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S GIO DC V O TO K THI TUYN SINH LP 10 THPT TP.HCM Nm hc: 2010
2011 CHNH THC MN: TON Thi gian lm bi: 120 pht Bi 1: (2 im)
Gii cc phng trnh v h phng trnh sau: a) 22 3 2 0x x = b)
4 16 2 9
x yx y
+ =
=
c) 4 24 13 3 0x x + = d) 22 2 2 1 0x x =
Bi 2: (1,5 im) a) V th (P) ca hm s
2
2xy = v ng thng (D): 1 1
2y x= trn cng
mt h trc to . b) Tm to cc giao im ca (P) v (D) bng php tnh. Bi
3: (1,5 im) Thu gn cc biu thc sau:
12 6 3 21 12 3A = + 2 2
5 35 2 3 3 5 2 3 3 52 2
B
= + + + + +
Bi 4: (1,5 im) Cho phng trnh 2 2(3 1) 2 1 0x m x m m + + + = (x
l n s)
a) Chng minh rng phng trnh lun lun c 2 nghim phn bit vi mi gi tr
ca m.
b) Gi x1, x2 l cc nghim ca phng trnh. Tm m biu thc sau t gi tr
ln nht: A = 2 21 2 1 23x x x x+ .
Bi 5: (3,5 im) Cho ng trn tm O ng knh AB=2R. Gi M l mt im bt k
thuc ng trn (O) khc A v B. Cc tip tuyn ca (O) ti A v M ct nhau ti
E. V MP vung gc vi AB (P thuc AB), v MQ vung gc vi AE (Q thuc
AE).
a) Chng minh rng AEMO l t gic ni tip ng trn v APMQ l hnh ch
nht.
b) Gi I l trung im ca PQ. Chng minh O, I, E thng hng. c) Gi K l
giao im ca EB v MP. Chng minh hai tam gic EAO v MPB
ng dng. Suy ra K l trung im ca MP. d) t AP = x. Tnh MP theo R v
x. Tm v tr ca M trn (O) hnh ch nht
APMQ c din tch ln nht.
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BI GII Bi 1: (2 im)
Gii cc phng trnh v h phng trnh sau: a) 22 3 2 0x x = (1) 9 16 25
= + =
(1) 3 5 1 3 5 24 2 4
x hay x + = = = =
b) 4 1 (1)6 2 9 (2)
x yx y
+ =
=
4 1 (1)14 7 ( (2) 2 (1))
x yx pt pt
+ =
= +
312
y
x
=
=
c) 4 24 13 3 0x x + = (3), t u = x2, phng trnh thnh : 4u2 13u +
3 = 0 (4) (4) c 2169 48 121 11 = = = 13 11 1 13 11(4) 3
8 4 8u hay u + = = = =
Do (3) 1 32
x hay x = =
d) 22 2 2 1 0x x = (5) ' 2 2 4 = + =
Do (5) 2 2 2 22 2
x hay x + = =
Bi 2: a) th: hc sinh t v Lu : (P) i qua O(0;0), ( )11; , 2;
2
2
.
(D) i qua ( )11; , 2; 22
Do (P) v (D) c 2 im chung l : ( )11; , 2; 22
.
b) PT honh giao im ca (P) v (D) l 2
21 1 2 02 2x
x x x
= + = 1 2x hay x = =
Vy to giao im cu (P) v (D) l ( )11; , 2; 22
.
Bi 3: 12 6 3 21 12 3A = + 2 2(3 3) 3(2 3) 3 3 (2 3) 3= + = +
3=
2 2
5 35 2 3 3 5 2 3 3 52 2
B
= + + + + +
2B = ( ) ( )2 25 4 2 3 6 2 5 5 4 2 3 6 2 5 3+ + + + +
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( ) ( )2 22 2 2 25 (1 3) ( 5 1) 5 ( 3 1) ( 5 1) 3= + + + + + = (
) ( )2 25 (1 3) ( 5 1) 5 ( 3 1) ( 5 1) 3+ + + + + = 5.3 5 20+ = B =
10.
Bi 4: a) ( )2 2 2 23 1 8 4 4 2 5 ( 1) 4 0m m m m m m m = + + = +
+ = + + >
Suy ra phng trnh lun lun c 2 nghim phn bit vi mi m. b) Ta c x1 +
x2 = 3m + 1 v x1x2 = 2m2 + m 1
A= 2 21 2 1 23x x x x+ ( )21 2 1 25x x x x= + 2 2(3 1) 5(2 1)m m
m= + + 2 21 16 6 ( )
4 2m m m= + + = + 2
25 1( )4 2
m=
Do gi tr ln nht ca A l : 254
. t c khi m = 12
Bi 5: a) Ta c gc EMO = 90O = EAO => EAOM ni tip. T gic APMQ c
3 gc vung : oEAO APM PMQ 90= = = => T gic APMQ l hnh ch nht b)
Ta c : I l giao im ca 2 ng cho AM v PQ ca hnh ch nht APMQ nn I l
trung im ca AM. M E l giao im ca 2 tip tuyn ti M v ti A nn theo nh
l ta c : O, I, E thng hng. c) Cch 1: hai tam gic AEO v MPB ng dng v
chng l 2 tam gic vung c 1 gc bng nhau l AOE ABM= ( v OE // BM )
=>
AO AEBP MP
= (1)
Mt khc, v KP//AE, nn ta c t s KP BPAE AB
= (2) T (1) v (2) ta c : AO.MP = AE.BP = KP.AB, m AB = 2.OA
=> MP = 2.KP Vy K l trung im ca MP.
Cch 2 : Ta c EK APEB AB
= (3) do AE // KP,
mt khc, ta c EI APEO AB
= (4) do 2 tam gic EOA v MAB ng dng
So snh (3) & (4), ta c : EK EIEB EO
= .
I K
B O
M Q
E
A P x
I
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Theo nh l o Thales => KI // OB, m I l trung im AM => K l
trung im MP. d) Ta d dng chng minh c :
abcd 4
a b c d4
+ + +
(*) Du = xy ra khi v ch khi a = b = c = d MP = 2 2 2 2 2MO OP R
(x R) 2Rx x = = Ta c: S = SAPMQ = 2 3MP.AP x 2Rx x (2R x)x= = S t
max 3(2R x)x t max x.x.x(2R x) t max
x x x
. . (2R x)3 3 3
t max
p dng (*) vi a = b = c = x3
Ta c : 4 4
4x x x 1 x x x R
. . (2R x) (2R x)3 3 3 4 3 3 3 16
+ + + =
Do S t max x (2R x)3
= 3
x R2
= .
Cch khc
x(2R x)x x x3MP.AP 3 x (2R x) 3 x 3 3 R3 2 3 3
+
= =
23 3 R4
. Do S t max x (2R x)3
= v x xR3 3
=
3x R
2= .
TS. Nguyn Ph Vinh (TT BDVH v LTH Vnh Vin)
