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book.mathvn.com 1 www.mathvn.com
VO LWWW.MATHVN.COM
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13
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12
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32
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512
412
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BD
AC
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1 1 1 1 1A= :
1- x 1 1 1 1x x x x
7 4 3
2 3 5 0x x
2 21 2
1 1
x x
2 21 2x x 3 3
1 2
1 1
x x
1 2x x
1 1 2:
2
a a a a a
aa a a a
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1 13
2 31
x y x y
x y x y
2 2 2
5 5 25
5 2 10 2 50
x x x
x x x x x
1 1 1 1 1
1 1 1 1 1
a a
a a a a a
AMB HMK
( ) 6
( ) 12
( ) 30
xy x y
yz y z
zx z x
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5
203
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)2;2
1
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C
CMD
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book.mathvn.com 13 www.mathvn.com
)..(2
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21 xx
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x
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13
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5
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3
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xxx
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231 xx
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2
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NB IB
53 3myxymx
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221
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2
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1
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25
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8
1622
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ymxmyx
232
12A
222
1B
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1C
32
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1ba
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1;
12
a
bx
b
a
2
2
x
21212 xxxx
22 11 xyyxS
ayxxy )1)(1( 22
xx 12
2
2x
y
21212 xxxx
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512
412
x
x
x
x
2 5
8152 xx
22
21 xx
33
6;
211
9ba
2
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7
522
xyyx
xyyx
BD
AC
DADCBCBA
CDCBADAB
..
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xyyxS
4
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4
2x
y
413 xx
011322
xx
1 1 1 1 1A= :
1- x 1 1 1 1x x x x
7 4 3
2 3 5 0x x
2 21 2
1 1
x x
2 21 2x x
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3 31 2
1 1
x x1 2x x
1 1 2:2
a a a a aaa a a a
1 13
2 31
x y x y
x y x y
2 2 2
5 5 25
5 2 10 2 50
x x x
x x x x x
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1 1 1 1 1
1 1 1 1 1
a a
a a a a a
AMB HMK
( ) 6
( ) 12
( ) 30
xy x y
yz y z
zx z x
2 3
5 4
x y
y x
3 1 4 4a > 0 ; a 4
42 2
a a a
aa a
3 3
1 2 0x x
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2
2
1
x m
x
1 2 5x x
1 1 2( 0; 0)
2 2 2 2 1
x xx x
x x x
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61
Cu 1.
1.Ch 9 4 2 2 2 1.
2.Rt g A 4 9 4 2 .Cu 2 nh 2x2 + 3x + 2m 1 = 01.Gi nh v2.Tm m trnh c hai nghi
Cu 3. M nh ch 1200m2
b i 4m thm 2 tu bCu 4 n tm O n tm A bnknh nh n (O) t D, c n cung nh
c a) Ch cc ti c n (A).b) Ch phn gic cc) Chd) Gi b.
Cu 5. Tm gi tr 2 + 3x + 4.
62
Cu 1. Tm hai s tb l 116.Cu 2 nh x2 7x + m = 0
a) Gi nh khi m = 1.b) G 1, x2 l cc nghi nh. Tnh S = x1
2 + x22.
c) Tm m nh c hai nghiCu 3. Cho tam gic DEF c D = 600, cc gc E, F l gc nh
n tm O. Cca) Tnh s
b) Ch
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c) Ch tm tdCu 4. Cho a, b l 2 s
2 22 2 2 2 a b a ba b a a b b
2
63
Cu 1.Th1
a) 2 6 4 3 5 2 8 .3 64
2 2b)
3 5 3 5
Cu 2 nh x2 2x 3m2 = 0 (1).
a) Gi nh khi m = 0.b) Tm m nh c hai nghic) Ch nh 3m2x2 + 2x 1 = 0 (m
phn bi m ngh nh (1).Cu 3. Cho tam gic ABC vung cn t trung tuy
hnh chi nAB, AC; H l hnh chi
a) T hnh g?b) Ch ng n n m n. Xc
tm c nc) Ch ng.
Cu 4. Tm nghi nh 2 3 3 x 3 y 3
64
Cu 1. Cho bia 3 a 2 a a 1 1
P :a 1 a 1 a 1a 2 a 1
a) Rt g
b) Tm a1 a 1
1P 8
Cu 2. M ng t ngB 72km, th ng t h ng l 15 pht. Tnh vt ng c dng n 4km/h.Cu 3. Tm t B c m s y = x2.G C l hnh chi B ln tr nh. Tnh di
tch t
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Cu 4 a OA v dy MN vunggc v y trn cung nh MN.
a) Chb) Tnh tch AH.AK theo R.
tnhgi trCu 5. Cho hai s nCh 2y2(x2 + y2) 2
65
Cu 1. Cho bix 1 2 x
P 1 : 1x 1 x 1 x x x x 1
a) Tm rt g
b) Tm cc gi tr nguyn c P x nh n.Cu 2.
a) Gi nh x4 4x3 2x2 + 4x + 1 = 0.
b) Gi2 2
2
x 3xy 2y 0
2x 3xy 5 0
Cu 3. Trong m nh2x
y2
. G
- 2) v c ha) Vi nh d
b) G hnh chi n tr nh.
ChCu 4 tit M D.Gminh r
b) B ng thu n.n c
d) Tm J c n ngocCu 5. Cho hai s n y tm gi tr
c 2 21 1
Ax y xy
.
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66
Cu 1.a) Gi nh 5x2 + 6 = 7x 2.
b) Gi nh 3x y 5x 2y 4
c) Tnh18 12
2 3
Cu 2. Cho (P) y = -2x2o thu
A(-1; -2); B(1 1
;2 2
); C( 2; 4)
b) Tm kc) Ch 2 + 1) khng thu
Cu 3. Cho tam gic ABC vung t. T
a) Ch AHD bb) Ch hai gc HCE v HAE bc) Chd) Che) Tnh gc BCA n
Cu 4.Cho hm s th n21f x 3f x
xv
67
Cu 1.
a) Tnh
9 1
2 1 5 : 1616 16
b) Gi3x y 2
x y 6
c) Ch 3 2 l nghi nh x2 6x + 7 = 0.
Cu 2. Cho (P): 21
y x3
.
1A 1; ; B 0; 5 ; C 3;1
3o thu
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b) Tm k nh y = kx 3 ti
c) Ch r 2 ctCu 3 tituy Q.
a) Chb) Chc) Ch
CD.
gic ABC.Cu 4. Tm gi tr 2 2A 2x 2xy y 2x 2y 1.
68
Cu 1.
1.Choa a a a
P 1 1 ; a 0, a 1a 1 1 a
a) Rt g
b) Tm a bi 2 .
c) Tm a bi a .
2.Ch 13 30 2 9 4 2 5 3 2
Cu 2 nh mx2 2(m-1)x + m = 0 (1)a) Gi nh khi m = - 1.b) Tm m nh (1) c 2 nghi
c) G x1 , x2. Hy l nh nh 1 22 1
x x;
x xlm
nghiCu 3.Cho tam gic nh n tm O,
P.a) Ch OP vung gc v
b) Tnh di rn l R v tgQAD =3
4.
Cu 4.a)Gi nh ax2 + bx + c = 0 c nghi 1. Ch
nh cx2 + bx + a = 0 c 2 v x1 + x2 0.
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b)Tm c n nh x2y + 2xy 4x + y = 0 sao cho y
69
Cu 1.1.Cho
2 2
2
1 2x 16x 1P ; x
1 4x 2
a) Ch2
P1 2x
b) Tnh P khi3
x2
2.Tnh2 5 24
Q12
Cu 2. Cho hai p nh2 2x x 2 0 (1); x 3b 2a x 6a 0 (2)
a) Gi nh (1).b) Tm a v b nhc) V m a nh (2) c nghi 1, x2 th n x1
2 + x22 =
7Cu 3. Cho tam gic ABC vung gc B ll trung tuy n tm H bn knh HA c
tha) Ch ng.b) Ch MAE DAE; MA DE .c) Ch n tm O. T
AMOH l hnh g?d) Cho gc ACB b 0 v AH = a. Tnh di
Cu 4.Gi nh2 2ax ax - a 4a 1
x 2a
. V
70
Cu 1.
1.Rt g 2 3 2 2 3 2 3 2 3 2 2 .
2.Choa b
xb a
v
a) Ch 2x 4 0 .
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b) Rt g 2F x 4 .
Cu 2 nh 2 2x 2 x 2mx 9 0 (*) ; x l tham s
a) Gi - 5.b) Tm m
Cu 3. Cho hm s y = - x2
(P); hm s (d).1.V (d) trn cng m m t (d).
-1; -2), b y cho bi nth
3.Tm nhCu 4. Cho tam gic nh hnh chith
1.Ch ti2.Gc DFE (D thu m phn gic trong v H l giao
ng.3.Tia DE c tam gic g th t
AFEK l hnh bnh hnh, l hnh thoi? GiCu 5. Hy tnh 1999 1999 1999F x y z
nh:
x y z a xy yz zx a xyz 0; a 0
71Cu 1.
1.Gi nh, h nh, ph nh2 2x 3y 12a) 2x 6 0 b) x x 6 0 c)
3x y 7
2.T a nghi nh, ph nh, hnh sau:
2 p 3 q 12a) 2 y 6 0 b) t t 6 0 c)
3 p q 7
Cu 2.
1.Ch2 2
1 2a 3 12a 2 2a .
2.Rt g
2 3 2 3 3 2 32 24 8 6
3 2 4 2 2 3 2 3 2 3
Cu 3. Cho tam gic ABC (AC > AB) c AM l trung tuy trnn (O)
n (O) c i
gc A t
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n (O) cminh hai t
3.Ch 2 = FI.FA.4.Ch
Cu 4. Rt g
2 2 2 2 2 2 2 2
1 1 1 1 1 1 1 1T 1 1 1 ... 12 3 3 4 4 5 1999 2000
72Cu 1.Gi nh sau
1) 4x 1 = 2x + 5 2) x2 8x + 15 = 0 3)2x 8x 15
02x 6
Cu 2.
1.Ch2
3 2 2 1 2 .
2.Rt g 3 2 2 .
3.Ch2 2
1 13 2 17 2 2 17
2 2 7 2 2 17
Cu 3 n (O)BC t
I.1.Ch2.G gc DEA
b3.Ch4.AT l ti ti ch o khi
Cu 4.1.Cho tam gic ABC c BC = a, AC = b, AB = c, G l tr
l kho
x y z
bc ac ab 2.Gi trnh
25 4 2025
x 1 y 3 z 24 104x 1 y 3 z 24
73
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Cu 1.Gi nh
2 2
2
x 2x y 0
x 2xy 1 0
Cu 2. Gi nh (x 1)(x + 2) < x2 + 4.
Cu 3.1.Rut g
1P 175 2 2
8 7.
2.V o c ph nh 2x2 4x m + 3 = 0 (m l tham sv nghiCu 4. Cho tam gic ABC c ba gc nh
n ngo c1.Ch BAM PQM; BPD BMA .2.Ch BA.DP.
3.Gi BPBM
theo a, b, m.
4.Gng.
74Cu 1.
1.Gi nh (x + 1)(x 4) < 0.2.Gi bi nh 1 x mx m v tham s
Cu 2. Gi nh
3 6 12x y x y
1 10
2x y x y
Cu 3. Tm gi tr 2 2P x 26y 10xy 14x 76y 59 . Khiu?
Cu 4. Cho hnh thoi ABCD c gc nh BAD . Vpha ngoi hnh thoi v tam gic
1.Tm tm c n
2.Ch r BD = 2a.sin2
.
3.Tnh gc ABK theo .4.Ch n m
Cu 5. Gi nh2
x x 2 1 1 x
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2. Tnh gi tr m s cc gi tr
Cu 4 AB v CD. Ti BD t E, F. G P v Q l trung
AF.
1.Ch Cd c v o th tam gic BPQc di y tnh di
77
Cu 1. Cho a, b, c l ba s
1 1 1x ; y ; zb c c a a b
Ch x + y = 2z.Cu 2 nh ph nh:
x2 (2a 1)x + 2(a tr
Cu 3. Gi nh:
2 2 2 2
2 2 2 2
x xy y x y 185
x xy y x y 65
Cu 4 n (O1) v (O2) c B. V 1) tixc v 2) t 2) ti 1) t
1. Ch2
2
BE AEBF AF
.
2.G v FBC.
3.Ch
78Cu 1.
1.Gi nh:
22
2 1 9 31
5 2 10 4a) b) 2x 1 5x 4x 1
22
2.Gi nh:x y 3 3x 2y 6z
a) b)xy 10 x y z 18
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3.G n ccti
Cu 4. Cho 2 2x x 1999 y y 1999 1999 . Tnh S = x + y.
80
Cu 1.
1.Cho2
1 1M 1 a : 1
1 a 1 a
a) Tm tb) Rt g
c) Tnh gi tr 3a 2 3 .
2.Tnh 40 2 57 40 2 57
Cu 2.nh (m + 2)x2 2(m 1) + 1 = 0 (1)
a) Gi nh khi m = 1.b) Tm m nh (1) c nghic) Tm m m h n h
khng ph o m.2.Cho ba s n a > 0; a2 = bc; a + b + c = abc. Ch2 2 2a) a 3, b 0, c 0. b) b c 2a
Cu 3. Cho (O) v m y trn cung l1.Nu cch d 1) qua M v ti n (O2) qua M
v ti2.G n (O1) v (O2). Ch
0AMB ANB 180 . C nh v3.Tia MN c hnh g?
Cu 4. Gi
ax+by=c
bx+cy=a
cx+ay=b
c nghi 3 + b3 + c3 = 3abc.
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.7
1;3
1
491
1694
223312
22
3
323
2
15120
4
156
2
1
2
2
2
xxx
xxxC
B
A
)(2
1 2 Pxy
.22
yx
yx
xy
221
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43xx
8
1'''
HC
HC
HB
HB
HA
HA
x
xxA
24
442
52
34
1211
yx
yx
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AEF
361122
xxx
1,0;11
11
aaa
aa
a
aaA
21
1
23
1
2
1
nn
1,0;1
1
1
1aa
aa
a
aaM
12
2522
xy
yx
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yxyxyx
xy
xyx
y
xyx
yS ,0,0;
2:
2
21xy
34
RSCDAB
211
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7,113
252
yxx
yxx
1,0;1
1xx
xx
x
xA
2
1x
2
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b) Chc) G n (O), (M), (N) theo th p cc tam gic ABC, DHB,
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T i ch chol 16. Tm hai ch
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HMP HMP KQN
1
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1
x
x x
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2
2 1 0x x
1 22
x y
2
2 1 12
21
x x xx
x
1
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,B A C A
2 2
1 1
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2 6 5 0x x
( 32 50 8) : 182 (2 1) 2 0mx m x m
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2
2
3 5
1
x
x
4 2 2002 2002x x
9 3 1 1:
93 3
x x x
xx x x x
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1 1 1
CN CD CP
4 8 1 2:
42 2
x x x
xx x x x
5
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1 2
11 1x y
8 51
1 1x y
21.12212
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xxx
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2
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12
22
m
m
SESDSK
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20052005 22ba
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)A
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1
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1. yxm
myx
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16
32a
1
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D
BAE
F
6 66
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ab bc cab c a
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2 28 2 4x x2 2
4 2 2 47
21x xy yx x y y
3 2
3 23 193 98
a abb ba
AB
2 2 2 2 2 21
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1 1 1 1 1 1
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a b cx y zx y z
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22 36
2 3
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OB
CN
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1 1 1 1
2 2 2 3MK MA MA MB MB MK
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2 2
2 22 5 2 0
4 0
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4 3 16or 5ba b cP
b c a a c b a b c
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ID
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2
2 22 32
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3 3 3
1 1 1 1 1 1 2
1
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x y z
1 1 1P
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2
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x x x xA
x x
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a b c a b c
b a c b a c b c c a a b
6 2
4
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Px x x x x
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2 2 1x y y 24 1 5 14x x x
2 2
3 3
4 4
35917
ax byax byax byax by
5 5A ax by
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1 1 1P
x y z
2004 6 6
2004 6 6
2004 6 6
222
x y zy z xz x y
2 2 3 3 1 3 4 1 2
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x x x x x xx
2 23
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4 3 2 3 2 11x x x
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4
4 4 41 ( )zP
z x y
259
O 10 T 2009
Kho thi ngy 26/6/2008 - ThCu I:1) Gi nh sau:
a) 5.x 45 0 b) x(x + 2) 5 = 0
2) Cho hm s2x
2
a) Tnh f(-1)M 2;1 c n m s sao ?
Cu II:1) Rt g
P = 4 a 1 a 11 .a a 2 a 2 v a 4.
Cu III:T
th s t b 23
s
Cu IV:
n tm O. L n (O),
n (O) tc n (O) tv
1)Ch2)G n (O). Ch
minh DM AC.3)Ch 2.
Cu V:Cho biB = (4x5 + 4x4 5x3 + 5x 2)2 + 2008.
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Tnh gi tr 1 2 1.2 2 1
260
S PT CHUYNQU -2009
Mn TONTh m bi 150 pht ( khng k
Bi 1 :
a) Th35
126320103 .
b) Tm gi tr 2008xx .Bi 2 :
Cho h nh:5myx3
2ymx
a) Gi nh khi 2m .b) Tm gi tr nh cho c nghi n
h3m
m1yx
2
2
.
Bi 3 :
a) Cho hm s 2x2
1y (P). Vi nh
N n n (P) l 2 v 1.b) Gi nh: 1xx2x3x3 22 .
Bi 4 :Cho hnh thang ABCD (AB // CD), giao
th BC l N.a) Ch : 1
AB
MO
CD
MO .
b) Ch .MN
2
CD
1
AB
1
c) Bi 2COD2
AOB nS;mS . Tnh ABCDS theo m v n (v CODAOB S,S ,
ABCDS l di gic COD, diABCD).
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Bi 5 n ( O; R ) v dy cung AB cn cung l BC lun song song. G
BD. Cha) T tb) OM BC.
song song v
Bi 6 :
a) Cho cc s yxx
y
y
x 22 .
b) Cho n l s n l n4 4n l h
261
S H NG
K- 2009
MN THI: TON
Th m bi 120 pht, khng k
Ch :
-
- H m bi vo t
Ph
1. Bi 21 4xx
gi tr o c
A. x 14
B. x 14
C. x 14
v x 0 D. x 0
2 o song song v - 2x?
A. y = 2x - 1. B. y = 2 (1- 2 x).
C. y = 2 - x. D. y = 2(1- 2x).
CHNH TH
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3. Hai h rnhx 3 3
1
k y
x yv
3x 3 3
1
y
x y
A. -3 B. 3 C. 1 D. -1
4 - 2 ; 1
2) thu m s o trong cc hm s
A. y = 22
x2 B. y = 222
x
C. y = 224
x D. y = - 224
x
5. Tam gic GEF vung t
i EF b
A. 13 B. 13 C. 2 13 D. 3 13
6. Tam gic ABC vung t 3
A. 32
a. B. 12
C. 32
D. 12
a
7. Cho tam gic ABC vungt n
ngo
A. 30 cm B. 15 2 cm C. 20 cm D. 15 cm
8. Cho tam gic ABC vung t
vng quanh c nh nn. Di tch ton ph nh nn
A. 96 cm2 B. 100 cm2
C. 144 cm2 D. 150 cm2
Ph
nh b x : x 2 4x + m + 1 = 0.1. Gi nh khi m = 3..2. V o c nh c nghi3. Tm gi tr nh cho c 2 nghi 1, x2 th
mn 12 + x2
2 = 10.
Gi nh:3 2 2 1
2 2 3
x y
x y
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M ng i 80 km trong m ng ng n
lng
Bi 4 : ( G m A, C B). Trn cng mn By cng vung gc vAB. Trn tia Ax l A), tia vung gc v
n1/ Ch
a/ T n. Xctrn
b/ AI.BK = AC.BC
c/ APB vung.2/ Cho A, I, B c m v
Bi 5 : ( Tm x n 1003x + 2y = 2008
------------------- H ------------------