thay vo (4) ta tm c b = 2 hoc b = - 2. Kt lun: Phng trnh c nghim (1, - 1, 3) v cc hon v. Cch 2: T (1) v (2) ta c 22 2 2 2 11x y z x y z xy yz xz Thay vo (3) ta c 3 3 3 27 5x y z

T (1) v (5) ta suy ra 3 3 3 30 3x y z x y z x y y z x z T suy ra trong 3 s th hai s c tng bng 0. Gi s x + y = 0, t (1) ta c z = 3. Thay vo (2) ta c x = 1, y = - 1 hoc x = - 1, y = 1.. Kt lun: H phng trnh c nghim (1, - 1, 3) v hon v (6 nghim)

Nhn xt.

a) Kh c bn v c nhiu cch gii, cc bn c th p dng cc ng thc quen thuc nh 3 3 3 2 2 23a b c abc a b c a b c ab ac bc hay 3 3 3 3 3a b c a b c a b b c a c . Tuy nhin kh di dng.

b) Bi h phng trnh i xng 3 n, kh phc tp. Nu cc bn tinh th gii nh cch 1, tuy nhin bi ny cng c nhiu cch gii, ngoi 2 cch trn cc tnh ra tch xzy = - 3 v dng phng php th gii tip. Nhng phi ra y 6 nghim th mi c im ti a. Khng nn dng Viet cho 3 bin v chc l khng c s dng.


Cu 2.

a) iu kin 21

2 02

xx x

x

. Vi iu kin trn phng trnh tng ng

vi 2 2 2 24 4 12 2 3 2x x x x x x x x (2)

t 2 2 0t x x t , phng trnh tr thnh:

2 13 2 02

tt t

t

Vi 1t ta gii c 1 13 1 13,2 2

x x

Vi t = 2 ta gii c 3, 2x x

Ta thy cc s 1 13 1 13, ,3, 22 2

u tha iu kin.

Vy phng trnh c 4 nghim v 1 13 1 133, 2, ,2 2

S

b) t a = AB, b = AC. Do tam gic ABC vung ti A nn ta c 2 2BC a b .

Ta cn chng minh 2 22 2 2a b a b Ta c 11 2

2ABCS ab ab

+ p dng bt ng thc Cauchy ta c 2 2 2 22 4 2a b ab a b + Ta c

2 2

2 2

22 2 2 2

2 2 2 2

22 2

2 2

2 2

4 4 2

4 4 0 2

2 0

a b a b

a b a b

a b a b a b

a b a b ab

a b

Bt ng thc sau cng ng, suy ra iu cn chng minh.

Nhn xt.

a) Cu ny thuc loi d i vi hc sinh lp chuyn. Cch hay nht l t n ph cho n gin trong vic gii, nu dng cch bnh phng ln th kh kh khn v a v phng trnh bc 4. Do bi ny cc nghim u nhn nn thy vic gii


iu kin khng lin quan nhiu, tuy th vn phi c iu kin v th li nu mun t im ti a.

b) Cu bt ng thc ny khng kh lm, ch dng phng php bin i tng ng v n gin. Nhng v chng minh th hai, nu khng chuyn s 2 qua m bnh phng ln th cc em phi chng minh v phi khng m.

Cu 3.

a) B s bn s 1, 3, 7, 9 tha mn bi. b) Ta chng minh trong 5 s nguyn dng phn bit th lun c 3 s c tng chia

ht cho 3 v ln hn 3. + V cc s l nguyn dng phn bit nn tng ba s bt k lun ln hn 3. Nu trong 5 s khi chia cho 3 c y cc s d 0, 1, 2 th tng 3 s c s d khc nhau s chia ht cho 3. Nu ch c nhiu nht hai s d th theo nguyn tc Dirichlet s c t nht 3 s c cng s d khi chia cho 3, vy tng 3 s s chia ht cho 3. Vy lun tn ti 3 s c tng chia ht cho 3 v ln hn 3 nn khng tn ti 5 s nguyn dng phn bit m tng ba s bt k u l s nguyn t.

Nhn xt.

a) Cu ny ch cn ch ra c 4 s m khng cn chng minh l lun g. Bi ny c nhiu p s, ngoi b trn cn nhiu b khc m cc em c th ch ra c nh: (1, 5, 7, 11), Nu tm ra b s ln th trong li gii nn cng li chng t l s nguyn t.

b) Cu ny l cu kh, kh nht trong nm nay. Nu khng bit c tng chng minh (tn ti tng 3 s chia ht 3) th xem nh b. Tuy vy nu mt im cu ny cng khng sao, v b chc l b chung.

Cu 4. Bi ny ta xt hai trng hp, tam gic ABC nhn hoc t. V cch chng minh l tng t nn xt tam gic ABC nhn. Khi A v K khc pha i vi BC.


V BC = R3 nn ta tnh c BAC = 600, suy ABE = AEB = 300 ( ABE cn ti A)

T gic ABKE ni tip, suy ra AKB = AEB = 300

Chng minh tng t ta cng c AKC = AFC = 300

T BKC = AKB + AKC = 600

Xt t gic OBKC c BOC + BKC = 1200 + 600 = 1800 nn l t gic ni tip.

Vy K thuc ng trn ngoi tip tam gic OBC c nh.

b)Ta c SKBC = 1\2 BC. KT (T l hnh chiu ca T trn BC)

Suy ra SKBC max KT max K l im chnh gia cung ln BC ca ng trn ngoi tip tam gic OBC. Khi A l im chnh gia cung ln BC ca (O).

Khi tam gic BCK u cnh BC = R3 nn c din tch l SBKC = 33R2/4.

c)Ta c AKC = AKE = 300 nn suy ra K, C, E thng hng.

T gic AHCE c AEH = ACH = 300 nn l t gic ni tip, suy ra AHE = ACE. T suy ra AHB = ACK

Xt ABH v ACK c ABH = AKC v AHB = ACK (cmt)

nn ABH ~ ACK (g.g)

Gi D l giao im ca AO v (O). Ta c ABC = ADC v BAH + ABC = DAC + ADC = 900, suy ra BAH = DAC

Hn na BAH = KAC (do ABH ~ ACK)

T KAC = OAC, suy ra A, K, O thng hng.

Cch khc.

T gic BOCK ni tip, suy ra OKB = OCK = 300

M AKB = 300 (cmt) do OKB = AKB, suy ra K, A, O thng hng. Vy AK lun i qua im O c nh.

Nhn xt. Bi hnh hc nm nay c v rc ri v hnh v, tuy nhin cng khng phi l bi ton qu kh v d d on v chng minh cc . C nhiu cch chng minh cho cu


a v cu c ca bi ton ny v c th lm cu c trc. Cc bn thng nhm ln v cha chng minh tnh thng hng ca ba im (K, C, E) v (K, B, F) v p dng lung tung. Ni tm li y l mt bi ton hay.

Cu 5. Gi cc i ln lt l A1, A2,,A12.

a) Gi s sau 4 vng u A1 vi A2,A3, A4,A5 v cha u vi 4 i cn li. Ta ch cn chng minh trong cc i cn li c hai i cha u vi nhau. Tht vy xt A6, sau 4 vng u A6 u nhiu nht l vi 4 trong 6 i A7, A12 cn li nn c t nht 1 i cha u vi A6. Vy i bng cng vi A6, A1 l 3 i cha u vi nhau.

b) Cu khng nh l khng. Ta xt mt lch nh sau: Chia lm hai nhm, mi nhm 6 i A1, A2, A6 v A7, A8, , A12. C mi vng cho hai i trong cng mt nhm u vi nhau. Th sau 5 vng hai i trong cng 1 nhm u vi nhau. Xt 3 i bt k, th theo nguyn l Dirichlet tn ti t nht hai i cng mt nhm, nn vi nhau. @

Nhn xt. Nm nay l nm Worldcup bi ra mt bi bng . Tuy nhin y cng khng phi bi ton qu quen thuc nh cc nm trc l tm im cc i. Cu a, xem ra li d lm hn cu b. Bi ton ny hay i hi suy lun tt v gn gng.

Nhn xt chung.

nm nay cho hay v i hi nhiu suy lun logic.

Cc cu c bn im l 1a, 2a. Cc cu phc tp hn l 1b, 2b, 3a, bi 4. Nu lm ng c cc cu trn th hy vng rt cao.

Cc cu kh phn loi l bi 5 v 3b, trong kh nht l 3b v 5b.

Hn gp cc bn trong nm hc mi ti trng Ph Thng Nng Khiu.

HNGDNGIITHITUYNSINHLP10NM2011 Mnthi:TON(chuyn)

Thigianlmbi:150pht,khngththigianpht. Cu I.Chophngtrnh ( + 3) + = 0,tronglthamssaochophngtrnhchainghimphnbit, a) Khi = 1,chngminhrngtachthc + = 2 + 2 + 6 b) Tmttcccgitr sao cho + = 5 c) Xtathc() = + + .Tmttccccps(, )saocho

tachthc() = ()vimigitrcathamsm Ligii. Ta c = (m+3)2 4m2 = 3(m+1)(3-m). > 0 -1 < m < 3.

a) Khi m = 1, phng trnh c hai nghim dng x1, x2. Theo nh l Vit ta c x1 + x2 = 4, x1.x2 = 1. Ta bin i tng ng

626622

626222622

2121214

2121

42

41

821

42

41

82

81

xxxxxxxxxx

xxxxxxxx

H thc cui cng ng do x1 + x2 = 4 v x1x2 = 1.

b) phng trnh c hai nghim khng m th 00 1 30

S mP

(*)

Theo nh l Viet th x1 + x2 = m+3 v x1x2 = m2. Ta c

.2||2523525 2212121 mmmmxxxxxx

Gii phng trnh ny, ta c hai nghim .2,32

mm So snh vi iu

kin (*) trn ta loi nghim m = -2.

Vy gi tr m 521 xx l 32

m . c) Ta c P(x1) = P(x2) x13 x23 + a(x12-x22) + b(x1-x2) = 0

(x1-x2)(x12+x1x2+x22+a(x1+x2)+b) = 0

(x1-x2)((x1 +x2)2 x1x2 + a(x1+x2)+b) = 0

(x1-x2)((m+3)2 m2 + a(m+3)+b) = 0

(6+a)m + 9 + 3a + b = 0 vi mi - 1 < m < 3

a = - 6, b = 9.

Vy (a, b) = (-6, 9) l cp s cn tm.

Cu II. a) Cho , lccsthcdng.Tmgitrnhnhtcabiuthc = 1 + 1 + 1 + b) Chox,y,zlccsthcthamniukin|| 1, || 1, || 1.

Chngminhrngtacbtngthc: 1 + 1 + 1 9 ( + + )

Ligii. a) Ta c + 2, suy ra 1 + + + 1 + 2 +

(1 + )(1 + ) (1 + ) 1 + .1 + 1 + (v 1+ ab > 0)

1 (v 1 + ab > 0)

Du = xy ra khi v ch khi a = b. Vy minP = 1 khi v ch khi a = b.

b) Bnh phng hai v ca bt ng thc, ta c bt ng thc tng ng 2222222222 )(9112112112111 zyxxzzyyxzyx

zxyzxyxzzyyx 111111111 222222 . hon tt php chng minh, ta ch cn chng minh xyyx 111 22 . Tht

vy, do 1 xy 0 nn (*) tng ng vi

(1 x2)(1 y2) (1 xy)2 (x y)2 0 (hin nhin ng) Cu III. ChotamgicABCcAB=b,AC=b.MlmtimthayitrncnhAB.ngtrnngoitiptamgicBMCctACtiN. a) ChngminhrngtamgicAMNngdngvitamgicACB.Tnhts

dintchtamgicAMNbngmtnadintchtamgicACB. b) GiIltmngtrnngoitiptamgicAMN.ChngminhrngIlun

thucmtngthngcnh. c) GiJltmngtrnngoitiptamgicBMC.ChngminhrngdiIJkhngi.

Ligii.

a) Theo tnh cht ca t gic ni tip ta c ANM = MBC = ABC. Mt

khc NAM = BAC. Suy ra hai tam gic AMN v ACB ng dng. T hai tam gic ng dng ny ta suy ra AM.AB = AN.AC. din tch tam gic AMN bng mt na din tch tam gic ACB th t s ng dng phi bng

21 , tc l

21

ACAM . Suy ra

2cAM . T y ta tnh c

2cbBM . Suy

ra .2 cbc

BMAM

b) Cch 1. Gi O l tm ng trn ngoi tip tam gic ABC v AH l ng cao ca tam gic ABC. Khi d thy rng OAC = BAH. T , do ANM = ACH nn t y ta suy ra OA vung gc MN, suy ra AO l ng cao trong tam gic ANM. T y, cng do NAO = MAH.

Mt khc do tam gic AMN v ABC ng dng nn suy ra IAM v OAC ng

dng, do IAM = CAO = MAH, nn tm I ng trn ngoi tip tam gic AMN nm trn AH c nh (pcm). (1 )

Cch 2 chng minh AO NM: V tip tuyn Ax ca (O), ta c OA Ax.

xAM = ACB = AMN, suy ra Ax//MN. Do OA MN

Cch 2

Ta c I l tm ng trn nn AIM = 2ANM = 2ABC.

M tam gic AIM cn ti I nn MAI = (1800- AIM) = 900 - ABC.

Suy ra MAI + ABC = 90 nn AI BC, I thuc ng cao AH ca tam gic ABC.

c) V hai ng trn ngoi tip hai tam gic ABC v BMC c chung dy cung BC nn OJ vung gc vi BC. Theo chng minh trn th AI vung gc vi BC. Suy ra AI // OJ. Tng t, IJ vung gc vi MN v AO vung gc vi MN theo cu a), suy ra IJ // AO. Suy ra AIJO l hnh bnh hnh. Suy ra IJ = AO = R khng i (pcm). Cu IV. Cho , , lccsnguynsaocho2 + , 2 + , 2 + ulccs

chnhphng(*). a) Bitrngctnhtmttrongbaschnhphngnitrnchiahtcho3.Chngminhrngtch( )( )( )chiahtcho27. b) Tntihaykhngccsnguyn,, thaiukin(*) sao cho ( )( )( )chiahtcho 27? Ligii.

a) Gi s 2a + b = m2, 2b + c = n2, 2c + a = p2. Cng ba ng thc li, ta c 3(a + b + c) = m2 + n2 + p2. Suy ra m2 + n2 + p2 chia ht cho 3. Ch rng bnh phng ca mt s nguyn chia 3 d 0 hoc 1. Do nu c 1 trong 3 s, chng hn m chia ht cho 3 th n2 + p2 chia ht cho 3 v nh th n2 v p2 cng phi chia ht cho 3.

Cui cng, ch rng nu 2a + b chia ht cho 3 th a b = 3a (2a+b) chia ht cho 3. Tng t ta c b c v c a chia ht cho 3, suy ra (a b)(b c)(c a) chia ht cho 27.

b) Tn ti. Chng hn c th ly a = 2, b = 0, c = 1. Cu V. ChohnhchnhtABCDcAB= 3, BC = 4. a) Chngminhrngt7imbtktronghnhchnhtABCDluntmchaiimmkhongcchgiachngkhnglnhn5 b) Chngminhrngkhngnhcua)vncnngvi6imbtknmtronghnhchnhtABCD.

Ligii. a) Chia hnh ch nht 3 x 4 thnh 6 hnh ch nht con 1 x 2. Theo nguyn l

Dirichlet, tn ti 2 trong 7 im cho thuc vo 1 hnh ch nht v do ng knh ca hnh ch nht ny bng 5 nn ta c iu phi chng minh.

b) Chia hnh ch nht thnh 5 phn nh hnh v.

Nhnxtvthi. - Cccudl1ab,3a(ngdng). - Cccumctrungbnhl2ab, 4b, 5a, 4b, 1c - Cccukh(phnloihsg)l4a, 3c, 5b.

TON CHUYN TUYN SINH VO LP 10 TRNG PTNK NM HC 2012 2013

Cu I.

1) Gii h phng trnh

2 2

2 2

2 2

( ) 2

( ) 2

( ) 2

x y z z

y z x x

z x y y

2) Cho hnh vung ABCD cnh a. M v N l hai im ln lt nm trn cc cnh AB v BC

sao cho AM CN xAB CB

vi 0 < x < 1. Cc ng thng qua M , N song song vi BD ln lt

ct AD ti Q v CD ti P. Tnh din tch t gic MNPQ theo a v x v tm x sao cho din tch ny ln nht. Cu II. S nguyn dng n c gi l s k diu nu nh tng cc bnh phng ca cc c ca n ( k c 1 v n ) ng bng .

a) Chng minh rng s 287 l s iu ha.

b) Chng minh rng s 3n p ( p nguyn t ) khng phi l s iu ha. c) Chng minh rng nu s n pq ( p,q l cc s nguyn t khc nhau) l s iu ha th n

+ 2 l s chnh phng. Cu III.

a) Tm cc gi tr x tha mn 2 5 4 2 1 0x x x b) Chng minh rng vi cc s khng m , ,x y z tha mn 3a b c ta c bt ng thc

a b c ab bc ca Cu IV.Cho tam gic ABC vung ti A. Trn ng thng vung gc vi AB ti B ta ly im D di ng nm cng pha vi C i vi ng thng AB .

a) Chng minh rng nu AC + BD < CD th trn cnh AB tn ti hai im M v N sao cho 90oCMD CND b) Gi s iu kin trn c tha mn. ng thng qua A song song vi MD ct ng thng qua B song song vi MC ti E. Chng minh rng ng thng DE lun i qua mt im c nh . Cu V.Cho a gic u n cnh . Dng 3 mu xanh , , vng t mu cc nh a gic mt cch ty ( mi nh c t bi mt mu v tt c cc nh u c t mu). Cho php thc hin thao tc sau y : chn hai nh k nhau bt k ( ngha l hai nh lin tip) khc mu v thay mu ca hai nh bng mu cn li.

a) Chng minh rng bng cch thc hin thao tc trn mt s ln ta lun lun lm cho cc nh ca a gic ch cn c t bi hai mu.

b) Chng minh rng vi n = 4 v n = 8, bng cch thc hin thao tc trn mt s ln ta c th lm cho cc nh ca a gic ch cn c t bi mt mu.

HNG DN GII Cu I. 1) Tr theo v hai phng trnh u ta c

2 2( 2 )( ) 2( ) ( )( 2)x y z x z x z x z x z x z Hay ( )( 2 2) 0 1x z y x z hoac y

Xt trng hp x = z

PT th ba cho ta 20

2 02

yy y

y

Vi y = 0 ta c 2 2 2

2 2

2 2 2

2

x z z z zx zz x x

Nghim (x,y,z) = ( 0,0,0) , (1,0,1)

Vi y = 2 ta c 2 2 2

2 2

( 2) 2 3 2 0

(2 ) 2

x z z x xx zz x x

Nghim (x,y,z) = (1,2,1) , (2,2,2) Xt trng hp y = 1. Ta c

2 2

2 2

2

( 1) 2

(1 ) 2

( ) 1

x z z

z x x

z x

Vi z x = 1 dn n 21

3 2 02

zz z

z

Nghim (x,y,z) = (0,1,1) , (1,1,2)

Vi z x = -1 dn n 20

01

zz z

z

Nghim (x,y,z) = (1,1,0), (2,1,1) Vy h c 8 nghim . 2) Chng minh c MNPQ l hnh ch nht

Ta c 1 1 (1 ) 2MN BM AB AM AM x MN x aAC BA AB AB

2MQ AM x MQ xaBD AB

(0,25)

T 2 2

2 2 1 1. 2 (1 ) 24 2 2

aS MN MQ a x x a x

Vy din tch t GTLN l 2

ax 2maS khi

12

x hay M l trung im AB.

Cu II. a)S n = 287 c cc c dng l 1 , 7, 41, 287.

ng thc 2 2 2 2 21 7 41 287 (287 3) chng t 287 l s iu ha.

b) D thy cc c dng ca 3n p l 2 31, , ,p p p .Gi s tri li, n l s k diu. Khi

2 2 4 6 3 2 6 3 2 4 31 ( 3) 6 9 6 8p p p p p p p p p

T suy ra rng p l c ca 8. Nh th p = 2 v ta gp iu mu thun. Vy 2n p khng th l s iu ha .

d) S n = pq ( p < q) c cc c dng l 1, p,q v pq. V n l iu ha nn

2 2 2 2 2 2 2 21 ( 3) 6 8 ( ) 4( 2)p q p q pq p q pq p q pq Do 4 l s chnh phng nn n + 2 l s chnh phng. Cu III. a) K 1x

Ta c 2 5 4 2 1 0 ( 1)( 4) 2 1 0 1 ( 4) 1 2 0x x x x x x x x x 21 ( 1) 1 3 1 2 0 1 1 2 1 1 0x x x x x x x Vy mi 1x u tha mn bt phng trnh cho.

b) T cu a) , t 1 0t x ta c BT 2 2 3 (0, 25)t t t (*) BT cn chng minh tng ng vi

2 2 2 2 2 22 2 2 2 2 2a b c a b c ab bc ca a b c

2 2 2 22 2 2 ( )a b c a b c a b c

2 2 2 2 2 2 3( )a b c a b c a b c

p dng (*) ta c 2 2 22 3 , 2 3 , 2 3a a a b b b c c c Cng cc BT trn ta c BT cn chng minh. Cu IV.

a) Xt ng trn ng knh CD c tm O l trung im CD. Gi I l trung im CD. Khi OI vung gc vi

AB v 2AC BDOI .

T gi thit ta c 2CD OI . Suy ra ng trn ng

knh CD ct on AB ti hai im M, N v do 90oCMD CND . b) Gi E l giao im ca ng thng qua A song song vi MD vi CD. Gi P,Q ln lt l giao im ca MD vi AC v MC vi BD.

T nh l Talet ta c 'CE CA

CD CP v

'CA BQ CE BQCD DQ CD DQ

T ta c BE // MC. Suy ra E E hay C,D,E thng hng. Vy ng thng DE lun i qua im C c nh. Cu V. a) Xt mt dy cc nh mu vng AV1V2VkB ( c th ch gm 1 nh) c gii hn bi 2 nh A v B ( c th trng nhau) khng phi mu vng . S dng thao tc cho ta i mu hai nh A v V1 thnh mu th ba ( hin nhin khng phi vng) . Tip tc nh th ta i mu cc nh |V2,V3,,Vk sang mu khng phi vng. Nh vy ta lm mt mu vng trong dy cc nh trn Bng cch thc hin nh trn i vi dy cc im mu vng khc ta suy ra c th lm cho cc nh ca a gic ch cn c t bi hai mu xanh v .

b)Theo cu a) ta ch xt trng hp cc nh a gic c t bi hai mu ,chng hn xanh v . Bng thao tc cho ta c hai kiu chuyn mu b 4 nh lin tip nh sau : ddxx --> dvvx---> xxvx ---->xddx--->vvvv v dxdx--->vvvv , dxxd ---> vvvv (1) Do tnh i xng nn suy ra nu mt b 4 nh m trong c hai nh cng mt mu v hai nh cn li cng mt mu khc th ta chuyn c 4 nh v mu th ba. Bng cch dng kiu bin i nh trn ta c dddx--->ddvv---> xxxx ( dng (1)) v ddxd---->dvvd---->xxxx (2) Ngha l nu c 3 nh cng mu th ta chuyn mu 3 nh v mu ca nh th t. Nh vy bng (1) v (2) ta c th chuyn mu ca mi b 4 nh lin tip v cng mt mu . iu ny chng minh cho trng hp n = 4. Vi n = 8, ta chia 8 nh thnh 2 b 4 nh. Nh chng minh trn, ta c th lm cho mi b 4 nh nh th c cng mu. Nu mu ca hai b l nh nhau th ta c PCM. Tri li ,chng hn ta c kiu t mu xxxxdddd. Ta c php bin i hai b 4 nh lin tip : xxxxdddd---->xxxvvddd---->xxxv|vddd---->vvvvvvvv ( dng (2)) Vy ta chng minh c cho trng hp n = 8. Nhn xt

- Nhn chung nm nay c nh gi l kh v hi di, khng c cu cho im (mc d 2a l d nhng nhiu bn b). Bn no phi thc s c nng lc th mi c th t c im 7 tr ln. Cch pht biu l, nn nhiu cu khng qu kh nhng cc bn vn b qua hoc lm lung tung.

- Cc cu d nht c th l 2a, 1b. Kh hn mt cht l 1a,3a, 2bc, 4a, 5b n = 4.

- Cc cu kh nht l 3b, 4b v 5.

- Nm nay l nm Euro nhng khng c cu bng , nhng cu t hp li kh hay v khng t c.

Nguyn Tng V Trng Ph Thng Nng Khiu

1

THI VO LP 10 CHUYN TON TRNG PH THNG NNG KHIU NM HC 2013 2014

Cu I. Cho phng trnh 2 24 2 1 0 x mx m m

(1) vi m l tham s . a)Tm m sao cho phng trnh (1) c hai nghim phn bit. Chng minh rng khi hai nghim khng th tri du. b) Tm m sao cho phng trnh (1) c hai nghim 1 2,x x tha mn iu kin

1 2 1 x x

Cu II. Gii h phng trnh

2

2

2

3 2 1 2 ( 2)

3 2 1 2 ( 2)

3 2 1 2 ( 2)

x y z x

y z x y

z x y z

Cu III. Cho ,x y l hai s khng m tha mn 3 3 x y x y a) Chng minh rng 1 y x

b) Chng minh rng 3 3 2 2 1 x y x y

Cu IV. Cho 2 3 1 M a a vi a l s nguyn dng. a) Chng minh rng mi c ca M u l s l. b) Tm a sao cho M chia ht cho 5. Vi nhng gi tr no ca a th M l ly tha ca 5 ? Cu V. Cho tam gic ABC c gc

060A , ng trn (I) ni tip tam gic (vi tm I) tip xc vi cc cnh BC,CA,AB ln lt ti D,E,F. ng thng ID ct EF ti K, ng thng qua K v song song vi BC ct AB,AC theo th t ti M,N. a) Chng minh rng cc t gic IFMK v IMAN ni tip . b) Gi J l trung im cnh BC.Chng minh rng ba im A,K,J thng hng. c) Gi r l bn knh ca dng trn (I) v S l din tch t gic IEAF.Tnh S theo r v chng minh

4IMN

SS ( IMNS

ch din tch tam gic IMN)

Cu VI. Trong mt k thi, 60 th sinh phi gii 3 bi ton. Khi kt thc k thi , ngi ta nhn thy rng: Vi hai th sinh bt k lun c t nht mt bi ton m c hai th sinh u gii c. Chng minh rng :

a)Nu c mt bi ton m mi th sinh u khng gii c th phi c mt bi ton khc m mi th sinh u gii c .

b)C mt bi ton m c t nht 40 th sinh gii c.

2

HNG DN GII

Cu I. a) Phng trnh c hai nghim phn bit:

2 2 21

' 4 2 1 3 2 1 0 13

mm m m m m

m

Khi tch hai nghim bng 2 22 1 ( 1) 0 m m m nn phng trnh khng th c hai nghim tri du.

b) Trc ht ta phi c iu kin 1 22

' 01, 0 4 03

2 1 0

x x S m m

P m m

Ta c 21 2 1 2 1 21 2 1 4 2 2 1 1 x x x x x x m m m 4 2 1 1 m m

Vi m 1 th ta c 14 2( 1) 1 2 12

m m m m ( loi v khng tha iu kin)

Vi m < 1 thi 14 2( 1) 1 6 32

m m m m (tha mn iu kin)

Vy 12

m

Cu II. Gii h phng trnh

2

2

2

3 2 1 2 ( 2)

3 2 1 2 ( 2)

3 2 1 2 ( 2)

x y z x

y z x y

z x y z

Cng ba phng trnh ta c:

2 2 2

2 2 2 2 2 2

3 2 2 2 3 2 2 2 4 4 4

1 1 1 01 1 1 0

1

x y z x y z xy yz xz x y z

x y y z x z x y zx y y z z x x y zx y z

Th li thy (1, 1,1 ) l nghim ca h phng trnh. Vy phng trnh c 1 nghim l (1, 1, 1) Cu III. a) Ta c 3 3x y x y 0 x y

Li c 3 3 3x y x y x 0 y x 1 .

b) T cu trn ta c

3 23 3 2 2

3 2

x x0 y x 1 x y x y

y y

3

Ta c

3 3 3 3

2 2

x y x y x y

x y x y x xy y

Nu x = y th ta c 3 3 2 20 0 0 1x y x y x y Nu x y , suy ra 2 2 2 21 1 0x xy y x y v xy Vy 3 3 2 2 1x y x y Cu IV. Ta c 2 3 1 1 2 1M a a a a a . M a(a+1) l tch hai s t nhin lin tip nn chia ht cho 2, suy ra 1 2 1M a a a l s l, do mi c ca M u l s l

b) Gi s M = a2 + 3a + 1 = chia ht cho 5. Khi v M = (a 1)2 +5a nn (a1)2 chia ht cho 5. Vy a 1 chia ht cho 5. Nh th a = 5k +1 (k l s t nhin)

Th li, d thy vi a = 5k + 1 th M chia ht cho 5.

By gi gi s M = (a 1)2 +5a= 5n. Do M chia ht cho 5 nn a = 5k + 1.

Vy M = (5k+1)2 + 3(5k+1) + 1 = 5(5k2 + 5k + 1). Ta c 5k2 + 5k + 1 khng chia ht cho 5 nn M khng chia ht cho 25.

Nu n 2 th M chia ht cho 25. iu ny v l nn n = 1. Khi k = 0 , a = 1 v A = 5 ( tha mn).

p s : a = 1.

Cu V.

4

a)Do MN // BC nn IK MN. Do IKN = IFM = 90o nn t gic IFMK l t gic ni tip.

Tam gic AEF u nn KFI = 30o. T IMN = KFI = IAN = 30o nn t gic IMAN ni tip.

b) Ta c IMN = INM = 300 nn tam gic IMN cn ti I.

Li do IK MN nn K l trung im ca MN.

Gi J l giao im ca AK v BC ta c MK AK NKBJ AJ CJ

m MK = NK nn BJ = CJ suy ra

J l trung im ca BC do J trng J. T ta c A, K, J thng hng.

c)Ta c AE = AF = 2IAF13 2 2. IF.AF 32

r S S r

D chng minh c EF14

IS S

Cc tam gic IEF v IMN l cn , c gc y l 30o nn ng dng.

T 2

IEFIEF

1( IF) SIF 4

IMNIMN

S IM Sdo IM SS

Du ng thc xy ra khi M F hay tam gic ABC u.

CuVI. a) K hiu cc bi ton l BT1,BT2,BT3.

T gi thit suy ra rng mi th sinh u gii c t nht mt bi ton.

Ta ga s , chng hn mi th sinh u khng gii c BT1. Khi mi th sinh u gii c BT2 hoc BT3. Nu c mt th sinh gii c ch mt bi ton, chng hn BT2 th theo u bi mi th sinh khc phi gii c BT2. Nh th mi th sinh u gii c BT2 ( pcm). Ngc li, mi th sinh u gii c c hai bi ton BT2 v BT3 v ta cng c pcm.

b) Ta xt hai trng hp

5

C mt th sinh no ch gii c ng mt bi ton. Khi mi th sinh khc cng gii c bi ton v ta c pcm.

Mi th sinh u gii c t nht hai bi ton.

Gi a l s th sinh gii c c 3 bi ton

b l s th sinh gii c BT1 v BT2

c l s th sinh gii c BT1 v BT3

d l s th sinh gii c BT2 v BT3

Ta c a + b + c + d = 60 (*)

Nh vy s th sinh khng gii c BT3 l b , s th sinh khng gii c BT2 l c v s th sinh khng gii c BT1 l d.

Nu b,c,d > 20 th b +c +d > 60 , mu thun vi (*). Do mt trong ba s b,c,d phi c t nht mt s 20. Ni cch khc , c mt bi ton m c nhiu nht l 20 th sinh khng gii c . Hay ni cch khc, c mt bi ton m c t nht 40 th sinh gii c (pcm).

Nguyn Tng V

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Đề và hướng dẫn giải Toán Chuyên PTNK 2008 - 2012

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