Professional Documents
Culture Documents
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 .
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
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).
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 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 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
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
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
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
E THI SO 4
Cau1( 2 ): Phan tch a thc sau thanh nhan t
A a 1 a 3 a 5 a 7 15
Kh a ta co : 0,25
mn = 10( m + n 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
Bi 2: (2 im)
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
= 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
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
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
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 )
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
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
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
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)
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
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
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)
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.
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
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
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 .
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)
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
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
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:
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:
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
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
= 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 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
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
S 20
Biu im - p n
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:
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 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 )
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):
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
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
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
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.
( 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:
http://satavina.com/Register.aspx?hrYmail=hoangngocc2tmy@gmail.com&hrID=66309
Website: http://violet.vn/nguyentuc2thanhmy