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Tuyn tp thi HSG Ton 8 Nm hc: 2011-2012

THI S 1
Cu 1: (4,0 im)
Phn tch cc a thc sau thnh nhn t :
a) 3x2 7x + 2; b) a(x2 + 1) x(a2 + 1).
Cu 2: (5,0 im)
Cho biu thc :
2 x 4 x2 2 x x 2 3x
A( ):( )
2 x x2 4 2 x 2 x 2 x3
a) Tm KX ri rt gn biu thc A ?
b) Tm gi tr ca x A > 0?
c) Tnh gi tr ca A trong trng hp : |x - 7| = 4.
Cu 3: (5,0 im)
a) Tm x,y,z tha mn phng trnh sau :
9x2 + y2 + 2z2 18x + 4z - 6y + 20 = 0.
x y z a b c x2 y 2 z 2
b) Cho 1 v 0 . Chng minh rng : 2 2 2 1 .
a b c x y z a b c

Cu 4: (6,0 im)
Cho hnh bnh hnh ABCD c ng cho AC ln hn ng cho BD. Gi E, F ln
lt l hnh chiu ca B v D xung ng thng AC. Gi H v K ln lt l hnh chiu ca
C xung ng thng AB v AD.
a) T gic BEDF l hnh g ? Hy chng minh iu ?
b) Chng minh rng : CH.CD = CB.CK
2
c) Chng minh rng : AB.AH + AD.AK = AC .

HNG DN CHM THI

Ni dung p n im
Bi 1
a 2,0
2 2
3x 7x + 2 = 3x 6x x + 2 = 1,0
= 3x(x -2) (x - 2) 0,5
= (x - 2)(3x - 1). 0,5
b 2,0
a(x2 + 1) x(a2 + 1) = ax2 + a a2x x = 1,0
= ax(x - a) (x - a) = 0,5

Gv: Nguyn Vn T 1 Trng THCS Thanh M


Tuyn tp thi HSG Ton 8 Nm hc: 2011-2012
= (x - a)(ax - 1). 0,5
Bi 2: 5,0
a 3,0
KX :
2 x 0
2
x 4 0 x 0
1,0
2 x 0 x 2
x 2 3x 0
x 3
2 x 2 x3 0
2 x 4 x2 2 x x 2 3x (2 x) 2 4 x 2 (2 x) 2 x 2 (2 x)
A( 2 ):( 2 3) . 1,0
2 x x 4 2 x 2x x (2 x)(2 x) x( x 3)
4 x2 8x x(2 x)
. 0,5
(2 x)(2 x) x 3
4 x( x 2) x(2 x) 4 x2
0,25
(2 x)(2 x)( x 3) x 3
4x 2
Vy vi x 0, x 2, x 3 th A . 0,25
x 3
b 1,0
2
4x
Vi x 0, x 3, x 2 : A 0 0 0,25
x 3
x 3 0 0,25
x 3(TMDKXD) 0,25
Vy vi x > 3 th A > 0. 0,25
c 1,0
x 7 4
x7 4 0,5
x 7 4
x 11(TMDKXD)
0,25
x 3( KTMDKXD)
121
Vi x = 11 th A = 0,25
2
Bi 3 5,0
a 2,5
9x2 + y2 + 2z2 18x + 4z - 6y + 20 = 0
2 2 2
(9x 18x + 9) + (y 6y + 9) + 2(z + 2z + 1) = 0 1,0
2 2 2
9(x - 1) + (y - 3) + 2 (z + 1) = 0 (*) 0,5
Do : ( x 1)2 0;( y 3)2 0;( z 1)2 0 0,5
Nn : (*) x = 1; y = 3; z = -1 0,25
Vy (x,y,z) = (1,3,-1). 0,25
b 2,5
a b c ayz+bxz+cxy
T : 0 0 0,5
x y z xyz
ayz + bxz + cxy = 0 0,25

Gv: Nguyn Vn T 2 Trng THCS Thanh M


Tuyn tp thi HSG Ton 8 Nm hc: 2011-2012
x y z x y z
Ta c : 1 ( )2 1 0,5
a b c a b c
2 2 2
x y z xy xz yz
2 2 2 2( ) 1 0,5
a b c ab ac bc
x 2
y 2
z 2
cxy bxz ayz
2 2 2 2 1 0,5
a b c abc
x2 y 2 z 2
2 2 2 1(dfcm) 0,25
a b c
Bi 4 6,0
H

B C
0,25
F
O

E
A
K
D

a 2,0
Ta c : BE AC (gt); DF AC (gt) => BE // DF 0,5
Chng minh : BEO DFO( g c g ) 0,5
=> BE = DF 0,25
Suy ra : T gic : BEDF l hnh bnh hnh. 0,25
b 2,0
Ta c: ABC ADC HBC KDC 0,5
Chng minh : CBH CDK ( g g ) 1,0
CH CK
CH .CD CK .CB 0,5
CB CD
b, 1,75
Chng minh : AFD AKC( g g ) 0,25
AF AK
AD. AK AF . AC 0,25
AD AC
Chng minh : CFD AHC( g g ) 0,25
CF AH
0,25
CD AC
CF AH
M : CD = AB AB. AH CF . AC 0,5
AB AC
Suy ra : AB.AH + AB.AH = CF.AC + AF.AC = (CF + AF)AC = AC2
0,25
(fcm).

Gv: Nguyn Vn T 3 Trng THCS Thanh M


Tuyn tp thi HSG Ton 8 Nm hc: 2011-2012
S 2
Cu1.
a. Phn tch cc a thc sau ra tha s:
x4 4
x 2 x 3 x 4 x 5 24
b. Gii phng trnh: x 30x 31x 30 0
4 2

a b c a2 b2 c2
c. Cho 1 . Chng minh rng: 0
bc ca ab bc ca ab

x 2 1 10 x 2
Cu2. Cho biu thc: A 2 :x 2
x 4 2x x2 x2
a. Rt gn biu thc A.
1
b. Tnh gi tr ca A , Bit x = .
2
c. Tm gi tr ca x A < 0.
d. Tm cc gi tr nguyn ca x A c gi tr nguyn.

Cu 3. Cho hnh vung ABCD, M l mt im tu trn ng cho BD. K ME AB, MF AD.


a. Chng minh: DE CF
b. Chng minh ba ng thng: DE, BF, CM ng quy.
c. Xc nh v tr ca im M din tch t gic AEMF ln nht.

Cu 4.
1 1 1
a. Cho 3 s dng a, b, c c tng bng 1. Chng minh rng: 9
a b c
b. Cho a, b d-ng v a2000 + b2000 = a2001 + b2001 = a2002 + b2002
Tinh: a2011 + b2011
HNG DN CHM THI HC SINH GII LP 8
Cu p n im
4 4 2 2
a. x + 4 = x + 4x + 4 - 4x
= (x4 + 4x2 + 4) - (2x)2
= (x2 + 2 + 2x)(x2 + 2 - 2x)

( x + 2)( x + 3)( x + 4)( x + 5) - 24


= (x2 + 7x + 11 - 1)( x2 + 7x + 11 + 1) - 24
= [(x2 + 7x + 11)2 - 1] - 24
= (x2 + 7x + 11)2 - 52
= (x2 + 7x + 6)( x2 + 7x + 16) (2 im)
Cu 1 = (x + 1)(x + 6) )( x2 + 7x + 16)
(6 im) b. x 30x
4
31x 30 0 <=>
2

x x 1 x 5 x 6 0 (*)
2

1 2 3
V x2 - x + 1 = (x - ) + >0 x
2 4
(*) <=> (x - 5)(x + 6) = 0
x 5 0 x 5

x 6 0 x 6 (2 im)
Gv: Nguyn Vn T 4 Trng THCS Thanh M
Tuyn tp thi HSG Ton 8 Nm hc: 2011-2012
a b c
c. Nhn c 2 v ca: 1
bc ca ab
vi a + b + c; rt gn pcm (2 im)
x 2 1 10 x 2
Biu thc: A 2
: x 2
x 4 2x x2 x2
1
a. Rt gn c kq: A
x2 (1.5 im)
1 1 1
Cu 2 b. x x hoc x
(6 im) 2 2 2

4 4
A hoc A
3 5 (1.5 im)
c. A 0 x 2 (1.5 im)
1
d. A Z Z ... x 1;3 (1.5 im)
x2
HV + GT + KL A E
B

(1 im)

F
M

D C
Cu 3 a. Chng minh: AE FM DF
(6 im) AED DFC pcm (2 im)
b. DE, BF, CM l ba ng cao ca EFC pcm (2 im)
c. C Chu vi hnh ch nht AEMF = 2a khng i
ME MF a khng i
S AEMF ME.MF ln nht ME MF (AEMF l hnh vung)
M l trung im ca BD. (1 im)
1 b c
a 1 a a

1 a c
a. T: a + b + c = 1 1
b b b
Cu 4: 1 a b
(2 im) c 1
c c (1 im)

1 1 1 a b a c b c
3
a b c b a c a c b
32229

Gv: Nguyn Vn T 5 Trng THCS Thanh M


Tuyn tp thi HSG Ton 8 Nm hc: 2011-2012
1
Du bng xy ra a = b = c =
3
b. (a2001 + b2001).(a+ b) - (a2000 + b2000).ab = a2002 + b2002
(a+ b) ab = 1
(a 1).(b 1) = 0
a = 1 hoc b = 1 (1 im)
Vi a = 1 => b2000 = b2001 => b = 1 hoc b = 0 (loi)
Vi b = 1 => a2000 = a2001 => a = 1 hoc a = 0 (loi)
Vy a = 1; b = 1 => a2011 + b2011 = 2

thi S 3
a 3 4a 2 a 4
Cu 1 : (2 im) Cho P=
a 3 7a 2 14a 8
a) Rt gn P
b) Tm gi tr nguyn ca a P nhn gi tr nguyn
Cu 2 : (2 im)
a) Chng minh rng nu tng ca hai s nguyn chia ht cho 3 th tng cc lp ph-ng ca chng
chia ht cho 3.
b) Tm cc gi tr ca x biu thc :
P=(x-1)(x+2)(x+3)(x+6) c gi tr nh nht . Tm gi tr nh nht .
Cu 3 : (2 im)
1 1 1 1
a) Gii ph-ng trnh : 2 2
x 9 x 20 x 11x 30 x 13x 42 18
2

b) Cho a , b , c l 3 cnh ca mt tam gic . Chng minh rng :


a b c
A= 3
bca a cb a bc
Cu 4 : (3 im)
Cho tam gic u ABC , gi M l trung im ca BC . Mt gc xMy bng 600 quay quanh im M
sao cho 2 cnh Mx , My lun ct cnh AB v AC ln l-t ti D v E . Chng minh :
BC 2
a) BD.CE=
4
b) DM,EM ln l-t l tia phn gic ca cc gc BDE v CED.
c) Chu vi tam gic ADE khng i.
Cu 5 : (1 im)
Tm tt c cc tam gic vung c s o cc cnh l cc s nguyn d-ng v s o din tch bng s
o chu vi .
p n thi hc sinh gii
Cu 1 : (2 )
a) (1,5) a3 - 4a2 - a + 4 = a( a2 - 1 ) - 4(a2 - 1 ) =( a2 - 1)(a-4)

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Tuyn tp thi HSG Ton 8 Nm hc: 2011-2012
=(a-1)(a+1)(a-4) 0,5
a3 -7a2 + 14a - 8 =( a3 -8 ) - 7a( a-2 ) =( a -2 )(a2 + 2a + 4) - 7a( a-2 )
=( a -2 )(a2 - 5a + 4) = (a-2)(a-1)(a-4) 0,5
Nu KX : a 1; a 2; a 4 0,25
a 1
Rt gn P= 0,25
a2
a23 3
b) (0,5) P= 1 ; ta thy P nguyn khi a-2 l -c ca 3,
a2 a2
m (3)= 1;1;3;3 0,25
T tm -c a 1;3;5 0,25
Cu 2 : (2)
a)(1) Gi 2 s phi tm l a v b , ta c a+b chia ht cho 3 . 0,25

Ta c a3+b3=(a+b)(a2-ab+b2)=(a+b) (a 2 2ab b 2 ) 3ab =

=(a+b) (a b) 2 3ab 0,5
V a+b chia ht cho 3 nn (a+b)2-3ab chia ht cho 3 ;

Do vy (a+b) (a b) 2 3ab chia ht cho 9 0,25
b) (1) P=(x-1)(x+6)(x+2)(x+3)=(x2+5x-6)(x2+5x+6)=(x2+5x)2-36 0,5
Ta thy (x2+5x)2 0 nn P=(x2+5x)2-36 -36 0,25
Do Min P=-36 khi (x2+5x)2=0
T ta tm -c x=0 hoc x=-5 th Min P=-36 0,25
Cu 3 : (2)
a) (1) x2+9x+20 =(x+4)(x+5) ;
x2+11x+30 =(x+6)(x+5) ;
x2+13x+42 =(x+6)(x+7) ; 0,25
KX : x 4; x 5; x 6; x 7 0,25
Ph-ng trnh tr thnh :
1 1 1 1

( x 4)( x 5) ( x 5)( x 6) ( x 6)( x 7) 18

1 1 1 1 1 1 1

x 4 x 5 x 5 x 6 x 6 x 7 18
1 1 1
0,25
x 4 x 7 18
18(x+7)-18(x+4)=(x+7)(x+4)
(x+13)(x-2)=0
T tm -c x=-13; x=2; 0,25

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Tuyn tp thi HSG Ton 8 Nm hc: 2011-2012
b) (1) t b+c-a=x >0; c+a-b=y >0; a+b-c=z >0
yz xz x y
T suy ra a= ;b ;c ; 0,5
2 2 2
yz xz x y 1 y x x z y z
Thay vo ta -c A= ( ) ( ) ( ) 0,25
2x 2y 2z 2 x y z x z y
1
T suy ra A (2 2 2) hay A 3 0,25
2
Cu 4 : (3 )
a) (1)
Trong tam gic BDM ta c : D 1 120 0 M 1

V M 2 =600 nn ta c : M 3 120 0 M 1 y
A
Suy ra D 1 M 3 x
E
Chng minh BMD CEM (1) 0,5
D 2
1
BD CM
Suy ra , t BD.CE=BM.CM B 1
2 3
C
BM CE M

BC BC 2
V BM=CM= , nn ta c BD.CE= 0,5
2 4
BD MD
b) (1) T (1) suy ra m BM=CM nn ta c
CM EM
BD MD

BM EM

Chng minh BMD MED 0,5

T suy ra D 1 D 2 , do DM l tia phn gic ca gc BDE


Chng minh t-ng t ta c EM l tia phn gic ca gc CED 0,5
c) (1) Gi H, I, K l hnh chiu ca M trn AB, DE, AC
Chng minh DH = DI, EI = EK 0,5
Tnh chu vi tam gic bng 2AH; Kt lun. 0,5
Cu 5 : (1)
Gi cc cnh ca tam gic vung l x , y , z ; trong cnh huyn l z
(x, y, z l cc s nguyn d-ng )
Ta c xy = 2(x+y+z) (1) v x2 + y2 = z2 (2) 0,25
2 2
T (2) suy ra z = (x+y) -2xy , thay (1) vo ta c :
z2 = (x+y)2 - 4(x+y+z)
z2 +4z =(x+y)2 - 4(x+y)
z2 +4z +4=(x+y)2 - 4(x+y)+4
(z+2)2=(x+y-2)2 , suy ra z+2 = x+y-2 0,25
Gv: Nguyn Vn T 8 Trng THCS Thanh M
Tuyn tp thi HSG Ton 8 Nm hc: 2011-2012
z=x+y-4 ; thay vo (1) ta -c :
xy=2(x+y+x+y-4)
xy-4x-4y=-8
(x-4)(y-4)=8=1.8=2.4 0,25
T ta tm -c cc gi tr ca x , y , z l :
(x=5,y=12,z=13) ; (x=12,y=5,z=13) ;
(x=6,y=8,z=10) ; (x=8,y=6,z=10) 0,25

E THI SO 4
Cau1( 2 ): Phan tch a thc sau thanh nhan t
A a 1 a 3 a 5 a 7 15

Cau 2( 2 ): Vi gia tr nao cua a va b th a thc:


x a x 10 1
phan tch thanh tch cua mot a thc bac nhat co cac he so nguyen
Cau 3( 1 ): tm cac so nguyen a va b e a thc A(x) = x4 3x3 ax b chia het cho a
thc B( x) x2 3x 4
Cau 4( 3 ): Cho tam giac ABC, ng cao AH,ve phan giac Hx cua goc AHB va
phan giac Hy cua goc AHC. Ke AD vuong goc vi Hx, AE vuong goc Hy.
Chng minh rangt giac ADHE la hnh vuong
Cau 5( 2 ): Chng minh rang
1 1 1 1
P 2
2 4 ... 1
2 3 4 1002
ap an va bieu iem
Cau ap an Bieu iem
1 A a 1 a 3 a 5 a 7 15
2 0,5

a 2 8a 7 a 2 8a 15 15 0,5
a
2 0,5
2
8a 22 a 2 8a 120 0,5
a
2
2
8a 11 1
a 8a 12 a
2 2
8a 10
a 2 a 6 a 2
8a 10
2 Gia s: x a x 10 1 x m x n ;(m, n Z ) 0,25
2 0,25
x 2 a 10 x 10a 1 x 2 m n x mn 0,25
m n a 10
m.n 10 a 1

Kh a ta co : 0,25
mn = 10( m + n 10) + 1 0,25

Gv: Nguyn Vn T 9 Trng THCS Thanh M


Tuyn tp thi HSG Ton 8 Nm hc: 2011-2012
mn 10m 10n 100 1 0,25
m(n 10) 10n 10) 1 0,25


0,25
v m,n nguyen ta co: m 101
n 101 v m 101
n 101

suy ra a = 12 hoac a =8
3 Ta co:
1 A(x) =B(x).(x2-1) + ( a 3)x + b + 4 0,5

e A( x) B( x) th ba3400 ba
3
4
0,5

4
3

0,25

0,25
0,25
T giac ADHE la hnh vuong 0,25
Hx la phan giac cua goc AHB ; Hy phan giac cua goc 0,25
0,5
AHC ma AHB va AHC la hai goc ke bu nen Hx va Hy
vuong goc
0,5
Hay DHE = 900 mat khac ADH AEH = 900
Nen t giac ADHE la hnh ch nhat ( 1) 0,25
AHB 900 0,25
AHD 450 0,25
2 2
AHC 900
Do AHE 450
2 2
AHD AHE
Hay HA la phan giac DHE (2)
T (1) va (2) ta co t giac ADHE la hnh vuong
5 1 1 1 1
2 P 2 2 4 ...
2 3 4 1002
1 1 1 1 0,5
...
2.2 3.3 4.4 100.100 0,5
1 1 1 1
...
1.2 2.3 3.4 99.100 0,5
1 1 1 1 1
1 ... 0,5
2 2 3 99 100
1 99
1 1
100 100

Gv: Nguyn Vn T 10 Trng THCS Thanh M


Tuyn tp thi HSG Ton 8 Nm hc: 2011-2012
THI S 5
Bi 1: (4 im)
Phn tch cc a thc sau thnh nhn t:
a) (x + y + z) 3 x3 y3 z3.
b) x4 + 2010x2 + 2009x + 2010.

Bi 2: (2 im)

Gii phng trnh:


x 241 x 220 x 195 x 166
10 .
17 19 21 23

Bi 3: (3 im)
Tm x bit:
2009 x 2009 x x 2010 x 2010
2 2
19
.
2009 x 2009 x x 2010 x 2010
2 2
49

Bi 4: (3 im)
2010x 2680
Tm gi tr nh nht ca biu thc A .
x2 1
Bi 5: (4 im)
Cho tam gic ABC vung ti A, D l im di ng trn cnh BC. Gi E, F ln lt l
hnh chiu vung gc ca im D ln AB, AC.
a) Xc nh v tr ca im D t gic AEDF l hnh vung.
b) Xc nh v tr ca im D sao cho 3AD + 4EF t gi tr nh nht.

Bi 6: (4 im)
Trong tam gic ABC, cc im A, E, F tng ng nm trn cc cnh BC, CA, AB sao
cho: AFE BFD, BDF CDE, CED AEF .
a) Chng minh rng: BDF BAC .
b) Cho AB = 5, BC = 8, CA = 7. Tnh di on BD.

Mt li gii:

Bi 1:
(x + y + z) 3 x3 y3 z3 = x y z x 3 y3 z3
3
a)

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

= y z 3x 2 3xy 3yz 3zx = 3 y z x x y z x y

Gv: Nguyn Vn T 11 Trng THCS Thanh M


Tuyn tp thi HSG Ton 8 Nm hc: 2011-2012
= 3 x y y z z x .

b) x4 + 2010x2 + 2009x + 2010 = x 4 x 2010x 2 2010x 2010

= x x 1 x 2 x 1 2010 x 2 x 1 = x 2 x 1 x 2 x 2010 .
Bi 2:
x 241 x 220 x 195 x 166
10
17 19 21 23
x 241 x 220 x 195 x 166
1 2 3 40
17 19 21 23
x 258 x 258 x 258 x 258
0
17 19 21 23
1 1 1 1
x 258 0
17 19 21 23
x 258
Bi 3:
2009 x 2009 x x 2010 x 2010
2 2
19
.
2009 x 2009 x x 2010 x 2010
2 2
49
KX: x 2009; x 2010 .
t a = x 2010 (a 0), ta c h thc:
a 1 a 1 a a 2 19 a 2 a 1 19
2

a 1 a 1 a a 2 49 3a 2 3a 1 49
2

49a 2 49a 49 57a 2 57a 19 8a 2 8a 30 0


3
a
2a 1 42 0 2a 3 2a 5 0 2 (tho K)
2

a 5
2
4023 4015
Suy ra x = hoc x = (tho K)
2 2
4023 4015
Vy x = v x = l gi tr cn tm.
2 2
Bi 4:
2010x 2680
A
x2 1
335x 2 335 335x 2 2010x 3015 335(x 3) 2
= 335 335
x2 1 x2 1
Vy gi tr nh nht ca A l 335 khi x = 3.

Bi 5:
Gv: Nguyn Vn T 12 Trng THCS Thanh M
Tuyn tp thi HSG Ton 8 Nm hc: 2011-2012
a) T gic AEDF l hnh ch nht (v E A F 90 )
o
C
t gic AEDF l hnh vung th AD l tia phn
gic ca BAC .
b) Do t gic AEDF l hnh ch nht nn AD = EF
Suy ra 3AD + 4EF = 7AD
3AD + 4EF nh nht AD nh nht D
D l hnh chiu vung gc ca A ln BC. F

Bi 6:
a) t AFE BFD , BDF CDE , CED AEF .
Ta c BAC 1800 (*) A E B

Qua D, E, F ln lt k cc ng thng vung gc vi BC, AC, AB ct nhau ti O.


Suy ra O l giao im ba ng phn gic ca tam gic DEF.
A
OFD OED ODF 90o (1) E
F
Ta c OFD OED ODF 270 (2)
o

O
(1) & (2) 180 (**)
o

(*) & (**) BAC BDF .


b) Chng minh tng t cu a) ta c:
B , C

AEF DBF DEC ABC
s
s
s

B D C
BD BA 5 5BF 5BF 5BF
BF BC 8 BD 8 BD 8 BD 8

CD CA 7 7CE 7CE 7CE
CD CD CD
CE CB 8 8 8 8
AE AB 5 7AE 5AF 7(7 CE) 5(5 BF) 7CE 5BF 24

AF AC 7

CD BD 3 (3)
Ta li c CD + BD = 8 (4)
(3) & (4) BD = 2,5

S 6
Bi 1(3 im): Tm x bit:
a) x2 4x + 4 = 25
x 17 x 21 x 1
b) 4
1990 1986 1004
c) 4x 12.2x + 32 = 0

1 1 1
0.
Bi 2 (1,5 im): Cho x, y, z i mt khc nhau v
x y z
yz xz xy
Tnh gi tr ca biu thc: A 2 2 2
x 2 yz y 2xz z 2xy

Gv: Nguyn Vn T 13 Trng THCS Thanh M


Tuyn tp thi HSG Ton 8 Nm hc: 2011-2012

Bi 3 (1,5 im): Tm tt c cc s chnh phng gm 4 ch s bit rng khi ta thm 1 n


v vo ch s hng nghn , thm 3 n v vo ch s hng trm, thm 5 n v vo ch s
hng chc, thm 3 n v vo ch s hng n v , ta vn c mt s chnh phng.

Bi 4 (4 im): Cho tam gic ABC nhn, cc ng cao AA, BB, CC, H l trc tm.
HA' HB' HC'
a) Tnh tng
AA' BB' CC'
b) Gi AI l phn gic ca tam gic ABC; IM, IN th t l phn gic ca gc AIC v gc
AIB. Chng minh rng: AN.BI.CM = BN. IC.AM.
(AB BC CA) 2
c) Tam gic ABC nh th no th biu thc t gi tr nh nht?
AA' 2 BB' 2 CC' 2

P N
Bi 1(3 im):
a) Tnh ng x = 7; x = -3 ( 1 im )
b) Tnh ng x = 2007 ( 1 im )
c) 4x 12.2x +32 = 0 2x.2x 4.2x 8.2x + 4.8 = 0 ( 0,25im )
x x x x x
2 (2 4) 8(2 4) = 0 (2 8)(2 4) = 0 ( 0,25im )
x 3 x 2 x 3 x 2
(2 2 )(2 2 ) = 0 2 2 = 0 hoc 2 2 = 0 ( 0,25im )
x 3 x 2
2 = 2 hoc 2 = 2 x = 3; x = 2 ( 0,25im )

Bi 2(1,5 im):
1 1 1 xy yz xz
0 0 xy yz xz 0 yz = xyxz ( 0,25im )
x y z xyz
x2+2yz = x2+yzxyxz = x(xy)z(xy) = (xy)(xz) ( 0,25im )

Tng t: y2+2xz = (yx)(yz) ; z2+2xy = (zx)(zy) ( 0,25im )

yz xz xy
Do : A ( 0,25im )
( x y)(x z) ( y x )( y z) (z x )(z y)

Tnh ng A = 1 ( 0,5 im )

Bi 3(1,5 im):
Gi abcd l s phi tm a, b, c, d N, 0 a, b, c, d 9, a 0 (0,25im)

Ta c: abcd k
2

vi k, m N, 31 k m 100
(a 1)(b 3)(c 5)(d 3) m 2 (0,25im)
abcd k 2

abcd 1353 m 2 (0,25im)

Do : m2k2 = 1353
Gv: Nguyn Vn T 14 Trng THCS Thanh M
Tuyn tp thi HSG Ton 8 Nm hc: 2011-2012
(m+k)(mk) = 123.11= 41. 33 ( k+m < 200 ) (0,25im)
m+k = 123 m+k = 41
mk = 11 hoc mk = 33
m = 67 m = 37
k = 56 hoc k= 4 (0,25im)
Kt lun ng abcd = 3136 (0,25im)
Bi 4 (4 im):
V hnh ng
(0,25im) A
1
.HA'.BC
S HBC 2 HA' C
a) ; B x
S ABC 1 AA' N
H
.AA'.BC M
2 I
(0,25im) B
A C

S HAB HC' SHAC HB'


Tng t: ;
D
S ABC CC' SABC BB'
(0,25im)
HA' HB' HC' SHBC SHAB SHAC
1 (0,25im)
AA' BB' CC' SABC SABC SABC
b) p dng tnh cht phn gic vo cc tam gic ABC, ABI, AIC:
BI AB AN AI CM IC
; ; (0,5im )
IC AC NB BI MA AI
BI AN CM AB AI IC AB IC
. . . . . 1 (0,5im )
IC NB MA AC BI AI AC BI (0,5im )
BI .AN.CM BN.IC.AM
c)V Cx CC. Gi D l im i xng ca A qua Cx (0,25im)
-Chng minh c gc BAD vung, CD = AC, AD = 2CC (0,25im)
- Xt 3 im B, C, D ta c: BD BC + CD (0,25im)
2 2 2
- BAD vung ti A nn: AB +AD = BD
2 2 2
AB + AD (BC+CD)
AB2 + 4CC2 (BC+AC)2
4CC2 (BC+AC)2 AB2 (0,25im)
Tng t: 4AA2 (AB+AC)2 BC2
4BB2 (AB+BC)2 AC2
-Chng minh c : 4(AA2 + BB2 + CC2) (AB+BC+AC)2
(AB BC CA) 2
4 (0,25im)
AA'2 BB'2 CC'2
ng thc xy ra BC = AC, AC = AB, AB = BC
AB = AC =BC ABC u
Kt lun ng (0,25im)
*Ch :Hc sinh c th gii cch khc, nu chnh xc th hng trn s im cu

Gv: Nguyn Vn T 15 Trng THCS Thanh M


Tuyn tp thi HSG Ton 8 Nm hc: 2011-2012
S 7

Bi 1 (4 im)
1 x3 1 x2
Cho biu thc A = x : 3 vi x khc -1 v 1.
1 x 1 x x x
2

a, Rt gn biu thc A.
2
b, Tnh gi tr ca biu thc A ti x 1 .
3
c, Tm gi tr ca x A < 0.
Bi 2 (3 im)
Cho a b b c c a 4. a b c ab ac bc .
2 2 2 2 2 2

Chng minh rng a b c .


Bi 3 (3 im)
Gii bi ton bng cch lp phng trnh.
Mt phn s c t s b hn mu s l 11. Nu bt t s i 7 n v v tng mu ln 4
n v th s c phn s nghch o ca phn s cho. Tm phn s .
Bi 4 (2 im)
Tm gi tr nh nht ca biu thc A = a4 2a3 3a2 4a 5 .
Bi 5 (3 im)
Cho tam gic ABC vung ti A c gc ABC bng 600, phn gic BD. Gi M,N,I theo
th t l trung im ca BD, BC, CD.
a, T gic AMNI l hnh g? Chng minh.
b, Cho AB = 4cm. Tnh cc cnh ca t gic AMNI.
Bi 6 (5 im)
Hnh thang ABCD (AB // CD) c hai ng cho ct nhau ti O. ng thng qua O
v song song vi y AB ct cc cnh bn AD, BC theo th t M v N.
a, Chng minh rng OM = ON.
1 1 2
b, Chng minh rng .
AB CD MN
c, Bit SAOB= 20082 (n v din tch); SCOD= 20092 (n v din tch). Tnh SABCD.

p n
Bi 1( 4 im )
a, ( 2 im )
Vi x khc -1 v 1 th : 0,5
1 x x x
3 2
(1 x)(1 x)
A= :
1 x (1 x)(1 x x 2 ) x(1 x)
(1 x)(1 x x 2 x) (1 x)(1 x) 0,5
= :
1 x (1 x)(1 2 x x 2 )

= (1 x 2 ) :
1 0,5
(1 x)
= (1 x )(1 x)
2
0,5
b, (1 im)
Gv: Nguyn Vn T 16 Trng THCS Thanh M
Tuyn tp thi HSG Ton 8 Nm hc: 2011-2012
2 5 5 2 5 0,25
Ti x = 1 = th A = 1 ( 3 ) 1 ( 3 )
3 3
25
= (1
5
)(1 )
0,25
9 3
34 8 272
. 10
2 0,5
9 3 27 27
c, (1im)
Vi x khc -1 v 1 th A<0 khi v ch khi (1 x 2 )(1 x) 0 (1) 0,25
V 1 x 2 0 vi mi x nn (1) xy ra khi v ch khi 1 x 0 x 1 0,5
KL 0,25

Bi 2 (3 im)
Bin i ng thc c 0,5
a b 2ab b c 2bc c a 2ac 4a 4b 4c 4ab 4ac 4bc
2 2 2 2 2 2 2 2 2

Bin i c (a 2 b 2 2ac) (b 2 c 2 2bc) (a 2 c 2 2ac) 0 0,5


Bin i c (a b) 2 (b c) 2 (a c) 2 0 (*) 0,5
V (a b) 2 0 ; (b c) 2 0 ; (a c) 2 0 ; vi mi a, b, c 0,5
nn (*) xy ra khi v ch khi (a b) 2 0 ; (b c) 2 0 v (a c) 2 0 ; 0,5
T suy ra a = b = c 0,5

Bi 3 (3 im)
0,5
Gi t s ca phn s cn tm l x th mu s ca phn s cn tm l x+11. Phn s
x
cn tm l (x l s nguyn khc -11)
x 11
x7 0,5
Khi bt t s i 7 n v v tng mu s 4 n v ta c phn s
x 15
(x khc -15)
x x 15 0,5
Theo bi ra ta c phng trnh =
x 11 x 7
Gii phng trnh v tm c x= -5 (tho mn) 1
T tm c phn s
5 0,5
6
Bi 4 (2 im)
0,5
Bin i c A= a 2 (a 2 2) 2a(a 2 2) (a 2 2) 3
= (a 2 2)(a 2 2a 1) 3 (a 2 2)(a 1) 2 3 0,5
V a 2 2 0 a v (a 1) 2 0a nn (a 2 2)(a 1) 2 0a do 0,5
(a 2 2)(a 1) 2 3 3a
Du = xy ra khi v ch khi a 1 0 a 1 0,25
KL 0,25
Bi 5 (3 im)

Gv: Nguyn Vn T 17 Trng THCS Thanh M


Tuyn tp thi HSG Ton 8 Nm hc: 2011-2012
B

M N

A
D I C

a,(1 im)
Chng minh c t gic AMNI l hnh thang 0,5
Chng minh c AN=MI, t suy ra t gic AMNI l hnh thang cn 0,5
b,(2im)
4 3 8 3 0,5
Tnh c AD = cm ; BD = 2AD = cm
3 3
1 4 3
AM = BD cm
2 3
4 3 0,5
Tnh c NI = AM = cm
3
8 3 1 4 3 0,5
DC = BC = cm , MN = DC cm
3 2 3
8 3 0,5
Tnh c AI = cm
3

A B
Bi 6 (5 im)
O
M N

C
a, (1,5 im) D

Lp lun c
OM OD
,
ON OC

0,5
AB BD AB AC
Lp lun c
OD OC

0,5
DB AC

OM ON
OM = ON
0,5
AB AB
b, (1,5 im)
Xt ABD c
OM DM
(1), xt ADC c
OM AM
(2) 0,5
AB AD DC AD
1 1 AM DM AD
T (1) v (2) OM.( ) 1
AB CD AD AD
Chng minh tng t ON. (
1 1
) 1
0,5
AB CD

Gv: Nguyn Vn T 18 Trng THCS Thanh M


Tuyn tp thi HSG Ton 8 Nm hc: 2011-2012
1
t c (OM + ON). (
1
)2
1

1

2 0,5
AB CD AB CD MN
b, (2 im)
S AOB OB S BOC OB S S 0,5
, AOB BOC S AOB .S DOC S BOC .S AOD
S AOD OD S DOC OD S AOD S DOC
Chng minh c S AOD S BOC 0,5
S AOB .S DOC (S AOD ) 2 0,5
Thay s c 20082.20092 = (SAOD)2 SAOD = 2008.2009
Do SABCD= 20082 + 2.2008.2009 + 20092 = (2008 + 2009)2 = 40172 (n v 0,5
DT)

S 8

Bi 1:
b2 c2 a 2 a 2 (b c) 2
Cho x = ;y=
2bc (b c) 2 a 2
Tnh gi tr P = x + y + xy
Bi 2:
Gii phng trnh:
1 1 1 1
a, = + + (x l n s)
ab x a b x

(b c)(1 a)2 (c a)(1 b)2 (a b)(1 c)2


b, + + =0
x a2 x b2 x c2
(a,b,c l hng s v i mt khc nhau)
Bi 3:
Xc nh cc s a, b bit:
(3x 1) a b
= +
( x 1) 3
( x 1) ( x 1) 2
3

Bi 4: Chng minh phng trnh:


2x2 4y = 10 khng c nghim nguyn.
Bi 5:
Cho ABC; AB = 3AC
Tnh t s ng cao xut pht t B v C

S 9
Bi 1: (2 im)
2 1 1 1 x 1
Cho biu thc: A 3 1 2 1 : 3
x
2
x 1 x x 2x 1 x
a/ Thu gn A
b/ Tm cc gi tr ca x A<1
c/ Tm cc gi tr nguyn ca x Ac gi tr nguyn
Bi 2: (2 im)

Gv: Nguyn Vn T 19 Trng THCS Thanh M


Tuyn tp thi HSG Ton 8 Nm hc: 2011-2012
a/ Phn tch a thc sau thnh nhn t ( vi h s l cc s nguyn):
x2 + 2xy + 7x + 7y + y2 + 10
b/ Bit xy = 11 v x2y + xy2 + x + y = 2010. Hy tnh x2 + y2
Bi 3 (1,5 im):
Cho a thc P(x) = x2+bx+c, trong b v c l cc s nguyn. Bit rng a thc
x4 + 6x2+25 v 3x4+4x2+28x+5 u chia ht cho P(x). Tnh P(1)
Bi 4 (3,5 im):
Cho hnh ch nht c AB= 2AD, gi E, I ln lt l trung im ca AB v CD. Ni D vi
E. V tia Dx vung gc vi DE, tia Dx ct tia i ca tia CB ti M.Trn tia i ca tia CE
ly im K sao cho DM = EK. Gi G l giao im ca DK v EM.
a/ Tnh s o gc DBK.
b/ Gi F l chn ng vung gc h t K xung BM. Chng minh bn im A, I, G, H
cng nm trn mt ng thng.
Bi 5 (1 im):
Chng minh rng: Nu ba s t nhin m, m+k, m+ 2k u l cc s nguyn t ln hn 3, th
k chia ht cho 6.

S 10

Bi 1: (3 im)
1 3 x2 1
Cho biu thc A 2 :
3 x 3x 27 3x
2
x 3
a) Rt gn A.
b) Tm x A < -1.
c) Vi gi tr no ca x th A nhn gi tr nguyn.

Bi 2: (2 im) Gii phng trnh:


1 6y 2
a) 2
3 y 10 y 3 9 y 1 1 3 y
2

x 3 x 6x 1
1 .
b) x 2 4 3 3 2
2 2
Bi 3: (2 im)
Mt xe p, mt xe my v mt t cng i t A n B. Khi hnh ln lt lc 5 gi,
6 gi, 7 gi v vn tc theo th t l 15 km/h; 35 km/h v 55 km/h.
Hi lc my gi t cch u xe p v xe p v xe my?
Bi 4: (2 im)
Cho hnh ch nht ABCD t im P thuc ng cho AC ta dng hnh ch nht
AMPN ( M AB v N AD). Chng minh:
a) BD // MN.
b) BD v MN ct nhau ti K nm trn AC.
Bi 5: (1 im)
Cho a = 111 (2n ch s 1), b = 444 (n ch s 4).
Chng minh rng: a + b + 1 l s chnh phng.
Gv: Nguyn Vn T 20 Trng THCS Thanh M
Tuyn tp thi HSG Ton 8 Nm hc: 2011-2012

S 11

Bi 1: (2im)
3x 2 y 1
a) Cho x 2xy 2y 2x 6y 13 0 .Tnh N
2 2

4xy
b) Nu a, b, c l cc s dng i mt khc nhau th gi tr ca a thc sau l s
dng: A a 3 b3 c3 3abc
Bi 2: (2 im)
Chng minh rng nu a + b + c = 0 th:
a b b c c a c a b
A 9
c a b a b b c c a
Bi 3: (2 im)
Mt t phi i qung ng AB di 60 km trong thi gian nht nh. Na qung
ng u i vi vn tc ln hn vn tc d nh l 10km/h. Na qung ng sau i vi
vn tc km hn vn tc d nh l 6 km/h.
Tnh thi gian t i trn qung ng AB bit ngi n B ng gi.
Bi 4: (3 im)
Cho hnh vung ABCD trn cnh BC ly im E. T A k ng thng vung gc vi AE
ct ng thng CD ti F. Gi I l trung im ca EF. AI ct CD ti M. Qua E dng ng
thng song song vi CD ct AI ti N.
a) Chng minh t gic MENF l hnh thoi.
b) Chng minh chi vi tam gic CME khng i khi E chuyn ng trn BC
Bi 5: (1 im)
Tm nghim nguyn ca phng trnh: x 6 3x 2 1 y4

S 12

Bi 1:
Phn tch thnh nhn t:
a, (x2 x +2)2 + (x-2)2
b, 6x5 +15x4 + 20x3 +15x2 + 6x +1
Bi 2:
a, Cho a, b, c tho mn: a+b+c = 0 v a2 + b2 + c2= 14.
Tnh gi tr ca A = a4+ b4+ c4
b, Cho a, b, c 0. Tnh gi tr ca D = x2011 + y2011 + z2011
x2 y 2 z 2 x2 y2 z 2
Bit x,y,z tho mn: = + +
a 2 b2 c 2 a 2 b2 c2
Bi 3:
1 1 4
a, Cho a,b > 0, CMR: +
a b ab
b, Cho a,b,c,d > 0
a d d b bc ca
CMR: + + + 0
d b bc ca ad

Gv: Nguyn Vn T 21 Trng THCS Thanh M


Tuyn tp thi HSG Ton 8 Nm hc: 2011-2012
Bi 4:
x 2 xy y 2
a, Tm gi tr ln nht: E = vi x,y > 0
x 2 xy y 2
x
b, Tm gi tr ln nht: M = vi x > 0
( x 1995) 2
Bi 5:
a, Tm nghim Z ca PT: xy 4x = 35 5y
b, Tm nghim Z ca PT: x2 + x + 6 = y2
Bi 6:
Cho ABC M l mt im min trong ca ABC . D, E, F l trung im AB, AC, BC;
A, B, C l im i xng ca M qua F, E, D.
a, CMR: ABAB l hnh bnh hnh.
b, CMR: CC i qua trung im ca AA

S 13

Bi 1: (2 im)
a) Phn tch a thc sau thnh nhn t:
a(b c) 2 (b c) b(c a) 2 (c a) c(a b) 2 (a b)
1 1 1
b) Cho a, b, c khc nhau, khc 0 v 0
a b c
1 1 1
Rt gn biu thc: N 2 2 2
a 2bc b 2ca c 2ab
Bi 2: (2im)
a) Tm gi tr nh nht ca biu thc:
M x 2 y 2 xy x y 1
b) Gii phng trnh: ( y 4,5) 4 ( y 5,5) 4 1 0
Bi 3: (2im)
Mt ngi i xe my t A n B vi vn tc 40 km/h. Sau khi i c 15 pht, ngi
gp mt t, t B n vi vn tc 50 km/h. t n A ngh 15 pht ri tr li B v gp
ngi i xe my ti mt mt a im cch B 20 km.
Tnh qung ng AB.
Bi 4: (3im)
Cho hnh vung ABCD. M l mt im trn ng cho BD. K ME v MF vung
gc vi AB v AD.
a) Chng minh hai on thng DE v CF bng nhau v vung gc vi nhau.
b) Chng minh ba ng thng DE, BF v CM ng quy.
c) Xc nh v tr ca im M t gic AEMF c din tch ln nht.
Bi 5: (1im)
Tm nghim nguyn ca phng trnh: 3x 5 y 345
2 2

S 14
Bi 1: (2,5im)
Phn tch a thc thnh nhn t

Gv: Nguyn Vn T 22 Trng THCS Thanh M


Tuyn tp thi HSG Ton 8 Nm hc: 2011-2012
a) x5 + x +1
b) x4 + 4
c) x x - 3x + 4 x -2 vi x 0
Bi 2 : (1,5im)
Cho abc = 2 Rt gn biu thc:
a b 2c
A
ab a 2 bc b 1 ac 2c 2
Bi 3: (2im)
Cho 4a2 + b2 = 5ab v 2a b 0
ab
Tnh: P
4a b 2
2

Bi 4 : (3im)
Cho tam gic ABC cn ti A. Trn BC ly M bt k sao cho BM CM. T N v
ng thng song song vi AC ct AB ti E v song song vi AB ct AC ti F. Gi N l
im i xng ca M qua E F.
a) Tnh chu vi t gic AEMF. Bit : AB =7cm
b) Chng minh : AFEN l hnh thang cn
c) Tnh : ANB + ACB = ?
d) M v tr no t gic AEMF l hnh thoi v cn thm iu kin ca ABC
cho AEMF l hnh vung.
Bi 5: (1im)
Chng minh rng vi mi s nguyn n th :
52n+1 + 2n+4 + 2n+1 chia ht cho 23.

S 15

Bi 1: (2 im)
a) Phn tch thnh tha s: (a b c) 3 a 3 b 3 c 3
2 x 3 7 x 2 12 x 45
b) Rt gn:
3x 3 19 x 2 33x 9
Bi 2: (2 im)
Chng minh rng: A n 3 (n 2 7) 2 36n chia ht cho 5040 vi mi s t nhin n.
Bi 3: (2 im)
a) Cho ba my bm A, B, C ht nc trn ging. Nu lm mt mnh th my bm A
ht ht nc trong 12 gi, my bm B ht htnc trong 15 gi v my bm C ht ht nc
trong 20 gi. Trong 3 gi u hai my bm A v C cng lm vic sau mi dng n my
bm B.
Tnh xem trong bao lu th ging s ht nc.
b) Gii phng trnh: 2 x a x 2a 3a (a l hng s).
Bi 4: (3 im)
Cho tam gic ABC vung ti C (CA > CB), mt im I trn cnh AB. Trn na mt
phng b AB c cha im C ngi ta k cc tia Ax, By vung gc vi AB. ng thng
vung gc vi IC k qua C ct Ax, By ln lt ti cc im M, N.
a) Chng minh: tam gic CAI ng dng vi tam gic CBN.
Gv: Nguyn Vn T 23 Trng THCS Thanh M
Tuyn tp thi HSG Ton 8 Nm hc: 2011-2012
b) So snh hai tam gic ABC v INC.
c) Chng minh: gc MIN = 900.
d) Tm v tr im I sao cho din tch IMN ln gp i din tch ABC.
Bi 5: (1 im)
Chng minh rng s:
22499
..........
9100..........
09 l s chnh phng. ( n 2 ).
...
n-2 s 9 n s 0

S 16:
Cu 1 : ( 2 iem ) Phn tch biu thc sau ra tha s
M = 3 xyz + x ( y2 + z2 ) + y ( x2 + z2 ) + z ( x2 + y2 )
Cu 2 : ( 4 iem ) nh a v b a thc A = x4 6 x3 + ax2 + bx + 1 l bnh phng ca
mt a thc khc .
Cu 3 : ( 4 iem ) Cho biu thc :
x2 6 1 10 x 2

P= 3
: x 2 x 2

x 4 x 6 3 x x 2
a) Rt gn p .
3
b) Tnh gi tr ca biu thc p khi /x / =
4
c) Vi gi tr no ca x th p = 7
d) Tm gi tr nguyn ca x p c gi tr nguyn .
Cu 4 : ( 3 iem ) Cho a , b , c tha mn iu kin a2 + b2 + c2 = 1
Chng minh : abc + 2 ( 1 + a + b + c + ab + ac + bc ) 0
Cu 5 : ( 3iem)
Qua trng tm G tam gic ABC , k ng thng song song vi AC , ct AB v BC ln
lt ti M v N . Tnh di MN , bit AM + NC = 16 (cm) ; Chu vi tam gic ABC bng 75
(cm)
Cu 6 : ( 4 iem ) Cho tam gic u ABC . M, N l cc im ln lt chuyn ng trn
hai cnh BC v AC sao cho BM = CN xc nh v tr ca M , N di on thng MN
nh nht .

Gv: Nguyn Vn T 24 Trng THCS Thanh M


Tuyn tp thi HSG Ton 8 Nm hc: 2011-2012

S 17

Bi 1: (2 im)
Phn tch a thc sau y thnh nhn t:
1. x2 7 x 6
2. x4 2008x2 2007 x 2008
Bi 2: (2im) Gii phng trnh:
1. x2 3x 2 x 1 0
2 2 2
1 1 1 1
8 x 4 x 2 2 4 x 2 2 x x 4
2
2.
x x x x
Bi 3: (2im) 1. CMR vi a,b,c,l cc s dng ,ta c: (a+b+c)( 1 1 1 ) 9
a b c
3. Tm s d trong php chia ca biu thc x 2 x 4 x 6 x 8 2008
cho a
thc x2 10x 21 .
Bi 4: (4 im)Cho tam gic ABC vung ti A (AC > AB), ng cao AH
(H BC). Trn tia HC ly im D sao cho HD = HA. ng vung gc vi BC ti
D ct AC ti E.
1. Chng minh rng hai tam gic BEC v ADC ng dng. Tnh di on
BE theo m AB .
2. Gi M l trung im ca on BE. Chng minh rng hai tam gic BHM v
BEC ng dng. Tnh s o ca gc AHM
3. Tia AM ct BC ti G. Chng minh: GB HD .
BC AH HC

Bi Cu Ni dung im
1
1. 2,0
1.1 (0,75 im)

Gv: Nguyn Vn T 25 Trng THCS Thanh M


Tuyn tp thi HSG Ton 8 Nm hc: 2011-2012
x 7 x 6 x x 6 x 6 x x 1 6 x 1
2 2 0.5

x 1 x 6 0,5
1.2 (1,25 im)
x4 2008x2 2007 x 2008 x4 x2 2007 x2 2007 x 2007 1 0,25
x x 1 2007 x x 1 x 1 x 2007 x x 1
4 2 2 2 2 2 2
0,25
x x 1 x x 1 2007 x x 1 x x 1 x x 2008
2 2 2 2 2
0,25
2. 2,0
2.1 x 2 3x 2 x 1 0 (1)
+ Nu x 1: (1) x 1 0 x 1 (tha mn iu kin x 1).
2
0,5
+ Nu x 1: (1) x 4 x 3 0 x x 3 x 1 0 x 1 x 3 0
2 2

x 1; x 3 (c hai u khng b hn 1, nn b loi) 0,5


Vy: Phng trnh (1) c mt nghim duy nht l x 1 .
2.2 1
2
1
2
1 1
2

8 x 4 x 2 2 4 x 2 2 x x 4 (2)
2

x x x x
iu kin phng trnh c nghim: x 0
1 1
2 2
1 1
(2) 8 x 4 x 2 2 x 2 2 x x 4
2
0,25
x x x x
2
1 1
8 x 8 x 2 2 x 4 x 4 16
2 2
0,5
x x
x 0 hay x 8 v x 0 . 0,25
Vy phng trnh cho c mt nghim x 8
3 2.0
3.1 Ta c:
1 1 1 a a b b c c
A= (a b c)( ) 1 1 1
a b c b c a c a b 0,5
a b a c c b
=3 ( ) ( ) ( )
b a c a b c
x y
M: 2 (BT C-Si)
y x 0,5
Do A 3 2 2 2 9. Vy A 9
3.2 Ta c:
P( x) x 2 x 4 x 6 x 8 2008
x 2 10 x 16 x 2 10 x 24 2008
0,5
t t x 10 x 21 (t 3; t 7) , biu thc P(x) c vit li:
2

P( x) t 5 t 3 2008 t 2 2t 1993
Do khi chia t 2 2t 1993 cho t ta c s d l 1993 0,5
4 4,0

Gv: Nguyn Vn T 26 Trng THCS Thanh M


Tuyn tp thi HSG Ton 8 Nm hc: 2011-2012
4.1 + Hai tam gic ADC v BEC
c:
Gc C chung.
CD CA
(Hai tam gic
CE CB
vung CDE v CAB ng 1,0
dng)

Do , chng dng dng (c.g.c).


Suy ra: BEC ADC 1350 (v tam gic AHD vung cn ti H theo gi thit).
Nn AEB 450 do tam gic ABE vung cn ti A. Suy ra: 0,5
BE AB 2 m 2
4.2 BM 1 BE 1 AD
Ta c: (do BEC ADC ) 0,5
BC 2 BC 2 AC
m AD AH 2 (tam gic AHD vung vn ti H)
BM 1 AD 1 AH 2 BH BH 0,5
nn (do ABH CBA )
BC 2 AC 2 AC AB 2 BE
Do BHM BEC (c.g.c), suy ra: BHM BEC 1350 AHM 450 0,5
4.3 Tam gic ABE vung cn ti A, nn tia AM cn l phn gic gc BAC.
GB AB AB ED AH HD
Suy ra: , m ABC DEC ED // AH 0,5
GC AC AC DC HC HC
GB HD GB HD GB HD 0,5
Do :
GC HC GB GC HD HC BC AH HC

Phng GD & T huyn Thng Tn


Trng THCS Vn T
Gv: Bi Th Thu Hin
S 18
bi:
Bi 1( 6 im): Cho biu thc:
2x 3 2x 8 3 21 2 x 8 x 2
P= 2 : 1
4 x 12 x 5 13x 2 x 20 2 x 1 4 x 4 x 3
2 2

a) Rt gn P
1
b) Tnh gi tr ca P khi x
2
c) Tm gi tr nguyn ca x P nhn gi tr nguyn.
d) Tm x P > 0.
Bi 2(3 im):Gii phng trnh:
15 x 1 1
a) 2 1 12
x 3x 4 x 4 3x 3
148 x 169 x 186 x 199 x
b) 10
25 23 21 19
c) x2 3 5
Gv: Nguyn Vn T 27 Trng THCS Thanh M
Tuyn tp thi HSG Ton 8 Nm hc: 2011-2012
Bi 3( 2 im): Gii bi ton bng cch lp phng trnh:
Mt ngi i xe gn my t A n B d nh mt 3 gi 20 pht. Nu ngi y tng vn tc
thm 5 km/h th s n B sm hn 20 pht. Tnh khong cch AB v vn tc d nh i ca
ngi .
Bi 4 (7 im):
Cho hnh ch nht ABCD. Trn ng cho BD ly im P, gi M l im i xng ca
im C qua P.
a) T gic AMDB l hnh g?
b) Gi E v F ln lt l hnh chiu ca im M ln AB, AD. Chng minh EF//AC v ba
im E, F, P thng hng.
c) Chng minh rng t s cc cnh ca hnh ch nht MEAF khng ph thuc vo v tr
ca im P.
PD 9
d) Gi s CP BD v CP = 2,4 cm, . Tnh cc cnh ca hnh ch nht ABCD.
PB 16
Bi 5(2 im): a) Chng minh rng: 20092008 + 20112010 chia ht cho 2010
b) Cho x, y, z l cc s ln hn hoc bng 1. Chng minh rng:
1 1 2

1 x2 1 y 2 1 xy

p n v biu im
Bi 1: Phn tch:
4x2 12x + 5 = (2x 1)(2x 5)
13x 2x2 20 = (x 4)(5 2x)
21 + 2x 8x2 = (3 + 2x)(7 4x)
4x2 + 4x 3 = (2x -1)(2x + 3) 0,5
1 5 3 7
iu kin: x ; x ; x ; x ;x 4 0,5
2 2 2 4
2x 3
a) Rt gn P = 2
2x 5
1 1 1
b) x x hoc x
2 2 2
1 1
+) x P =
2 2
1 2
+) x P = 1
2 3
2x 3 2
c) P = = 1
2x 5 x 5
Ta c: 1 Z
2
Vy P Z khi Z
x 5
x5 (2)
M (2) = { -2; -1; 1; 2}
Gv: Nguyn Vn T 28 Trng THCS Thanh M
Tuyn tp thi HSG Ton 8 Nm hc: 2011-2012
x 5 = -2 x = 3 (TMK)
x 5 = -1 x = 4 (KTMK)
x5=1 x = 6 (TMK)
x5=2 x = 7 (TMK)
KL: x {3; 6; 7} th P nhn gi tr nguyn. 1
2x 3 2
d) P= = 1 0,25
2x 5 x 5
Ta c: 1 > 0
2
P > 0 th >0 x5>0 x>5 0,5
x5
Vi x > 5 th P > 0. 0,25
Bi 2:
15 x 1 1
1 12
a)
x 2 3x 4 x 4 3x 3
15 x 1 1
1 12 x 4; x 1
x 4 x 1 x 4 3 x 1
K:

3.15x 3(x + 4)(x 1) = 3. 12(x -1) + 12(x + 4)

3x.(x + 4) = 0
3x = 0 hoc x + 4 = 0
+) 3x = 0 => x = 0 (TMK)
+) x + 4 = 0 => x = -4 (KTMK)
S = { 0} 1
148 x 169 x 186 x 199 x
b)
10
25 23 21 19
148 x 169 x 186 x 199 x
1 2 3 4 0
25 23 21 19
1 1 1 1
(123 x) = 0
25 23 21 19
1 1 1 1
Do > 0
25 23 21 19
Nn 123 x = 0 => x = 123
S = {123} 1

c) x2 3 5
Ta c: x 2 0x => x2 3 >0

nn x2 3 x2 3
PT c vit di dng:

Gv: Nguyn Vn T 29 Trng THCS Thanh M


Tuyn tp thi HSG Ton 8 Nm hc: 2011-2012
x2 3 5
x2 =53
x2 =2
+) x - 2 = 2 => x = 4
+) x - 2 = -2 => x = 0
S = {0;4} 1
Bi 3(2 )
Gi khong cch gia A v B l x (km) (x > 0) 0,25
Vn tc d nh ca ngi xe gn my l:
x 3x
(km / h) 1
h )
1 10 (3h20 = 3 0,25
3 3
3
Vn tc ca ngi i xe gn my khi tng ln 5 km/h l:
3x
5 km / h 0,25
10
Theo bi ta c phng trnh:

3x
5 .3 x 0,5
10
x =150 0,5
Vy khong cch gia A v B l 150 (km) 0,25
3.150
Vn tc d nh l: 45 km / h
10
Bi 4(7)
V hnh, ghi GT, KL ng 0,5

D C
P

M
O
I F

E A B

a) Gi O l giao im 2 ng cho ca hnh ch nht ABCD.


PO l ng trung bnh ca tsm gic CAM.
AM//PO
t gic AMDB l hnh thang. 1
b) Do AM //BD nn gc OBA = gc MAE (ng v)
Tam gic AOB cn O nn gc OBA = gc OAB

Gv: Nguyn Vn T 30 Trng THCS Thanh M


Tuyn tp thi HSG Ton 8 Nm hc: 2011-2012
Gi I l giao im 2 ng cho ca hnh ch nht AEMF th tam gic AIE cn I nn
gc IAE = gc IEA.
T chng minh trn : c gc FEA = gc OAB, do EF//AC (1) 1
Mt khc IP l ng trung bnh ca tam gic MAC nn IP // AC (2)
T (1) v (2) suy ra ba im E, F, P thng hng. 1
c) MAF DBA g g nn
MF AD
khng i. (1)
FA AB
PD 9 PD PB
d) Nu th k PD 9k , PB 16k
PB 16 9 16
CP PB
Nu CP BD th CBD DCP g g 1
PD CP
do CP2 = PB.PD
hay (2,4)2 = 9.16 k2 => k = 0,2
PD = 9k = 1,8(cm)
PB = 16k = 3,2 (cm) 0,5d
BD = 5 (cm)
C/m BC2= BP.BD = 16 0,5
do BC = 4 (cm)
CD = 3 (cm) 0,5

Bi 5:
a) Ta c: 20092008 + 20112010 = (20092008 + 1) + ( 20112010 1)
V 20092008 + 1 = (2009 + 1)(20092007 - )
= 2010.() chia ht cho 2010 (1)
20112010 - 1 = ( 2011 1)(20112009 + )
= 2010.( ) chia ht cho 2010 (2) 1
T (1) v (2) ta c pcm.
1 1 2

b)
1 x2 1 y 2 1 xy (1)
1 1 1 1

0
1 x 2
1 xy 1 y 2
1 xy
x y x y x y
0
1 x2 1 xy 1 y 2 1 xy
y x xy 1 0 2
2


1 x2 1 y 2 1 xy
V x 1; y 1 => xy 1 => xy 1 0
=> BT (2) ng => BT (1) ng (du = xy ra khi x = y) 1

Gv: Nguyn Vn T 31 Trng THCS Thanh M


Tuyn tp thi HSG Ton 8 Nm hc: 2011-2012
S 19

Bi 1: (3) a) Phn tch a thc x3 5x2 + 8x 4 thnh nhn t


b) Tm gi tr nguyn ca x A B bit
A = 10x2 7x 5 v B = 2x 3 .
c) Cho x + y = 1 v x y 0 . Chng minh rng
x y 2 x y
3 2 2 0
y 1 x 1 x y 3
3

Bi 2: (3) Gii cc phng trnh sau:


a) (x2 + x)2 + 4(x2 + x) = 12
x 1 x 2 x 3 x 4 x 5 x 6
b)
2008 2007 2006 2005 2004 2003
Bi 3: (2) Cho hnh vung ABCD; Trn tia i tia BA ly E, trn tia i tia CB ly F sao cho AE = CF
a) Chng minh EDF vung cn
b) Gi O l giao im ca 2 ng cho AC v BD. Gi I l trung im EF. Chng minh O, C, I
thng hng.
Bi 4: (2)Cho tam gic ABC vung cn ti A. Cc im D, E theo th t di chuyn trn AB, AC sao cho
BD = AE. Xc nhv tr im D, E sao cho:
a/ DE c di nh nht
b/ T gic BDEC c din tch nh nht.

H-ng dn chm v biu im


Bi 1: (3 im)
a) ( 0,75) x3 - 5x2 + 8x - 4 = x3 - 4x2 + 4x x2 + 4x 4 (0,25)
= x( x2 4x + 4) ( x2 4x + 4) (0,25)
=(x1)(x2)2 (0,25)
b) (0,75) Xt A 10x 7x 5 5x 4 (0,25)
2
7
B 2x 3 2x 3
7
Vi x Z th A B khi Z 7 ( 2x 3) (0,25)
2x 3
M (7) = 1;1; 7;7 x = 5; - 2; 2 ; 1 th A B (0,25)
x y
3 = x 3 x y3 y
4 4
c) (1,5) Bin i
y 1 x 1 (y 1)(x 1)
3

= x y4 (x y) ( do x + y = 1 y - 1= -x v x - 1= - y) (0,25)
4

xy(y 2 y 1)(x 2 x 1)
= x y x y x 2 y2 (x y) (0,25)
xy(x 2 y 2 y 2 x y 2 yx 2 xy y x 2 x 1)
= x y (x 2 y2 1) (0,25)
xy x 2 y 2 xy(x y) x 2 y 2 xy 2

= x y (x 2
x y 2 y) = x y x(x 1) y(y 1) (0,25)
xy x 2 y 2 (x y) 2 2 xy(x 2 y 2 3)

= x y x( y) y(x) = x y (2xy) (0,25)


xy(x 2 y 2 3) xy(x 2 y 2 3)
=
2(x y) Suy ra iu cn chng minh (0,25)
x 2 y2 3

Gv: Nguyn Vn T 32 Trng THCS Thanh M


Tuyn tp thi HSG Ton 8 Nm hc: 2011-2012
Bi 2: (3 )a) (1,25)
(x2 + x )2 + 4(x2 + x) = 12 t y = x2 + x
y2 + 4y - 12 = 0 y2 + 6y - 2y -12 = 0 (0,25)
(y + 6)(y - 2) = 0 y = - 6; y = 2 (0,25)
* x2 + x = - 6 v nghim v x2 + x + 6 > 0 vi mi x (0,25)
* x2 + x = 2 x2 + x - 2 = 0 x2 + 2x - x - 2 = 0 (0,25)
x(x + 2) (x + 2) = 0 (x + 2)(x - 1) = 0 x = - 2; x = 1 (0,25)
Vy nghim ca phng trnh x = - 2 ; x =1
x 1 x 2 x 3 x 4 x 5 x 6 x 1 x2 x 3 x4 x5 x6
b) (1,75) ( 1) ( 1) ( 1) ( 1) ( 1) ( 1)
2008 2007 2006 2005 2004 2003 2008 2007 2006 2005 2004 2003

x 2009 x 2009 x 2009 x 2009 x 2009 x 2009 x 2009 x 2009 x 2009 x 2009 x 2009 x 2009
0
2008 2007 2006 2005 2004 2003 2008 2007 2006 2005 2004 2003
(0,25)
1 ; 1 1 ; 1 1
( x 2009)( 1

1

1

1

1

1
) 0 (0,5) V 1

2008 2007 2006 2005 2004 2003 2008 2005 2007 2004 2006 2003

Do : 1 1 1 1 1 1 0 (0,25) Vy x + 2009 = 0 x = -2009


2008 2007 2006 2005 2004 2003
E I
2
Bi 3: (2 im) 1
1
a) (1) B C 2 F
Chng minh EDF vung cn
Ta c ADE = CDF (c.g.c) EDF cn ti D
Mt khc: ADE = CDF (c.g.c) E 1 F2 O
M E 1 E 2 F1 = 900 F2 E 2 F1 = 900 A D
EDF = 90 . Vy EDF vung cn
0

b) (1) Chng minh O, C, I thng


Theo tnh cht ng cho hnh vung CO l trung trc BD
M EDF vung cn DI = EF
1
2
1 B
Tng t BI = EF DI = BI
2
I thuc dng trung trc ca DB I thuc ng thng CO
Hay O, C, I thng hng
D

Bi 4: (2 im) C
a) (1) A
DE c di nh nht E
t AB = AC = a khng i; AE = BD = x (0 < x < a)
p dng nh l Pitago vi ADE vung ti A c:
DE2 = AD2 + AE2 = (a x)2 + x2 = 2x2 2ax + a2 = 2(x2 ax) a2 (0,25)
a2 2 a2 a2
= 2(x ) + (0,25)
4 2 2
a
Ta c DE nh nht DE2 nh nht x = (0,25)
2

Gv: Nguyn Vn T 33 Trng THCS Thanh M


Tuyn tp thi HSG Ton 8 Nm hc: 2011-2012
a
BD = AE = D, E l trung im AB, AC (0,25)
2
b) (1)
T gic BDEC c din tch nh nht.
1 1 1 1
Ta c: SADE = AD.AE = AD.BD = AD(AB AD)= (AD2 AB.AD) (0,25)
2 2 2 2
2 2 2
1 AB AB AB 1 AB 2 AB AB2
= (AD2 2 .AD + )+ = (AD ) + (0,25)
2 2 4 8 2 4 2 8
AB2 AB2 3
Vy SBDEC = SABC SADE = AB2 khng i (0,25)
2 8 8
3
Do min SBDEC = AB2 khi D, E ln lt l trung im AB, AC (0,25)
8

S 20

Bi 1: Phn tch a thc thnh nhn t:


a) x2 y2 5x + 5y
b) 2x2 5x 7

Bi 2: Tm a thc A, bit rng:


4 x 2 16 A

x2 2 x
5x 5
Bi 3: Cho phn thc:
2x 2 2x
a) Tm iu kin ca x gi tr ca phn thc c xc nh.
b) Tm gi tr ca x gi tr ca phn thc bng 1.
x2 1 2
Bi 4: a) Gii phng trnh :
x 2 x x( x 2)
b) Gii bt phng trnh: (x-3)(x+3) < (x=2)2 + 3

Bi 5: Gii bi ton sau bng cch lp phng trnh:


Mt t sn xut lp k hoch sn xut, mi ngy sn xut c 50 sn
phm. Khi thc hin, mi ngy t sn xut c 57 sn phm. Do hon thnh
trc k hoch mt ngy v cn vt mc 13 sn phm. Hi theo k hoch t phi sn
xut bao nhiu sn phm v thc hin trong bao nhiu ngy.

Bi 6: Cho ABC vung ti A, c AB = 15 cm, AC = 20 cm. K ng cao AH v


trung tuyn AM.
a) Chng minh ABC ~ HBA
b) Tnh : BC; AH; BH; CH ?
c) Tnh din tch AHM ?

Biu im - p n

Gv: Nguyn Vn T 34 Trng THCS Thanh M


Tuyn tp thi HSG Ton 8 Nm hc: 2011-2012
p n Biu im
Bi 1: Phn tch a thc thnh nhn t:
a) x2 y2 5x + 5y = (x2 y2) (5x 5y) = (x + y) (x y) 5(x
y)
= (x - y) (x + y 5) (1 im)
b) 2x2 5x 7 = 2x2 + 2x 7x 7 = (2x2 + 2x) (7x + 7) = 2x(x +1)
7(x + 1)
= (x + 1)(2x 7). (1 im)
Bi 2: Tm A (1 im)
A=
x(4 x 2 16 x[(2 x) 2 4 2 x(2 x 4)(2 x 4) x.2( x 2).2( x 2)
4( x 2) 4 x 8
x 2 2x x 2 2x x( x 2) x( x 2)
Bi 3: (2 im)
a) 2x2 + 2x = 2x(x + 1) 0
2x 0 v x + 1 0
x 0 v x -1 (1 im)
b) Rt gn:
5x 5 5( x 1) 5
(0,5 im)
2 x 2 x 2 x( x 1) 2 x
2

5 5
1 5 2x x (0,25 im)
2x 2
5 5
V tho mn iu kin ca hai tam gic nn x (0,25 im)
2 2
Bi 4: a) iu kin xc nh: x 0; x 2
x(x 2) - (x - 2) 2
- Gii: x2 + 2x x +2 = 2;
x( x 2) x( x 2) 1
x= 0 (loi) hoc x = - 1. Vy S = 1
b) x2 9 < x2 + 4x + 7
x2 x2 4x < 7 + 9 - 4x < 16 x> - 4 1
Vy nghim ca phng trnh l x > - 4
Bi 5: Gi s ngy t d nh sn xut l : x ngy
iu kin: x nguyn dng v x > 1 0,5
Vy s ngy t thc hin l: x- 1 (ngy)
- S sn phm lm theo k hoch l: 50x (sn phm) 0,5
- S sn phm thc hin l: 57 (x-1) (sn phm)
Theo bi ta c phng trnh: 57 (x-1) - 50x = 13 0,5
57x 57 50x = 13
7x = 70 0,5
x = 10 (tho mn iu kin)
Vy: s ngy d nh sn xut l 10 ngy. 1
S sn phm phi sn xut theo k hoch l: 50 . 10 = 500 (sn phm)
Bi 6: a) Xt ABC v HBA, c:

Gv: Nguyn Vn T 35 Trng THCS Thanh M


Tuyn tp thi HSG Ton 8 Nm hc: 2011-2012
Gc A = gc H = 900; c gc B chung
ABC ~ HBA ( gc. gc) 1
b) p dng pitago trong vung ABC
1
ta c : BC = AB 2 AC 2 = 15 2 20 2 = 625 = 25 (cm)
AB AC BC 15 20 25
v ABC ~ HBA nn hay 1
HB HA BA HB HA 15
20.05
AH = 12 (cm)
25
15.15
BH = 9 (cm)
25 1
HC = BC BH = 25 9 = 16 (cm)
BC 25
c) HM = BM BH = BH 9 3,5(cm)
2 2
1 1
SAHM = AH . HM = . 12. 3,5 = 21 (cm2)
2 2 1
- V ng hnh: A

B H M C 1

S 21
Bi 1(3 im): Tm x bit:
a) x2 4x + 4 = 25
x 17 x 21 x 1
b) 4
1990 1986 1004
c) 4x 12.2x + 32 = 0

1 1 1
0.
Bi 2 (1,5 im): Cho x, y, z i mt khc nhau v
x y z
yz xz xy
Tnh gi tr ca biu thc: A 2 2 2
x 2 yz y 2xz z 2xy

Bi 3 (1,5 im): Tm tt c cc s chnh phng gm 4 ch s bit rng khi ta thm 1 n


v vo ch s hng nghn , thm 3 n v vo ch s hng trm, thm 5 n v vo ch s
hng chc, thm 3 n v vo ch s hng n v , ta vn c mt s chnh phng.

Bi 4 (4 im): Cho tam gic ABC nhn, cc ng cao AA, BB, CC, H l trc tm. a)
HA' HB' HC'
Tnh tng
AA' BB' CC'
b) Gi AI l phn gic ca tam gic ABC; IM, IN th t l phn gic ca gc AIC v gc
AIB. Chng minh rng: AN.BI.CM = BN.IC.AM.
Gv: Nguyn Vn T 36 Trng THCS Thanh M
Tuyn tp thi HSG Ton 8 Nm hc: 2011-2012
(AB BC CA) 2
c) Chng minh rng: 4.
AA'2 BB'2 CC'2
P N THI CHN HC SINH GII

Bi 1(3 im):
a) Tnh ng x = 7; x = -3 ( 1 im )
b) Tnh ng x = 2007 ( 1 im )
c) 4x 12.2x +32 = 0 2x.2x 4.2x 8.2x + 4.8 = 0 ( 0,25im )
x x x x x
2 (2 4) 8(2 4) = 0 (2 8)(2 4) = 0 ( 0,25im )
x 3 x 2 x 3 x 2
(2 2 )(2 2 ) = 0 2 2 = 0 hoc 2 2 = 0 ( 0,25im )
x 3 x 2
2 = 2 hoc 2 = 2 x = 3; x = 2 ( 0,25im )

Bi 2(1,5 im):
1 1 1 xy yz xz
0 0 xy yz xz 0 yz = xyxz ( 0,25im )
x y z xyz
x2+2yz = x2+yzxyxz = x(xy)z(xy) = (xy)(xz) ( 0,25im )

Tng t: y2+2xz = (yx)(yz) ; z2+2xy = (zx)(zy) ( 0,25im )

yz xz xy
Do : A ( 0,25im )
( x y)(x z) ( y x )( y z) (z x )(z y)

Tnh ng A = 1 ( 0,5 im )

Bi 3(1,5 im):
Gi abcd l s phi tm a, b, c, d N, 0 a, b, c, d 9, a 0 (0,25im)

Ta c: abcd k
2

vi k, m N, 31 k m 100
(a 1)(b 3)(c 5)(d 3) m 2 (0,25im)
abcd k 2

abcd 1353 m 2 (0,25im)

Do : m2k2 = 1353
(m+k)(mk) = 123.11= 41. 33 ( k+m < 200 ) (0,25im)
m+k = 123 m+k = 41
hoc
mk = 11 mk = 33
m = 67 m = 37
k = 56 hoc k= 4 (0,25im)
Kt lun ng abcd = 3136 (0,25im)

Bi 4 (4 im):

Gv: Nguyn Vn T 37 Trng THCS Thanh M


Tuyn tp thi HSG Ton 8 Nm hc: 2011-2012
V hnh ng (0,25im)
1 A
.HA'.BC
S HBC 2 HA'
a) S ; C (0,25im)
1 AA' x
ABC .AA'.BC H
B
2 N
M
S HAB HC' SHAC HB'
Tng t: ;
I A C
(0,25im)
S ABC CC' SABC BB' B
D
HA' HB' HC' SHBC SHAB SHAC
1 (0,25im)
AA' BB' CC' SABC SABC SABC
b) p dng tnh cht phn gic vo cc tam gic ABC, ABI, AIC:
BI AB AN AI CM IC
; ; (0,5im )
IC AC NB BI MA AI
BI AN CM AB AI IC AB IC
. . . . . 1 (0,5im )
IC NB MA AC BI AI AC BI (0,5im )
BI .AN.CM BN.IC.AM
c)V Cx CC. Gi D l im i xng ca A qua Cx (0,25im)
-Chng minh c gc BAD vung, CD = AC, AD = 2CC (0,25im)
- Xt 3 im B, C, D ta c: BD BC + CD (0,25im)
2 2 2
- BAD vung ti A nn: AB +AD = BD
2 2 2
AB + AD (BC+CD) (0,25im)
2 2 2
AB + 4CC (BC+AC)
4CC2 (BC+AC)2 AB2
Tng t: 4AA2 (AB+AC)2 BC2
4BB2 (AB+BC)2 AC2 (0,25im)
2 2 2 2
-Chng minh c : 4(AA + BB + CC ) (AB+BC+AC)
(AB BC CA) 2
4 (0,25im)
AA'2 BB'2 CC'2
(ng thc xy ra BC = AC, AC = AB, AB = BC AB = AC =BC
ABC u)

S 22
Cu 1: (5im) Tm s t nhin n :
a, A=n3-n2+n-1 l s nguyn t.
n 4 3n 3 2n 2 6n 2
b, B = C gi tr l mt s nguyn.
n2 2
c, D= n5-n+2 l s chnh ph-ng. (n 2)
Cu 2: (5im) Chng minh rng :
a b c
a, 1 bit abc=1
ab a 1 bc b 1 ac c 1

b, Vi a+b+c=0 th a4+b4+c4=2(ab+bc+ca)2

Gv: Nguyn Vn T 38 Trng THCS Thanh M


Tuyn tp thi HSG Ton 8 Nm hc: 2011-2012
2 2 2
a b c c b a
c, 2
2 2
b c a b a c
Cu 3: (5im) Gii cc ph-ng trnh sau:
x 214 x 132 x 54
a, 6
86 84 82
b, 2x(8x-1)2(4x-1)=9
c, x2-y2+2x-4y-10=0 vi x,ynguyn d-ng.
Cu 4: (5im). Cho hnh thang ABCD (AB//CD), 0 l giao im hai -ng cho.Qua 0 k
-ng thng song song vi AB ct DA ti E,ct BCti F.
a, Chng minh :Din tch tam gic AOD bng din tch tam gic BOC.
1 1 2
b. Chng minh:
AB CD EF
c, Gi Kl im bt k thuc OE. Nu cch dng -ng thng i qua Kv chia i din tch
tam gic DEF.

Cu Ni dung bi gii im
a, (1im) A=n3-n2+n-1=(n2+1)(n-1) 0,5
A l s nguyn t th n-1=1 n=2 khi A=5 0,5
2
b, (2im) B=n2+3n- 2 0,5
n 2
B c gi tr nguyn 2 n2+2 0,5
Cu 1 n2+2 l -c t nhin ca 2 0,5
n2+2=1 khng c gi tr tho mn 0,5
(5im)
Hoc n2+2=2 n=0 Vi n=0 th B c gi tr nguyn.
c, (2im) D=n5-n+2=n(n4-1)+2=n(n+1)(n-1)(n2+1)+2
=n(n-1)(n+1) n 2 4 5 +2= n(n-1)(n+1)(n-2)(n+2)+5 n(n-
0,5
0,5
1)(n+1)+2
M n(n-1)(n+1)(n-2)(n+2 5 (tich 5s t nhin lin tip) 0,5
V 5 n(n-1)(n+1 5 Vy D chia 5 d- 2
Do s D c tn cng l 2 hoc 7nn D khng phi s chnh 0,5
ph-ng
Vy khng c gi tr no ca n D l s chnh ph-ng
a, (1im)
a b c ac abc c

ab a 1 bc b 1 ac c 1 abc ac c abc abc ac ac c 1
2
0,5
ac abc c abc ac 1
= 1
1 ac c c 1 ac ac c 1 abc ac 1 0,5

Gv: Nguyn Vn T 39 Trng THCS Thanh M


Tuyn tp thi HSG Ton 8 Nm hc: 2011-2012

b, (2im) a+b+c=0 a2+b2+c2+2(ab+ac+bc)=0 a2+b2+c2= -


2(ab+ac+bc) 0.5
a4+b4+c4+2(a2b2+a2c2+b2c2)=4( a2b2+a2c2+b2c2)+8abc(a+b+c) V
a+b+c=0 0.5
Cu 2 a4+b4+c4=2(a2b2+a2c2+b2c2) (1) 0.5
(5im) Mt khc 2(ab+ac+bc)2=2(a2b2+a2c2+b2c2)+4abc(a+b+c) . V
a+b+c=0 0.5
2(ab+ac+bc)2=2(a2b2+a2c2+b2c2) (2)
T (1)v(2) a4+b4+c4=2(ab+ac+bc)2

c, (2im) p dng bt ng thc: x2+y2 2xy Du bng khi 0,5


x=y 0,5
a2 b2 a b a a2 c2 a c c 0,5
2
2 2. . 2. ; 2
2 2. . 2. ;
b c b c c b a b a b
2 2
c b c b b
2
2 2. . 2. 0,5
a c a c a
Cng tng v ba bt ng thc trn ta c:
a 2 b2 c2 a c b
2( 2 2 2 ) 2( )
b c a c b a
2 2 2
a b c a c b
2 2 2
b c a c b a

x 214 x 132 x 54
a, (2im) 6
86 84 82
x 214 x 132 x 54 1,0
( 1) ( 2) ( 3) 0
86 84 82
x 300 x 300 x 300 0,5
0
86 84 82
1 1
0 x-300=0 x=300 Vy S = 300
1 0,5
(x-300)
86 84 82
b, (2im) 2x(8x-1)2(4x-1)=9
(64x2-16x+1)(8x2-2x)=9 (64x2-16x+1)(64x2-16x) = 72 0,5
Cu 3 t: 64x2-16x+0,5 =k Ta c: (k+0,5)(k-0,5)=72 k2=72,25 0,5
(5im) k= 8,5
Vi k=8,5 tac ph-ng trnh: 64x2-16x-8=0 (2x-1)(4x+1)=0; 0,5
1 1
x= ; x 0,5
2 4
Vi k=- 8,5 Ta c ph-ng trnh: 64x -16x+9=0 (8x-1) +8=0 v
2 2

nghim.
1 1
Vy S = ,
2 4

Gv: Nguyn Vn T 40 Trng THCS Thanh M


Tuyn tp thi HSG Ton 8 Nm hc: 2011-2012
c, (1im) x -y +2x-4y-10 = 0 (x +2x+1)-(y2+4y+4)-7=0
2 2 2
0,5
(x+1) -(y+2) =7 (x-y-1)(x+y+3) =7 V x,y nguyn
2 2

d-ng 0,5
Nn x+y+3>x-y-1>0 x+y+3=7 v x-y-1=1 x=3 ; y=1
Ph-ng trnh c nghim d-ng duy nht (x,y)=(3;1)
a,(1im) V AB//CD S DAB=S CBAA B 0,5
(cng y v cng -ng cao) 0,5
S DAB SAOB = S CBA- SAOB K O
E F
Hay SAOD = SBOC I
M 0,5
N
C
D 1,0

0,5
EO AO
b, (2im) V EO//DC Mt khc AB//DC 1,0
Cu 4 DC AC
(5im) AB AO AB AO AB AO EO AB
1,0
DC OC AB BC AO OC AB BC AC DC AB DC
EF AB AB DC 2 1 1 2

2 DC AB DC AB.DC EF DC AB EF
c, (2im) +Dng trung tuyn EM ,+ Dng EN//MK (N DF) +K
-ng thng KN l -ng thng phi dng
Chng minh: SEDM=S EMF(1).Gi giao ca EM v KN l I th
SIKE=SIMN
(cma) (2) T (1) v(2) SDEKN=SKFN.

KIM TIN QUA MNG VIT NAM


Qu thy c v bn hy dnh thm mt cht thi gian c bi gii thiu sau ca ti v hy tri n
ngi ng ti liu ny bng cch dng Email v m s ngi gii thiu ca ti theo hng dn sau.
N s mang li li ch cho chnh thy c v cc bn, ng thi tri n c vi ngi gii thiu mnh:

Knh cho qu thy c v cc bn.


Li u tin cho php ti c gi ti qu thy c v cc bn li chc tt p nht. Khi thy c v
cc bn c bi vit ny ngha l thy c v cc bn c thin hng lm kinh doanh
Ngh gio l mt ngh cao qu, c x hi coi trng v tn vinh. Tuy nhin, c l cng nh ti thy
rng ng lng ca mnh qu hn hp. Nu khng phi mn hc chnh, v nu khng c dy thm, liu
rng tin lng c cho nhng nhu cu ca thy c. Cn cc bn sinh vinvi bao nhiu th phi trang
tri, tin gia nh gi, hay i gia s kim tin thm liu c ?
Bn thn ti cng l mt gio vin dy mn TON v vy thy c s hiu tin lng mi thng thu v
s c bao nhiu. Vy lm cch no kim thm cho mnh 4, 5 triu mi thng ngoi tin lng.

Gv: Nguyn Vn T 41 Trng THCS Thanh M


Tuyn tp thi HSG Ton 8 Nm hc: 2011-2012
Thc t ti thy rng thi gian thy c v cc bn lt web trong mt ngy cng tng i nhiu.
Ngoi mc ch kim tm thng tin phc v chuyn mn, cc thy c v cc bn cn su tm, tm hiu thm
rt nhiu lnh vc khc. Vy ti sao chng ta khng b ra mi ngy 5 n 10 pht lt web kim cho
mnh 4, 5 triu mi thng.
iu ny l c th?. Thy c v cc bn hy tin vo iu . Tt nhin mi th u c gi ca n. qu
thy c v cc bn nhn c 4, 5 triu mi thng, cn i hi thy c v cc bn s kin tr, chu kh v
bit s dng my tnh mt cht. Vy thc cht ca vic ny l vic g v lm nh th no? Qu thy c v
cc bn hy c bi vit ca ti, v nu c hng th th hy bt tay vo cng vic ngay thi.
Thy c chc nghe nghiu n vic kim tin qua mng. Chc chn l c. Tuy nhin trn internet
hin nay c nhiu trang Web kim tin khng uy tn
( l nhng trang web nc ngoi, nhng trang web tr th lao rt cao...). Nu l web nc ngoi th
chng ta s gp rt nhiu kh khn v mt ngn ng, nhng web tr th lao rt cao u khng uy tn, chng
ta hy nhn nhng g tng xng vi cng lao ca chng ta, l s tht.
Vit Nam trang web tht s uy tn l : http://satavina.com .Lc u bn thn ti cng thy
khng chc chn lm v cch kim tin ny. Nhng gi ti hon ton tin tng, n gin v ti c
nhn tin t cng ty.( thy c v cc bn c tch ly c 50.000 thi v yu cu satavina thanh ton bng
cch np th in thoi l s tin ngay).Tt nhin thi gian u s tin kim c chng bao nhiu, nhng sau
s tin kim c s tng ln. C th thy c v cc bn s ni: l v vn, chng ai t nhin mang tin
cho mnh. ng chng ai cho khng thy c v cc bn tin u, chng ta phi lm vic, chng ta phi
mang v li nhun cho h. Khi chng ta c qung co, xem video qung co ngha l mang v doanh thu
cho Satavina, ng nhin h n cm th chng ta cng phi c cho m n ch, khng th ai di g m lm
vic cho h.

Vy chng ta s lm nh th no y. Thy c v cc bn lm nh ny nh:


1/ Satavina.com l cng ty nh th no:
l cng ty c phn hot ng trong nhiu lnh vc, tr s ti ta nh Femixco, Tng 6, 231-233 L
Thnh Tn, P.Bn Thnh, Q.1, TP. H Ch Minh.
GPKD s 0310332710 - do S K Hoch v u T TP.HCM cp. Giy php ICP s 13/GP-STTTT do S
Thng Tin & Truyn Thng TP.HCM cp.qun 1 Thnh Ph HCM.
Khi thy c l thnh vin ca cng ty, thy c s c hng tin hoa hng t vic c qung co v
xem video qung co( tin ny c trch ra t tin thu qung co ca cc cng ty qung co thu trn
satavina)
2/ Cc bc ng k l thnh vin v cch kim tin:
ng k lm thnh vin satavina thy c lm nh sau:
Bc 1:
Nhp a ch web: http://satavina.com vo trnh duyt web( Dng trnh duyt firefox, khng nn dng trnh
duyt explorer)
Giao din nh sau:

Gv: Nguyn Vn T 42 Trng THCS Thanh M


Tuyn tp thi HSG Ton 8 Nm hc: 2011-2012

nhanh chng qu thy c v cc bn c th coppy


ng linh sau:
http://satavina.com/Register.aspx?hrYmail=hoangngocc2tmy@gmail.com&hrID=66309

( Thy c v cc bn ch in thng tin ca mnh l c. Tuy nhin, chc nng ng k thnh vin mi ch
c m vi ln trong ngy. Mc ch l thy c v cc bn tm hiu k v cng ty trc khi gii thiu
bn b )

Bc 2:
Click chut vo mc ng k, gc trn bn phi( c th s khng c giao din bc 3 v thi gian
ng k khng lin tc trong c ngy, thy c v cc bn phi tht kin tr).
Bc 3:
Nu c giao din hin ra. thy c khai bo cc thng tin:

Gv: Nguyn Vn T 43 Trng THCS Thanh M


Tuyn tp thi HSG Ton 8 Nm hc: 2011-2012

Thy c khai bo c th cc mc nh sau:


+ Mail ngi gii thiu( l mail ca ti, ti l thnh vin chnh thc):
hoangngocc2tmy@gmail.com
+ M s ngi gii thiu( Nhp chnh xc) : 66309
Hoc qu thy c v cc bn c th coppy Link gii thiu trc tip:
http://satavina.com/Register.aspx?hrYmail=hoangngocc2tmy@gmail.com&hrID=66309

+ a ch mail: y l a ch mail ca thy c v cc bn. Khai bo a ch tht cn vo kch hot ti


khon nu sai thy c v cc bn khng th l thnh vin chnh thc.
+ Nhp li a ch mail:.....
+ Mt khu ng nhp: nhp mt khu khi ng nhp trang web satavina.com
+ Cc thng tin mc:

Gv: Nguyn Vn T 44 Trng THCS Thanh M


Tuyn tp thi HSG Ton 8 Nm hc: 2011-2012
Thng tin ch ti khon: thy c v cc bn phi nhp chnh xc tuyt i, v thng tin ny ch c nhp
1 ln duy nht, khng sa c. Thng tin ny lin quan n vic giao dch sau ny. Sai s khng giao dch
c.
+ Nhp m xc nhn: nhp cc ch, s c bn cnh vo trng
+ Click vo mc: ti c k hng dn.....
+ Click vo: NG K
Sau khi ng k web s thng bo thnh cng hay khng. Nu thnh cng thy c v cc bn vo hm th
khai bo kch hot ti khon. Khi thnh cng qu thy c v cc bn vo web s c y thng tin
v cng ty satavina v cch thc kim tin. Hy tin vo li nhun m satavina s mang li cho thy c. Hy
bt tay vo vic ng k, chng ta khng mt g, ch mt mt cht thi gian trong ngy m thi.
Knh chc qu thy c v cc bn thnh cng.
Nu qu thy c c thc mc g trong qu trnh tch ly tin ca mnh hy gi trc tip hoc mail cho
ti:

Ngi gii thiu: Nguyn Vn T Email ngi


gii thiu: hoangngocc2tmy@gmail.com
M s ngi gii thiu: 66309
Qu thy c v cc bn c th coppy Link gii thiu trc tip:
http://satavina.com/Register.aspx?hrYmail=hoangngocc2tmy@gmail.com&hrID=66309

2/ Cch thc satavina tnh im quy ra tin cho thy c v cc bn:


+ im ca thy c v cc bn c tch ly nh vo c qung co v xem video qung co.
Nu ch tch ly im t chnh ch cc thy c v cc bn th 1 thng ch c khong 1tr.Nhng tng
im thy c cn pht trin mng li bn b ca thy c v cc bn.
3/ Cch thc pht trin mng li:
- Xem 1 qung co video: 10 im/giy. (c hn 10 video qung co, mi video trung bnh 1 pht)
- c 1 tin qung co: 10 im/giy. (hn 5 tin qung co)
_Tr li 1 phiu kho st.:100,000 im / 1 bi.
_Vit bi....
Trong 1 ngy bn ch cn dnh t nht 5 pht xem qung co, bn c th kim c: 10x60x5= 3000 im,
nh vy bn s kim c 300ng .
- Bn gii thiu 10 ngi bn xem qung co (gi l Mc 1 ca bn), 10 ngi ny cng dnh 5 pht xem
qung co mi ngy, cng ty cng chi tr cho bn 300ng/ngi.ngy.
- Cng tng t nh vy 10 Mc 1 ca bn gii thiu mi ngi 10 ngi th bn c 100 ngi (gi l mc
2 ca bn), cng ty cng chi tr cho bn 300ng/ngi.ngy.
- Tng t nh vy, cng ty chi tr n Mc 5 ca bn theo s sau :
- Nu bn xy dng n Mc 1, bn c 3.000ng/ngy
90.000 ng/thng.
- Nu bn xy dng n Mc 2, bn c 30.000ng/ngy
900.000 ng/thng.
- Nu bn xy dng n Mc 3, bn c 300.000ng/ngy
9.000.000 ng/thng.
- Nu bn xy dng n Mc 4, bn c 3.000.000ng/ngy
90.000.000 ng/thng.
- Nu bn xy dng n Mc 5, bn c 30.000.000ng/ngy
900.000.000 ng/thng.
Tuy nhin thy c v cc bn khng nn m t n mc 5. Ch cn c gng 1thng c 1=>10 triu l
qu n ri.
Nh vy thy c v cc bn thy satavina khng cho khng thy c v cc bn tin ng khng. Vy hy
ng k v gii thiu mng li ca mnh ngay i.
Lu : Ch khi thy c v cc bn l thnh vin chnh thc th thy c v cc bn mi c php gii thiu
ngi khc.
Gv: Nguyn Vn T 45 Trng THCS Thanh M
Tuyn tp thi HSG Ton 8 Nm hc: 2011-2012
Hy gii thiu n ngi khc l bn b thy c v cc bn nh ti gii thiu v hy quan tm n
nhng ngi m bn gii thiu v chm sc h( khi l thnh vin thy c v cc bn s c m s
ring).Khi gii thiu bn b hy thay ni dung mc thng tin ngi gii thiu l thng tin ca thy c v
cc bn. Chc qu thy c v cc bn thnh cng v c th kim c 1 khon tin cho ring mnh.
Ngi gii thiu: Nguyn Vn T Email ngi gii
thiu: hoangngocc2tmy@gmail.com
M s ngi gii thiu: 66309
Qu thy c v cc bn c th coppy Link gii thiu trc tip:

http://satavina.com/Register.aspx?hrYmail=hoangngocc2tmy@gmail.com&hrID=66309

Website: http://violet.vn/nguyentuc2thanhmy

HY KIN NHN BN S THNH CNG


Chc bn thnh cng!

Gv: Nguyn Vn T 46 Trng THCS Thanh M


Tuyn tp thi HSG Ton 8 Nm hc: 2011-2012

Gv: Nguyn Vn T 47 Trng THCS Thanh M

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