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THI TH I HC CAO NG 2012
Mn thi: TON
I. PHN CHUNG DNH CHO TT C CC TH SINH (7,0 im)
Cu 1: Cho hm s
2x 1
y
x 2
+
.
1. Kho st s bin thin v v th (C) ca hm s cho.
2. Vit phng trnh tip tuyn ca th (C), bit h s gc ca tip tuyn bng 5.
Cu 2:
1. Gii phng trnh: 25
x
6.5
x
+ 5 = 0
2. Tnh tch phn sau
0
I x(1 cos x)dx
.
3. Tm gi tr nh nht v gi tr ln nht ca hm s f(x) = x
2
ln(1 2x) trn on [2; 0].
Cu 3: Cho hnh chp S.ABC c mt bn SBC l tam gic u cnh a, cnh bn SA vung gc vi mt phng
y. Bit gc BAC = 120
o
, tnh th tch ca khi chp S.ABC theo a.
Cu 4: Cho x, y, z l cc s dng tho
1 1 1
1
x y z
+ +
. CMR:
1 1 1
1
2x y z x 2y z x y 2z
+ +
+ + + + + +
.
II. PHN RING
A. Theo chng trnh Chun
Cu 5a: Trong khng gian Oxyz, cho mt cu (S): (x 1)
2
+ (y 2)
2
+(z 2)
2
= 36 v mt phng (P): x + y +
2z + 18 = 0.
1. Xc nh ta tm T v tnh bn knh ca mt cu (S). Tnh khong cch t T n mp (P).
2. Vit phng trnh ng thng d i qua T v vung gc vi (P). Tm ta giao im ca d v (P).
Cu 6a: Gii phng trnh: 8z
2
4z + 1 = 0 trn tp s phc.
B. Theo chng trnh Nng cao
Cu 5b: Cho im A(1; 2; 3) v ng thng d c phng trnh
x 1 y 2 z 3
2 1 1
+ +

1. Vit phng trnh tng qut ca mt phng i qua im A v vung gc vi ng thng d.

2. Tnh khong cch t im A n d. Vit phng trnh mt cu tm A, tip xc vi d.
Cu 6b: Gii phng trnh
2
2z iz 1 0 + trn tp s phc.
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Mn thi: TON
A. PHN CHUNG CHO TT C CC TH SINH: (8 im)
Cu 1: Cho hm s y = 4x
3
+ mx
2
3x.
1. Kho st v v th (C) hm s khi m = 0.
2. Tm m hm s c hai cc tr ti x
1
v x
2
tha x
1
= 4x
2
.
Cu 2:
1. Gii h phng trnh:
x 2y xy 0
x 1 4y 1 2
'
+
2. Gii phng trnh: cosx = 8sin

3
(x + /6)
Cu 3: (2 im)
1. Cho hnh chp S.ABC c SA vung gc vi mt phng (ABC), tam gic ABC vung ti C; M, N l hnh
chiu ca A trn SB, SC. Bit MN ct BC ti T. Chng minh rng tam gic AMN vung v AT tip xc vi
mt cu ng knh AB.
2. Tnh tch phn A =
2
e
e
dx
x ln x ln ex
Cu 4: (2 im)
1. Trong khng gian vi h trc ta Oxyz, cho bn im A(4; 5; 6); B(0; 0; 1); C(0; 2; 0); D(3; 0; 0).
Chng minh cc ng thng AB v CD cho nhau. Vit phng trnh ng thng (D) vung gc vi mt
phngOxy v ct c cc ng thngAB; CD.
2. Cho ba s thc dng a, b, c tha
3 3 3
2 2 2 2 2 2
a b c
1
a ab b b bc c c ca a
+ +
+ + + + + +
Tm gi tr ln nht ca biu thc S = a + b + c
B. PHN T CHN
Cu 5a: Theo chng trnh chun (2 im)
1. Trong khng gian vi h trc ta Oxyz, cho im A(4; 5; 6). Vit phng trnh mt phng (P) qua A; ct
cc trc ta ln lt ti I; J; K m A l trc tm ca tam gic IJK.
2. Bit (D) v (D) l hai ng thng song song. Ly trn (D) 5 im v trn (D) n im v ni cc im ta
c cc tam gic. Tm n s tam gic lp c bng 45.
Cu 5b: Theo chng trnh nng cao (2 im)
1. Trong mt phng vi h trc ta Oxy, cho ng thng (D): x 3y 4 = 0 v ng trn (C): x
2
+ y
2

4y = 0. Tm M thuc (D) v N thuc (C) sao cho chng i xng qua A(3; 1).
2. Tm m bt phng trnh: 5
2x
5
x+1
2m5
x
+ m
2
+ 5m > 0 tha vi mi s thc x.
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Mn thi: TON
PHN CHUNG CHO TT C CC TH SINH: (7 im)
Cu I. Cho hm s y = f(x) = x
4
2x
2
.
1. Kho st v v th (C) ca hm s.
2. Trn (C) ly hai im phn bit A v B c honh ln lt l a v b. Tm iu kin i vi a v b hai
tip tuyn ca (C) ti A v B song song vi nhau.
Cu II.
1. Gii phng trnh lng gic:
( ) 2 cos x sin x
1
tan x cot 2x cot x 1
+
2. Gii bt phng trnh:
( )
2
3 1 1
3 3
1
log x 5x 6 log x 2 log x 3
2
+ + > +
Cu III (1 im) Tnh tch phn
( )
2
4 4
0
I cos 2x sin x cos x dx
Cu IV (1 im) Cho mt hnh tr trn xoay v hnh vung ABCD cnh a c hai nh lin tip A, B nm trn
ng trn y th nht ca hnh tr, hai nh cn li nm trn ng trn y th hai ca hnh tr. Mt
phng (ABCD) to vi y hnh tr gc 45
0
. Tnh din tch xung quanh v th tch ca hnh tr.
Cu V (1 im) Cho phng trnh ( ) ( )
3
4
x 1 x 2m x 1 x 2 x 1 x m + +
Tm m phng trnh c mt nghim duy nht.
PHN RING (3 im)
A. Theo chng trnh chun.
Cu VIa (2 im)
1. Trong mt phng Oxy, cho ng trn (C): x
2
+ y
2
4x 2y = 0 v ng thng (): x + 2y 12 = 0. Tm
im M trn sao cho t M v c vi (C) hai tip tuyn lp vi nhau mt gc 60
o
.
2. Trong khng gian vi h ta Oxyz, cho t din ABCD vi A(2; 1; 0), B(1; 1; 3), C(2; 1; 3), D(1; 1; 0).
Tm ta tm v bn knh ca mt cu ngoi tip t din ABCD.
Cu VIIa (1 im) C 10 vin bi c bn knh khc nhau, 5 vin bi xanh c bn knh khc nhau v 3 vin bi
vng c bn knh khc nhau. Hi c bao nhiu cch chn ra 9 vin bi c ba mu?
B. Theo chng trnh nng cao.
Cu VIb (2 im)
1. Trong mt phng ta Oxy, cho hnh ch nht ABCD c din tch bng 12, tm I thuc ng thng (d):
x y 3 = 0 v c honh x
I
= 9/2, trung im ca mt cnh l giao im ca (d) v trc Ox. Tm ta cc
nh ca hnh ch nht.
2. Trong h ta Oxyz, cho mt cu (S): x
2
+ y
2
+ z
2
4x + 2y 6z + 5 = 0 v mt phng (P): 2x + 2y z +
16 = 0. im M di ng trn (S) v im N di ng trn (P). Tnh di ngn nht ca on thng MN. Xc
nh v tr ca M, N tng ng.
Cu VIIb: Cho
a, b, c
l nhng s dng tha mn a
2
+ b
2
+ c
2
= 3. Chng minh
2 2 2
1 1 1 4 4 4
a b b c c a a 7 b 7 c 7
+ + + +
+ + + + + +
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Mn thi: TON
PHN CHUNG CHO TT C CC TH SINH: (7 im)
Cu I. Cho hm s y = f(x) = mx
3
+ 3mx
2
(m 1)x 1, m l tham s
1. Kho st s bin thin v v th ca hm s trn khi m = 1.
2. Xc nh cc gi tr ca m hm s y = f(x) khng c cc tr.
Cu II. Gii phng trnh:
1.
4 4
sin x cos x tan x cot x
sin 2x 2
+ +
2.
2 3
4 8
2
log (x 1) 2 log 4 x log (4 x) + + + +
Cu III. Tnh tch phn
3/ 2
2
1/ 2
dx
I
x 1 x
Cu IV. Cho hnh nn c nh S, y l ng trn tm O, SA v SB l hai ng sinh, bit SO = 3, khong

cch t O n mt phng SAB bng 1, din tch tam gic SAB bng 18. Tnh th tch v din tch xung quanh
ca hnh nn cho.
Cu V. Tm m h bt phng trnh sau c nghim
2
2
x 7x 6 0
x 2(m 1)x m 3 0
+
'
+ +
PHN RING (3 im)

Cu VIa (2 im)
1. Cho tam gic ABC bit cc cnh AB, BC ln lt l 4x + 3y 4 = 0; x y 1 = 0. Phn gic trong ca gc
A nm trn ng thng x + 2y 6 = 0. Tm ta cc nh ca tam gic ABC.
2. Cho hai mt phng (P): x + 2y 2z + 5 = 0; (Q): x + 2y 2z 13 = 0. Vit phng trnh ca mt cu (S) i
qua gc ta O, qua im A(5; 2; 1) v tip xc vi c (P) v (Q).
Cu VIIa (1 im) Tm s nguyn dng n tha mn cc iu kin sau
4 3 2
n 1 n 1 n 2
n 4 3
n 1 n 1
5
C C A
4
7
C A
15

+ +
<
'

Cu VIb (2 im)
1. Cho ng thng d: x 5y 2 = 0 v ng trn (C): x
2
+ y
2
+ 2x 4y 8 = 0. Xc nh ta cc giao
im A, B ca ng trn (C) v ng thng d (im A c honh dng). Tm ta C thuc ng trn
(C) sao cho tam gic ABC vung B.
2. Cho mt phng (P): x 2y + 2z 1 = 0 v cc ng thng
1 2
x 1 y 3 z x 5 y z 5
d : ; d :
2 3 2 6 4 5
+

. Tm cc im M, N ln lt thuc d
1
, d
2
sao cho MN // (P) v
cch (P) mt khong bng 2.
Cu VIIb: Tnh o hm f(x) ca hm s
3
1
f (x) ln
(3 x)
v gii bt phng trnh

2
0
f '(x)
1 6 t
sin dt
x 2 2
>
+
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Mn thi: TON
Bi 1: Cho hm s y = x
4
+ mx
3
2x
2
3mx + 1 (1)
1. Kho st s bin thin v v th (C) ca hm s (1) khi m = 0.
2. nh m hm s (1) c hai cc tiu.
Bi 2:
1. Gii phng trnh:
3 3
2 3 2
cos3xcos x sin3xsin x
8
+

2. Gii phng trnh:
( )
2 2
2x 1 x x 2 x 1 x 2x 3 0 + + + + + + +
Bi 3: Cho cc im A(1; 1; 0), B(1; 1; 2), C(2; 2; 1), D(1; 1; 1).
1. Vit phng trnh ca mt phng cha AB v song song vi CD. Tnh gc gia AB, CD.
2. Gi s mt phng () i qua D v ct ba trc ta ti cc im M, N, P khc gc O sao cho D l trc tm
ca tam gic MNP. Hy vit phng trnh ca ().
Bi 4: Tnh tch phn
( )
2
0
I x 1 sin 2xdx
.
Bi 5: Gii phng trnh 4
x
2
x+1
+ 2(2
x
1)sin(2
x
+ y 1) + 2 = 0.
Bi 6: Gii bt phng trnh
2 2
x x 1 x x 2
9 1 10.3
+ +
+
.
Bi 7:
1. Cho tp A gm 50 phn t khc nhau. Xt cc tp con khng rng cha mt s chn cc phn t rt ra t
tp A. Hy tnh xem c bao nhiu tp con nh vy.
2. Cho s phc
1 3
z i
2 2
+ . Hy tnh: 1 + z + z
2
.
Bi 8: Cho lng tr ABC.ABC c A.ABC l hnh chp tam gic u cnh y AB = a, cnh bn AA = b.
Gi l gc gia hai mt phng (ABC) v (ABC). Tnh tan v th tch ca khi chp A.BBCC.
Bi 9: Trong mt phng vi h ta Oxy cho im C(2; 0) v elip (E):
2 2
x y
1
4 1
+ . Tm ta cc im A,
B thuc (E), bit rng hai im A, B i xng vi nhau qua trc honh v tam gic ABC l tam gic u.
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Mn thi: TON
PHN CHUNG CHO TT C CC TH SINH (7 im)
Cu I. Cho hm s y = 8x
4
9x
2
+ 1.
1. Kho st s bin thin v v th (C) ca hm s.
2. Da vo th (C) hy bin lun theo m s nghim ca phng trnh
8cos
4
x 9cos
2
x + m = 0 vi x [0; ].
Cu II. Gii phng trnh, h phng trnh
1.
( )
3
log x
1
x 2 x x 2
2
_

,
2.
2 2
2 2
x y x y 12
y x y 12
+ +
'

Cu III. Tnh din tch ca min phng gii hn bi cc ng

2
y x 4x
v y = 2x.
Cu IV. Cho hnh chp ct tam gic u ngoi tip mt hnh cu bn knh r cho trc. Tnh th tch hnh chp
ct bit rng cnh y ln gp i cnh y nh.
Cu V nh m phng trnh sau c nghim
2
4sin3xsinx + 4cos 3x cos x + cos 2x + m 0
4 4 4
_ _ _
+

, , ,
PHN RING (3 im)
Cu VIa (2 im)
1. Trong mt phng Oxy, cho ABC c nh A(1; 2), ng trung tuyn BM: 2x + y + 1 = 0 v phn gic
trong CD: x + y 1 = 0. Vit phng trnh ng thng BC.
2. Trong khng gian Oxyz, cho ng thng (D) c phng trnh:
x 2 t
y 2t
z 2 2t
+

'
. Gi l ng thng qua
im A(4; 0; 1) song song vi (D) v I(2; 0; 2) l hnh chiu vung gc ca A trn (D). Trong cc mt
phng qua , hy vit phng trnh ca mt phng c khong cch n (D) l ln nht.
Cu VIIa (1 im) Cho x, y, z l 3 s thc thuc (0; 1]. Chng minh rng
1 1 1 5
xy 1 yz 1 zx 1 x y z
+ +
+ + + + +
Cu VIb (2 im)
1. Cho hnh bnh hnh ABCD c din tch bng 4. Bit A(1; 0), B(0; 2) v giao im I ca hai ng cho
nm trn ng thng y = x. Tm ta nh C v D.
2. Cho hai im A(1; 5; 0), B(3; 3; 6) v ng thng c phng trnh tham s
x 1 2t
y 1 t
z 2t
+

'
. Mt im M
thay i trn ng thng , tm im M chu vi tam gic MAB t gi tr nh nht.
Cu VIIb (1 im) Cho a, b, c l ba cnh tam gic. Chng minh
1 1 2 b c
a 2
3a b 3a c 2a b c 3a c 3a b
_
+ + + + <

+ + + + + +
,
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Mn thi: TON
I. PHN CHUNG CHO TT C CC TH SINH (7,0 im)
Cu I: Cho hm s y = x
3
+ 2mx
2
+ (m + 3)x + 4 c th l (C
m
).
1. Kho st s bin thin v v th (C
1
) ca hm s trn khi m = 1.
2. Cho ng thng (d): y = x + 4 v im K(1; 3). Tm cc gi tr ca tham s m sao cho (d) ct (C
m
) ti ba
im phn bit A(0; 4), B, C sao cho tam gic KBC c din tch bng
8 2
.
Cu II:
1. Gii phng trnh cos2x + 5 = 2(2 cosx)(sinx cosx)
2. Gii h phng trnh
2
2
x 1 y(x y) 4y
(x 1)(x y 2) y
+ + +
'
+ +
Cu III:
1. Tnh tch phn
2
2
6
1
I sin x sin x dx
2
2. Tm cc gi tr ca tham s thc m sao cho phng trnh sau c nghim.

2 2
1 1 x 1 1 x
9 (m 2)3 2m 1 0
+ +
+ + +
Cu IV: Cho hnh chp S.ABC c gc ((SBC), (ACB)) = 60
0
, ABC v SBC l cc tam gic u cnh a. Tnh
theo a khong cch t B n mt phng (SAC).
II. PHN RING (3 im)
Cu Va:
1. Cho parabol (P): y = x
2
2x v elip (E):
2
2
x
y 1
9
+ . Chng minh rng (P) giao (E) ti 4 im phn bit
cng nm trn mt ng trn. Vit phng trnh ng trn i qua 4 im .
2. Cho mt cu (S): x
2
+ y
2
+ z
2
2x + 4y 6z 11 = 0 v mt phng () c phng trnh 2x + 2y z + 17 =
0. Vit phng trnh mt phng () song song vi () v ct (S) theo giao tuyn l ng trn c chu vi bng
6.
Cu Via:
Tm h s ca s hng cha x
2
trong khai trin nh thc
n
4
1
x
2 x
_
+

,
. Bit rng n l s nguyn dng tha
mn:
2 3 n 1
0 1 2 n
n n n n
2 2 2 6560
2C C C C
2 3 n 1 n 1
+
+ + + +
+ +
L
Cu Vb:
1. Cho im A(10; 2; 1) v ng thng d c phng trnh
x 1 y z 1
2 1 3

. Lp phng trnh mt phng
(P) i qua A, song song vi d v khong cch t d ti (P) l ln nht.
2. Cho im A(2; 3), B(3; 2), ABC c din tch bng 3/2; trng tm G ca ABC thuc ng thng (d):
3x y 8 = 0. Tm bn knh ng trn ni tip ABC.
Cu VIb:
Tm cc s thc b, c phng trnh z
2
+ bz + c = 0 nhn s phc z = 1 + i lm mt nghim.
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Mn thi: TON
I. PHN CHUNG CHO TT C TH SINH (7,0 im)
Cu I (2,0 im) Cho hm s ( ) ( )
3 2
1
y m 1 x mx 3m 2 x
3
+ + (1)
1. Kho st s bin thin v v th hm s (1) khi m = 2.
2. Tm cc gi tr ca tham s m hm s (1) ng bin trn tp xc nh ca n.
Cu II (2,0 im)
1. Gii phng trnh: (2cosx 1)(sinx + cosx) = 1
2. Gii phng trnh:
( ) ( ) ( )
2 3 3
1 1 1
4 4 4
3
log x 2 3 log 4 x log x 6
2
+ + +
Cu III (1,0 im) Tnh tch phn
2
2
0
cos x
I dx
sin x 5sin x 6
Cu IV (1,0 im) Cho lng tr ng ABC.ABC c y l tam gic u. Mt phng ABC to vi y mt

gc 30
o
v tam gic ABC c din tch bng 8. Tnh th tch khi lng tr.
Cu V (1,0 im) Cho x, y l hai s dng tha iu kin x + y = 5/4.
Tm gi tr nh nht ca biu thc:
4 1
S
x 4y
+
II. PHN RING (3 im)
Cu VIa (2.0 im)
1. Trong mt phng Oxy. Vit phng trnh ng thng
( )
i qua im M(3; 1) v ct trc Ox, Oy ln lt
ti B v C sao cho tam gic ABC cn ti A vi A(2; 2).
2. Cho im A(4; 0; 0) v im B(x
o
; y
o
; 0) (x
o
> 0, y
o
> 0) sao cho OB = 8 v gc AOB = 60
o
. Xc nh ta
im C trn trc Oz th tch t din OABC bng 8.
Cu VIIa (1 im)
T cc ch s 0; 1; 2; 3; 4; 5 c th lp c bao nhiu s t nhin m mi s c 6 ch s khc nhau v ch
s 2 ng cnh ch s 3.
Cu VIb (2 im)
1. Vit phng trnh ng thng i qua im M(4; 1) v ct cc tia Ox, Oy ln lt ti A v B sao cho gi
tr ca tng OA + OB nh nht.
2. Cho t din ABCD c ba nh A(2; 1; 1), B(3; 0; 1), C(2; 1; 3), cn nh D nm trn trc Oy. Tm ta
nh D nu t din c th tch V = 5.
Cu VIIb (1 im)
T cc s 0; 1; 2; 3; 4; 5. Hi c th thnh lp c bao nhiu s c 3 ch s khng chia ht cho 3 m cc ch
s trong mi s l khc nhau.
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THI TH I HC, CAO NG 2012
Mn thi: TON
Cu I. Cho hm s y = x
3
3(m + 1)x
2
+ 9x + m 2 (1) c th l (C
m
)
1. Kho st v v th hm s (1) vi m = 1.
2. Xc nh m (C
m
) c cc i, cc tiu v hai im cc tr i xng vi nhau qua ng thng y = x/2.
Cu II.
1. Gii phng trnh: ( )
( )
3
sin 2x cos x 3 2 3cos x 3 3cos2x 8 3 cos x sinx 3 3 0 + +
.
2. Gii bt phng trnh: ( )
2
2 1
2
1 1
log x 4x 5 log
2 x 7
_
+ >

+
,
.
3. Tnh din tch hnh phng gii hn bi cc ng y = xsin2x, y = 2x, x = /2.
Cu III.
1. Cho hnh lng tr ABC.ABC c y ABC l tam gic u cnh a, cnh bn hp vi y mt gc l 45
o
.
Gi P l trung im BC, chn ng vung gc h t A xung (ABC) l H sao cho
1
AP AH
2
uuur uuur
. gi K l
trung im AA, () l mt phng cha HK v song song vi BC ct BB v CC ti M, N. Tnh t s th tch
ABCKMN v ABCKMN.
2. Gii h phng trnh sau trong tp s phc:
( )
2
2
2 2 2 2
6
a a 5
a a
a b ab b a a 6 0
+
'
+ + +
Cu IV.
1. Cho m bng hng trng v n bng hng nhung khc nhau. Tnh xc sut ly c 5 bng hng trong
c t nht 3 bng hng nhung? Bit m, n l nghim ca h sau:
m 2 2 1
m n 3 m
n 1
9 19
C C A
2 2
P 720
+ + <
'
2) Cho Elip c phng trnh chnh tc

2 2
x y
1
25 9
+ (E), vit phng trnh ng thng song song Oy v ct
(E) ti hai im A, B sao cho AB = 4.
Cho hai ng thng d
1
v d
2
ln lt c phng trnh:
1
x 2 t
d : y 2 t
z 3 t
+
+
'
v
2
x 1 y 2 z 1
d :
2 1 5

Vit phng trnh mt phng cch u hai ng thng d
1
v d
2
?
Cu V: Cho a, b, c khng m v
2 2 2
a b c 3 + + . Tm gi tr nh nht ca biu thc
3 3 3
2 2 2
a b c
P
1 b 1 c 1 a
+ +
+ + +
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Mn thi: TON
I.PHN CHUNG CHO TT C TH SINH. (7 im)
3
+ mx + 2 (1)
1. Kho st s bin thin v v th ca hm s (1) khi m = 3.
2. Tm m th hm s (1) ct trc hanh ti mt im duy nht.
Cu II.
1. Gii h phng trnh:
3 3
2 2 3
x y 1
x y 2xy y 2
+
'
+ +
2. Gii phng trnh:

2 2
2sin (x ) 2sin x tan x
4
.
Cu III. Tnh tch phn
2 2
1
4 x
I dx
x
Cu IV. Cho hnh chp S.ABCD c y ABCD l hnh vung cnh a, SA = h l ng cao, M l im thay
i trn CD. K SH vung gc BM. Xc nh v tr M th tch t din S.ABH t gi tr ln nht. Tnh gi
tr ln nht .
Cu V. Tm m phng trnh sau c nghim thc
4 2
x 1 x m +
II. PHN RING (3 im)
Th sinh ch c lm mt trong hai phn
Cu VI a.
1. Trong h ta Oxy, cho hai ng thng d
1
: x 2y + 3 = 0, d
2:
4x + 3y 5 = 0. Lp phng trnh ng
trn (C) c tm I trn d
1
, tip xc d
2
v c bn knh R = 2.
2. Cho hai ng thng d
1

x y z
1 1 2
, d
2
:
x 1 2t
y t
z 1 t

'
v mt phng (P): x y z = 0. Tm ta hai im

M trn d
1
, N trn d
2
sao cho MN song song (P) v MN = 6
Cu VII a. Tm s phc z tha mn
4
z i
1
z i
+ _
,
Cu VI b.
1. Cho hnh ch nht ABCD c cnh AB: x 2y 1 = 0, ng cho BD: x 7y + 14 = 0 v ng cho AC
qua im M(2; 1). Tm ta cc nh ca hnh ch nht.
2. Cho ba im O(0; 0; 0), A(0; 0; 4), B(2; 0; 0) v mp(P): 2x + 2y z + 5 = 0. Lp p.tr m.cu (S) i qua ba
im O, A, B v c khang cch t tm I n mt phng (P) bng 5/3.
Cu VII b. Gii bt phng trnh
x x
3
log 3 log 3 <
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Mn thi: TON
CU I. Cho hm s
3 2 3
3 1
y x mx m
2 2
+
1. Kho st v v th hm s khi m = 1.
2. Xc nh m th hm s c cc i, cc tiu i xng nhau qua t y = x
CU II.
1. Gii phng trnh:
2 2 3 3
tan x tan xsin x cos x 1 0 +
2. Cho PT:
2
5 x x 1 5 6x x m + + +
(1)
a) Tm m pt(1)c nghim.
b) Gii PT khi ( )
m 2 1 2 +
CU III.
1. Tnh tch phn: I =
( )
4
3
4
1
dx
x x 1 +
2. Tnh cc gc ca tam gic ABC bit: 2A = 3B;

2
a b
3
CU IV.
1. Vit phng trnh mt phng (P) qua O, vung gc vi mt phng (Q): x + y + z = 0 v cch im M(1; 2;
1 ) mt khong bng
2
.
2. C 6 hc sinh nam v 3 hc sinh n xp hng dc i vo lp. Hi c bao nhiu cch xp c ng 2HS
nam ng xen k 3HS n
CU V.
1. Cho ng thng (d):
x 2 4t
y 3 2t
z 3 t
+
+
'
v mt phng (P): x + y + 2z + 5 = 0
Vit phng trnh .thng () nm trong (P), song song vi (d) v cch (d) mt khong l
14
2. Gii PT:
2x 1 x 1 x x 1
5.3 7.3 1 6.3 9 0
+
+ +
CU VI. Gii h pt:
1 2 3
1 2 3
1 2 3
z z z 4 2i
2z z z 2 5i
z 2z 3z 9 2i
+ + +
+ +
'
+ + +
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Mn thi: TON
I. Phn CHUNG cho tt c th sinh (7 im)
Cu I. Cho hm s
2x 1
y
x 2
+
+
c th l (C)
1. Kho st s bin thin v v th ca hm s
2. Chng minh ng thng d: y = x + m lun lun ct th (C) ti hai im phn bit A, B. Tm m
on AB c di nh nht.
Cu II.
1. Gii phng trnh: 9sinx + 6cosx 3sin2x + cos2x = 8
2. Gii bt phng trnh
2 2 2
2 2 4
log x log x 3 5(log x 3) >
Cu III. Tm nguyn hm
3 5
dx
I
sin x.cos x
Cu IV. Cho lng tr tam gic ABC.A

1
B
1
C
1
c tt c cc cnh bng a, gc to bi cnh bn v mt phng y
bng 30
o
. Hnh chiu H ca im A trn mt phng (A
1
B
1
C
1
) thuc ng thng B
1
C
1
. Tnh khong cch gia
hai ng thng AA
1
v B
1
C
1
theo a.
Cu V. Xt ba s thc khng m a, b, c tha mn a
2012
+ b
2012
+ c
2012
= 3. Tm gi tr ln nht ca biu thc P =
a
4
+ b
4
+ c
4
II. Phn ring (3 im)
A. Theo chng trnh chun
Cu VIa.
1. Cho ng trn (C) c phng trnh (x 1)
2
+ (y + 2)
2
= 9 v ng thng d: x + y + m = 0. Tm m trn -
ng thng d c duy nht mt im A m t k c hai tip tuyn AB, AC ti ng trn (C) (B, C l hai
tip im) sao cho tam gic ABC vung.
x 1 2t
y t
z 1 3t
+
'
. Lp phng trnh mt phng (P) i

qua A, song song vi d v khong cch t d ti (P) l ln nht.
Cu VIIa.
1. C bao nhiu s t nhin c 4 ch s khc nhau v khc 0 m trong mi s lun lun c mt hai ch s
chn v hai ch s l.
B. Theo chng trnh nng cao (3 im)
Cu VIb (2 im)
1. Cho ng trn (C): x
2
+ y
2
2x + 4y 4 = 0 v ng thng d: x + y + m = 0. Tm m trn ng thng d c
duy nht mt im A m t k c hai tip tuyn AB, AC ti ng trn (C) (B, C l hai tip im) sao
cho tam gic ABC vung.
x 1 y z 1
2 1 3

. Lp phng trnh mt phng (P)
i qua A, song song vi d v khong cch t d ti (P) l ln nht.
Cu VIIb. C bao nhiu s t nhin c 5 ch s khc nhau m trong mi s lun lun c mt hai ch s chn
v ba ch s l.
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Mn thi: TON
I. PHN CHUNG (7 im)
Cu 1: Cho hm s y = x
3
+ 3x
2
+ mx + 1 c th (C
m
).
1. Kho st v v th khi m = 3.
2. Xc nh m (C
m
) ct ng thng y = 1 ti 3 im phn bit C(0, 1), D, E sao cho cc tip tuyn ca (C
m
)
ti D v E vung gc vi nhau.
Cu 2:
1. Gii phng trnh: 2cos3x + 3 sinx + cosx = 0
2. Gii h phng trnh
2 2
2 2
x 91 y 2 y
y 91 x 2 x
+ +
'
+ +
Cu 3: Cho s thc b ln2. Tnh J =

ln10 x
3 x
b
e dx
e 2
v tm
b ln2
lim J.
Cu 4: Tnh th tch ca hnh chp S.ABC, bit y ABC l mt tam gic u cnh a, mt bn (SAB) vung
gc vi y, hai mt bn cn li cng to vi y gc 90
o
.
Cu 5: Ch x, y, z dng tho
1 1 1
2012
x y z
+ +
. Tm GTLN ca biu thc
P =
1 1 1
2x y z x 2y z x y 2z
+ +
+ + + + + +
II. PHN T CHN
Phn 1: Theo chng trnh chun
Cu 6.1a.
1. Phng trnh hai cnh ca mt tam gic trong mt phng ta l 5x 2y + 6 = 0; 4x + 7y 21 = 0. Vit
phng trnh cnh th ba ca tam giac , bit rng trc tm ca n trng vi gc ta O.
2. Tm trn Ox im A cch u ng thng (d):
x 1 y z 2
1 2 2
+
v mp(P): 2x y 2z = 0.
Cu 6.2a. Cho tp hp X = {0, 1 , 2, 3, 4, 5, 6, 7}. C th lp c bao nhiu s t nhin gm 5 ch s khc
nhau i mt t X, sao cho mt trong ba ch s u tin phi bng 1.
Phn 2: Theo chng trnh nng cao
Cu 6.1b.
1. Cho ng trn (C): x
2
+ y
2
6x + 5 = 0. Tm M thuc trc tung sao cho qua M v c hai tip tuyn ca
(C) m gc gia hai tip tuyn bng 60
o
.
2. Cho hai ng thng (d
1
):
x 2t
y t
z 4

'
; (d
2
):
x 3 t
y t
z 0

'
. CM (d
1
) v (d
2
) cho nhau. Vit phng trnh mt cu
(S) c ng knh l on vung gc CHUNG ca (d
1
) v (d
2
).
Cu 6.2b. Gii phng trnh sau trong tp s phc C: z
4
z
3
+ 6z
2
8z 16 = 0
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Mn thi: TON
PHN CHUNG CHO TT C CC TH SINH
Cu I (2 im):
1. Kho st v v th (C) ca hm s:
3x 4
y
x 2
. Tm cc im thuc (C) cch u 2 tim cn.

2. Tm cc gi tr ca m phng trnh sau c 2 nghim trn on [0, 2/3].
sin
6
x + cos
6
x = m (sin
4
x + cos
4
x)
Cu II (2 im):
1. Tm cc nghim trn (0, 2) ca phng trnh:
sin3x sin x
sin 2x cos2x
1 cos2x
2. Gii phng trnh:

3 3
x 34 x 3 1 +
Cu III (1 im): Cho chp S.ABC c y ABC l tam gic vung ti C, AC = 2, BC = 4. Cnh bn SA = 5
vung gc vi y. Gi D l trung im cnh AB.
1. Tnh gc gia AC v SD;
2. Tnh khong cch gia BC v SD.
Cu IV (2 im):
1. Tnh tch phn: I =
2
0
sin x cosx 1
dx
sin x 2cosx 3
+
+ +
2. Gii phng trnh sau trn tp s phc C:

z iz 1 2i
3. Hy xc nh tp hp cc im trong mt phng phc biu din cc s phc z tho mn
1 z 1 2 < <
PHN T CHN: Th sinh chn cu V.a hoc cu V.b
Cu V.a. (2 im) Theo chng trnh Chun
1. Vit phng trnh cc cnh ca tam gic ABC bit B(2; 1), ng cao v ng phn gic trong qua nh
A, C ln lt l (d
1
): 3x 4y + 27 = 0 v (d
2
): x + 2y 5 = 0
2. Cho cc ng thng ( )
1
x 1
d : y 4 2t
z 3 t

+
'
v ( )
2
x 3u
d : y 3 2u
z 2

+
'
a. Chng minh rng (d

1
) v (d
2
) cho nhau.
b. Vit phng trnh mt cu (S) c ng knh l on vung gc CHUNG ca (d
1
) v (d
2
).
3. Mt hp cha 30 bi trng, 7 bi v 15 bi xanh. Mt hp khc cha 10 bi trng, 6 bi v 9 bi xanh. Ly
ngu nhin t mi hp bi mt vin bi. Tm xc sut 2 bi ly ra cng mu.
Cu V.b. (2 im) Theo chng trnh Nng cao
1. Cho tam gic ABC vung ti A, p.trnh t BC l: 3 x y 3 = 0, cc nh A v B thuc Ox v bn knh
.trn ni tip tam gic ABC bng 2. Tm ta trng tm G ca tam gic ABC.
2. Cho .thng (d):
x z 0
y 1
+
'

v 2 mp (P): x + 2y + 2z + 3 = 0 v (Q): x + 2y + 2z + 7 = 0
a. Vit phng trnh hnh chiu ca (d) trn (P)
b. Lp ptr mt cu c tm I thuc ng thng (d) v tip xc vi hai mt phng (P) v (Q)
3. Chn ngu nhin 5 con bi trong b t l kh. Tnh xc sut sao cho trong 5 qun bi c ng 3 qun bi
thuc cng b bn (v d 3 con K).
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Mn thi: TON
Cu I. Cho hm s
2
x 2x 5
y
x 1
+
(C)
1. Kho st v v th hm s
2. Tm M thuc (C) tng cc khong cch t M n 2 tim cn l nh nht
Cu II.
x 2 x 2
3.25 3x 10 5 x 3

+
2. Gii h phng trnh:
sin x sin y 2
cos x cos y 2
'
+
Cu III.
1. Gii phng trnh:
( ) ( )
x 1
x
log cos x sin x log cos x cos 2x 0 + +
.
2. Gii bt phng trnh: ( ) ( )
3 2
x 1 x 1 3x x 1 0 + + + + + >
3. C bao nhiu s t nhin gm 5 ch s sao cho trong mi s cc ch s ng trc u ln hn ch s
ng lin sau n.
Cu IV.
1. Trong h ta Oxyz cho 2 im A(0; 0; 3); B(2, 0, 1) v mp (P): 3x 8y + 7z 1 = 0
Tm ta im C trn (P) sao cho ABC l tam gic u.
2. Cho t din ABCD c AB = CD = a, AC = BD = b, AD = BC = c. Hy xc nh cc gc hp bi cc cnh
i din ca t din .
Cu V.
1. Tnh:
/ 4 1
2
3
0 0
xsin x
I dx; J x x 2x 2dx
cos x
+

2. Cho 3 s dng a, b, c. Chng minh rng:
2 2 2
1 1 1 a b c
.
a bc b ac c ab 2abc
+ +
+ +
+ + +
3. Cho z =
1 3
i
2 2
+ , Hy tnh
2 3 2
z ; (z) ;1 z z + +
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Mn thi: TON
I. PHN CHUNG
Cu 1:
1. Kho st v v th (C) ca hm s y =
2x 4
x 1
+
2. Tm trn (C) hai im i xng nhau qua ng thng MN bit M(3; 0) v N(1; 1)
Cu 2:
1. Gii phng trnh
4
1 3x 7
4cos x cos 2x cos 4x cos
2 4 2
+
2. Gii phng trnh: 3
x
.2x = 3
x
+ 2x + 1
Cu 3: Tnh tch phn K =
2
x
0
1 sinx
e dx
1+cosx
+ _

,
Cu 4: Cho hnh chp tam gic u S.ABC di cnh bn bng 1. Cc mt bn hp vi mt phng y mt

gc . Tnh th tch hnh cu ni tip hnh chp S.ABC.
Cu 5: Cho ng thng (d):
x 2 y z 4
3 2 2

v hai im A(1; 2; 1), B(7; 2; 3). Tm trn (d) nhng

im M sao cho khong cch t n A v B l nh nht
II. PHN RING
A. Theo cng trnh chun
Cu 6a:
1. Nm on thng c di 2cm, 4cm, 6cm, 8cm, 10cm. Ly ngu nhin ba on thng trong nm on
thng trn. Tm xc sut ba on thng ly ra lp thnh mt tam gic.
2. Gii h phng trnh:
x x 8 y x y y
x y 5
'

Cu 7a: Tm gi tr nh nht y =
2
cosx
sin x(2cosx -sinx)
vi 0 < x /3
B. Theo chng trnh nng cao
Cu 6b:
1. Tm cc gi tr x trong khai trin nh thc Newton:
( )
x
n
5 lg(10 3 ) (x 2)lg3
2 2

+ bit rng s hng th 6 ca
khai trin bng 21 v
1 3 2
n n n
C C 2C +
2. Cho
2 2
3 cos sin
3 3
_
+

,
. Tm cc s phc sao cho
3
=
Cu 7b: Gi a, b, c l ba cnh ca mt tam gic c chu vi bng 2. Chng minh rng
2 2 2
52
a b c 2abc 2
27
+ + + <
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Mn thi: TON
Cu I: (2 im) Cho hm s y = x
3
3x
2
9x + m, trong
m
l tham s thc.
1. Kho st s bin thin v v th ca hm s cho khi m = 0.
2. Tm tt c cc gi tr ca tham s
m
th hm s cho ct trc honh ti 3 im phn bit c honh
lp thnh cp s cng.
Cu II: (2 im)
1. Gii phng trnh:
2 2
1 x 1 x
cos sin
4 3 2 2
+ .
2. Gii phng trnh:
8
4 8
2
1 1
log (x 3) log (x 1) 3log (4x)
2 4
+ + .
Cu III: (1 im)
Tnh tch phn:
4
2
6
tan x
I dx
cos x 1 cos x
.
Cu IV: (1 im) Tnh th tch ca khi hp ABCD.ABCD theo a. Bit rng AABD l t din u cnh
a.
Cu V: (1 im) Tm cc gi tr ca m phng trnh sau c nghim duy nht thuc on [1/2; 1].
2 3 2
3 1 x 2 x 2x 1 m + +
Cu VI: (2 im)
1. Trong mt phng Oxy, cho ng thng (d): 2x y 5 = 0 v hai im A(1; 2); B(4; 1). Vit phng trnh
ng trn c tm thuc ng thng (d) v i qua hai im A, B.
2. Trong khng gian Oxyz, cho hai im A(1; 1; 2), B(2; 0; 2).
a. Tm qu tch cc im M sao cho MA
2
MB
2
= 5.
b. Tm qu tch cc im cch u hai mt phng (OAB) v (Oxy).
Cu VII: (1 im)
1. Vi n l s t nhin, chng minh ng thc
0 1 2 3 n 1 n n 1
n n n n n n
C 2.C 3.C 4.C ... n.C (n 1).C (n 2).2

+ + + + + + + + .
2. Gii h phng trnh:
x iy 2z 10
x y 2iz 20
ix 3iy (1 i)z 30
+
+
'
+ +
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Mn thi: TON
CU I:
Cho hm s:
2 2 3
mx (m 1)x 4m m
y
x m
+ + + +
+
(C
m
)
1. Kho st s bin thin v v th ca hm s khi m = 1
2. Tm cc gi tr ca tham s m th (C
m
) c 1 im cc tr thuc gc phn t th II v 1 im cc tr
thuc gc phn t th IV ca mt phng to
CU II:
1. Gi D l min c gii hn bi cc ng y = 3x + 10, y = 1, y = x
2
(x > 0) v D nm ngoi parabol y =
x
2
. Tnh th tch vt th trn xoay c to nn khi D quay xung quang trc Ox.
2. Cho k v n l cc s nguyn tha 0 k n. Chng minh rng:
n n n 2
2n k 2n k 2n
C .C (C )
+

CU III:
2 2 2
x 3x 2 x 4x 3 2. x 5x 4 + + + +
2. Cho phng trnh:
2 2 2 2
4 1 2
2log (2x x 2m 4m ) log (x mx 2m ) 0 + + +
(2)
Xc nh tham s m phng trnh (2) c 2 nghim x
1
, x
2
tha
2 2
1 2
x x 1 + >
CU IV:
1. Xc nh cc gi tr ca tham s a phng trnh sau c nghim:
6 6
sin x cos x a sin2x +
2. Cho tam gic ABC tha:
a cos A bcos B ccos C 2p
a sin B bsin C csin A 9R
+ +
+ +
vi a = BC, b = CA, c = AB; p l na chu vi; R l bn knh ng trn ngoi tip ca tam gic. Chng t tam
gic ABC l tam gic u.
CU V: Trong mt phng Oxy cho elip
2 2
x y
(E) : 1
9 4
+ v hai ng thng (d): ax by = 0; (d): bx + ay =
0; vi a
2
+ b
2
> 0. Gi M, N l cc giao im ca (d) vi (E); P, Q l cc giao im ca (d) vi (E).
1. Tnh din tch t gic MNPQ theo a v b
2. Tm iu kin i vi a, b din tch t gic MNPQ nh nht.
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Mn thi: TON
PHN CHUNG CHO TT C CC TH SINH (7.0 im)
Cu I. (2.0 im) Cho hm s
x
y
x 1
(C)
1. Kho st s bin thin v v th hm s (C)
2. Vit phng trnh tip tuyn vi th (C), bit rng khong cch t tm i xng ca th (C) n tip
tuyn l ln nht.
Cu II. (2.0 im)
1. Gii phng trnh 2cos 6x 2cos 4x 3 cos 2x sin 2x 3 + +
2. Gii h phng trnh
2
2 2
1
2x x 2
y
y y x 2y 2
'
Cu III. (1.0 im)

Tnh tch phn
1
2 3
0
x
(x sin x )dx
1 x
+
+
Cu IV. (1.0 im)

Cho x, y, z l cc s thc dng ln hn 1 v tho mn iu kin
1 1 1
2
x y z
+ +
Tm gi tr ln nht ca biu thc A = (x 1)(y 1)(z 1).
Cu V. (1.0 im)
Cho hnh chp S.ABCD y ABCD l hnh thoi. SA = x (0 < x < 3 ) cc cnh cn li u bng 1.
Tnh th tch ca hnh chp S.ABCD theo x
PHN RING (3.0 im)
Th sinh ch c lm mt trong hai phn A hoc B (Nu th sinh lm c hai phn s khng dc chm im).
A. Theo chng trnh nng cao
Cu VIa. (2.0 im)
1. Trong mt phng Oxy cho hai ng thng (d
1
): 4x 3y 12 = 0 v (d
2
): 4x + 3y 12 = 0.
Tm ta tm v bn knh ng trn ni tip tam gic c 3 cnh nm trn (d
1
), (d
2
), trc Oy.
2. Cho hnh lp phng ABCD.ABCD

c cnh bng 2. Gi M l trung im ca on AD, N l
tm hnh vung CCDD. Tnh bn knh mt cu i qua cc im B, C, M, N.
Cu VIIa. (1.0 im)
Gii bt phng trnh
2 3
3 4
2
log (x 1) log (x 1)
0
x 5x 6
+ +
>

B. Theo chng trnh chun
Cu VIb. (2.0 im)
1. Cho im A(1; 0), B(1; 2) v ng thng (d): x y 1 = 0. Lp phng trnh ng trn i qua 2
im A, B v tip xc vi ng thng (d).
2. Trong khng gian Oxyz cho im A(1; 0; 1), B(2; 1; 2) v mt phng (Q): x + 2y + 3z + 3 = 0. Lp phng
trnh mt phng (P) i qua A, B v vung gc vi (Q).
Cu VIIb. (1.0 im) Gii phng trnh
x x 1 x 2 2x 3
x x x x 2
C 2C C C

+
+ +
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Mn thi: TON
Cu I. (2 im) Cho hm s y = x
3
3x
2
+ mx + 4, trong m l tham s thc.
1. Kho st s bin thin v v th ca hm s cho, vi m = 0.
2. Tm tt c cc gi tr ca tham s m hm s cho nghch bin trn khong (0; +).
Cu II. (2 im) Gii cc phng trnh sau
1. 3 (2cos
2
x + cosx 2) + (3 2cosx)sinx = 0
2.
2
2 4 2
log (x 2) log (x 5) log 8 0 + +
Cu III. (1 im) Tnh din tch hnh phng gii hn bi th hm s y =
x
e 1 +
, trc honh v hai ng
thng x = ln3, x = ln8.
Cu VI. (1 im) Cho hnh chp S.ABCD c y ABCD l hnh vung cnh a, SA = SB = a, mt phng (SAB)
vung gc vi mt phng (ABCD). Tnh bn knh mt cu ngoi tip hnh chp S.ABCD.
Cu V. (1 im) Xt cc s thc dng x, y, z tha mn iu kin x + y + z = 1. Tm gi tr nh nht ca biu
thc:
2 2 2
x (y z) y (z x) z (x y)
P
yz zx xy
+ + +
+ +
Cu VI. (2 im)
1. Trong mt phng vi h ta Oxy, cho ng trn (C) c phng trnh: x
2
+ y
2
6x + 5 = 0. Tm im M
thuc trc tung sao cho qua M k c hai tip tuyn vi (C) m gc gia hai tip tuyn bng 60
o
.
2. Trong khng gian vi h ta Oxyz, cho im M(2; 1; 0) v ng thng (d):
x 1 y 1 z
2 1 1
+

Vit phng trnh chnh tc ca ng thng i qua im M, ct v vung gc vi ng thng d.

Cu VII. (1 im) Tm h s ca x
2
trong khai trin thnh a thc ca biu thc P = (x
2
+ x 1)
6
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Mn thi: TON
I. PHN CHUNG CHO TT C CC TH SINH (7,0 im)
Cu I. (2,0 im) Cho hm s
x 2
y
x 2
+
, c th l (C)
1. Kho st v v (C)
2. Vit phng trnh tip tuyn ca (C), bit tip tuyn i qua im A(6; 5)
Cu II. (2,0 im)
1. Gii phng trnh:
cos x cos3x 1 2 sin 2x
4
_
+ + +

,
.
2. Gii h phng trnh:
3 3
2 2 3
x y 1
x y 2xy y 2
+
'
+ +
Cu III. (1,0 im) Tnh tch phn

ln3 2x
x x
ln2
e dx
I
e 1 e 2
Cu VI. (1,0 im)

Hnh chp t gic u SABCD c khong cch t A n mt phng (SBC) bng 2. Vi gi tr no ca gc
gia mt bn v mt y ca chp th th tch ca chp nh nht?
Cu V. (1,0 im) Cho ba s dng a, b, c sao cho abc = 1. Chng minh rng
1 1 1
1
a b 1 b c 1 c a 1
+ +
+ + + + + +
II. PHN T CHN (3,0 im)
Cu VIa. (2,0 im)
1. Trong mt phng Oxy cho cc im A(1; 0); B(2;4);C(1; 4); D(3; 5) v ng thng (d): 3x y 5 = 0.
Tm im M trn d sao cho hai tam gic MAB, MCD c din tch bng nhau.
2. Vit phng trnh ng vung gc chung ca hai ng thng sau
1 2
x 1 2t
x y 1 z 2
d : ; d : y 1 t
2 1 1
z 3
+
+
+
'
Cu VIIa. (1,0 im) Tm s thc x, y tha mn ng thc: x(3 + 5i) + y(1 2i)
3
= 7 + 32i
Cu VIb. (2,0 im)
1. Trong mt phng vi h ta Oxy cho ng thng d: x 2y 2 = 0 v im A(0; 1); B(3; 4). Tm ta
im M trn ng thng d sao cho 2MA
2
+ MB
2
l nh nht.
2. Trong khng gian vi h ta Oxyz cho hai im A(1; 7; 1), B(4; 2; 0) v mt phng (P): x + 2y 2z + 1
= 0. Vit phng trnh hnh chiu ca ng thng AB trn mt phng (P)
Cu VIIb. (1,0 im) Cho s phc z = 1 + 3 i. Hy vit dng lng gic ca s phc z
5
.
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Mn thi: TON
Cu I (2 im) Cho hm s y = x
3
3x
2
+ 4
2. Gi d l ng thng i qua im A(3; 4) v c h s gc l m. Tm m d ct (C) ti 3 im phn bit A,
M, N sao cho hai tip tuyn ca (C) ti M v N vung gc vi nhau.
Cu II (2im)
1. Gii h phng trnh:
2
2
x 1 y(x y) 4y
(x 1)(x y 2) y
+ + +
'
+ +
2. Gii phng trnh: 2 2 sin(x ) cos x 1

12

1
2
0
I = xln(x + x +1)dx
Cu IV (1 im) Cho hnh lng tr ABC.ABC c y l tam gic u cnh a, hnh chiu vung gc ca A
ln mt phng (ABC) trng vi tm O ca tam gic ABC. Mt mt phng (P) cha BC v vung gc vi AA,
ct lng tr theo mt thit din c din tch bng
2
a 3
8
. Tnh th tch khi lng tr ABC.ABC.
CuV (1 im) Cho a, b, c l ba s thc dng tha mn abc = 1. Tm GTLN ca biu thc
2 2 2 2 2 2
1 1 1
P = + +
a + 2b +3 b +2c +3 c +2a +3
.
Cu VIa (2 im)
1. Trong mp vi h trc ta Oxy cho parabol (P): y = x
2
2x v elip (E):
2
2
x
+ y =1
9
. Chng minh rng (P)
giao (E) ti 4 im phn bit cng nm trn mt ng trn. Vit phng trnh ng trn i qua 4 im .
2. Trong khng gian vi h trc ta Oxyz cho mt cu (S) c phng trnh x
2
+ y
2
+ z
2
2x + 4y 6z 11
= 0 v mt phng (): 2x + 2y z + 17 = 0. Vit phng trnh mt phng () song song vi () v ct (S) theo
giao tuyn l ng trn c chu vi bng 6.
Cu VIIa (1 im): Tm h s ca s hng cha x
2
trong khai trin nh thc Niutn ca
n
4
1
x +
2 x
_

,
, bit
rng n l s nguyn dng tha mn:
2 3 n+1
0 1 2 n
n n n n
2 2 2 6560
2C + C + C +...... + C =
2 3 n +1 n +1
Cu VIb (2 im)
1. Trong mt phng Oxy cho hai ng thng (d
1
): x + y + 5 = 0, (d
2
): x + 2y 7 = 0 v tam gic ABC c A(2;
3), trng tm G(2; 0), im B thuc d
1
v

im C thuc d
2
. Vit phng trnh ng trn ngoi tip tam gic
ABC.
2. Trong khng gian vi h trc ta Oxyz cho tam gic ABC vi A(1; 2; 5), B(1; 4; 3), C(5; 2; 1) v mt
phng (P): x y z 3 = 0. Gi M l mt im thay i trn mt phng (P). Tm gi tr nh nht ca biu
thc MA
2
+ MB
2
+ MC
2
.
Cu VIIb (1 im) Tm cc gi tr ca tham s thc m sao cho phng trnh (m 3)
x
+ (2 m)x + 3 m =
0 c nghim thc
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Mn thi: TON
Cu I (2 im): Cho hm s y =
2x 3
x 2
c th l (C)
1. Kho st s bin thin v v th (C) ca hm s trn.
2. Tm trn (C) nhng im M sao cho tip tuyn ti M ca (C) ct 2 tim cn ca (C) ti A, B sao cho AB
ngn nht.
Cu II (2 im)
1. Gii phng trnh:
3 3
sin x.sin3x cos x cos3x 1
8
tan x tan x
6 3
+

_ _
+

, ,
2. Gii h phng trnh:
3 3 3
2 2
8x y 27 18y
4x y 6x y
+
'
+

2
2
6
1
I sin x sin xdx
2
Cu IV (1 im) Cho hnh chp S.ABC c gc ((SBC), (ACB)) = 60

o
, ABC v SBC l cc tam gic u cnh
a. Tnh theo a khong cch t B n mt phng (SAC).
Cu V (1 im): Cho x, y, z l cc s thc dng.Tm gi tr ln nht ca biu thc
A =
x y z
x (x y)(x z) y (y x)(y z) z (z x)(z y)
+ +
+ + + + + + + + +
Cu VIa (2 im)
1. Cho ABC c B(1; 2), phn gic trong gc A c phng trnh (): 2x + y 1 = 0; khong cch t C n
() bng 2 ln khong cch t B n (). Tm A, C bit C thuc trc tung.
2. Trong khng gian Oxyz cho mp (P): x 2y + z 2 = 0 v hai ng thng (d
1
):
x 1 3 y z 2
1 1 2
+ +
; (d
2
):
x 1 y 2 z 1
2 1 1

. Vit phng trnh tham s ca ng thng nm trong mp (P) v ct c 2 ng
thng (d
1
), (d
2
).
Cu VIIa (1im) T cc s 0, 1, 2, 3, 4, 5, 6, lp c bao nhiu s c 5 ch s khc nhau m nht thit phi
c ch s 5.
Cu Vb (2im)
1. Cho ABC c din tch bng 3/2; A(2; 3), B(3; 2), trng tm G thuc ng thng (d): 3x y 8 = 0.
Tm bn knh ng trn ni tip ABC.
2. Trong khng gian Oxyz cho ng thng (d) l giao tuyn ca 2 mt phng (P): 2x 2y z + 1 = 0,
(Q): x + 2y 2z 4 = 0 v mt cu (S): x
2
+ y
2
+ z
2
+ 4x 6y + m = 0. Tm tt c cc gi tr ca m (S) ct
(d) ti 2 im MN sao cho MN = 8.
Cu VIIb (1 im): Gii h phng trnh
x-y x y
x y
e e 2(x 1)
e x y 1
+
+
+ +
'
+
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Mn thi: TON
Cu I (2 im): Cho hm s
2x 1
y
x 1
(C)
2. Tm m ng thng d: y = x + m ct (C) ti hai im phn bit A, B sao cho OAB vung ti O.
Cu II (2 im)
1. Gii phng trnh:
2
cos x(cos x 1)
2(1 sin x)
sin x cos x
+
+
2. Gii h phng trnh:
2 2
2 2
x y xy 3
x 1 y 1 4
+
'
+ + +
Cu III (1 im) Tnh tch phn:

2
cos x
0
I (e sin x) sin 2xdx
Cu IV (1im) Cho hnh chp S.ABCD c y ABCD l hnh vung cnh a. SA vung gc vi y v SA =
a. Gi M, N ln lt l trung im AD, SC.
1. Tnh th tch t din BDMN v khong cch t D n mp (BMN).
2. Tnh gc gia hai ng thng MN v BD
Cu V (1 im) Chng minh rng:
2
x
x
e cos x 2 x , x R
2
+ +
Cu VIa (2 im)
1. Lp phng trnh ng thng d i qua im A(1; 2) v ct ng trn (C) c phng trnh (x 2)
2
+ (y +
1)
2
= 25 theo mt dy cung c di bng 8.
2. Chng t rng phng trnh x
2
+ y
2
+ z
2
+ 2(cos)x 2(sin)y + 4z 4 4sin
2
= 0 lun l phng trnh
ca mt mt cu. Tm bn knh mt cu l ln nht.
Cu VIIa (1 im) Lp s t nhin c 5 ch s khc nhau t cc ch s {0; 1; 2; 3; 4; 5; 6; 7}. Hy tnh xc
sut lp c s t nhin chia ht cho 5.
Cu VIb (2 im)
1. Cho ABC bit B(2; 1), ng cao qua A c phng trnh d
1
: 3x 4y + 27 = 0, phn gic trong gc C c
phng trnh d
2
: x + 2y 5 = 0. Tm ta im A.
2. Trong khng gian Oxyz, cho im A(3; 4; 2); (d):
y z 1
x
2 3
v m.phng (P): 4x + 2y + z 1 = 0
a) Tm ta im H l hnh chiu vung gc ca im A ln mt phng (P).
b) Vit phng trnh mt phng () cha (d) v vung gc vi mt phng (P).
Cu VIIb (1 im): Tnh tng
0 1 2 1006
2012 2012 2012 2012
S C C C ... C + + + + .
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Mn thi: TON
Cu I. (2,0 im) Cho hm s y = x
3
3(m + 1)x
2
+ 9x m, vi
m
l tham s thc.
1. Kho st s bin thin v v th ca hm s cho ng vi m = 1.
2. Xc nh m hm s cho t cc tr ti x
1
, x
2
sao cho
1 2
x x 2
.
Cu II. (2,0 im)
1. Gii phng trnh:
1 sin 2x
cot x 2sin(x )
sin x cos x 2 2

+ +
+
.
2. Gii phng trnh: 3
5
5
2log (3x 1) 1 log (2x 1) + +
.
Cu III. (1,0 im) Tnh tch phn
5 2
1
x 1
I dx
x 3x 1
+
.
Cu IV. (1,0 im) Cho hnh lng tr tam gic u ABC.ABC c AB = 1, CC = m > 0. Tm
m
bit rng
gc gia hai ng thng AB v BC bng 60
o
.
Cu V. (1,0 im) Cho cc s thc khng m x, y, z tho mn x
2
+ y
2
+ z
2
= 3. Tm gi tr ln nht ca biu
thc
5
A xy yz zx
x y z
+ + +
+ +
.
Cu VIa. (2,0 im)
1. Trong mt phng vi h ta Oxy cho tam gic ABC c A(4; 6), phng trnh cc ng thng cha
ng cao v trung tuyn k t nh C ln lt l 2x y + 13 = 0 v 6x 13y + 29 = 0. Vit phng trnh
ng trn ngoi tip tam gic ABC.
2. Trong khng gian vi h ta Oxyz cho hnh vung MNPQ c M(5; 3; 1), P(2; 3; 4). Tm ta nh
Q bit rng nh N nm trong mt phng (): x + y z 6 = 0.
Cu VIIa. (1,0 im) Cho tp E = {0, 1, 2, 3, 4, 5, 6}. T cc ch s ca tp E lp c bao nhiu s t nhin
chn gm 4 ch s i mt khc nhau?
Cu VIb. (2,0 im)
1. Trong mt phng vi h ta Oxy xt elp (E) i qua im M(2; 3) v c phng trnh mt ng
chun l x + 8 = 0. Vit phng trnh chnh tc ca (E).
2. Trong khng gian vi h ta Oxyz cho cc im A(1; 0; 0), B(0; 1; 0), C(0; 3; 2) v mt phng (): x +
2y + 2 = 0. Tm ta ca im M cch u cc im A, B, C v mt phng ().
Cu VIIb. (1,0 im) Khai trin v rt gn biu thc 1 x + 2(1 x)
2
+ ... + n(1 x)
n
thu c a thc P(x) =
a
o
+ a
1
x + ... + a
n
x
n
. Tnh h s a
8
bit rng n l s nguyn dng tho mn
2 3
n n
1 7 1
C C n
+
.
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Mn thi: TON
Cu I (2 im)
1. Kho st v v th hm s y = x
4
4x
2
+ 3
2. Tm m phng trnh
4 2
2
x 4x 3 log m +
c ng 4 nghim.
Cu II (2 im)
1. Gia i bt phng tri nh:
( ) ( )
3
x x x
2
5 1 5 1 2 0
+
+ +
2. Gia i phng tri nh:
2
x (x 2) x 1 x 2 +
Cu III (1 im)
Tnh gii hn sau:
x 1 2
3
x 1
e tan(x 1) 1
lim
x 1
Cu IV (1 im)
Cho hnh chp S.ABCD c y l hnh thoi, gc BAD = . Hai mt bn (SAB) v (SAD) cng vung gc vi
mt y, hai mt bn cn li hp vi y mt gc . Cnh SA = a. Tnh din tch xung quanh v th tch khi
chp S.ABCD.
Cu V (1 im) Cho tam gic ABC vi cc cnh l a, b, c. Chng minh rng
a
3
+ b
3
+ c
3
+ 3abc a(b
2
+ c
2
) + b(c
2
+ a
2
) + c(a
2
+ b
2
)
Cu VIa (2 im)
1. Trong mt phng ta Oxy cho ng thng : x + 2y 3 = 0 v hai im A(1; 0), B(3; 4). Hy tm trn
ng thng mt im M sao cho
MA 3MB +
uuuur uuur
nh nht.
2. Trong khng gian vi h ta Oxyz cho hai ng thng
1
x 1 t
d : y 2t
z 2 t

'
v
2
x t
d : y 1 3t
z 1 t

+
'
. Lp phng
trnh ng thng i qua M(1; 0; 1) v ct c hai ng thng d
1
v d
2
.
Cu VIIa (1 im) Tm s phc z tha mn:
2
z 2z 0 +
Cu VIb (2im)
1.Trong mt phng ta cho hai ng trn (C
1
): x
2
+ y
2
= 13 v (C
2
): (x 6)
2
+ y
2
= 25 ct nhau ti
A(2; 3). Vit phng trnh ng thng i qua A v ct (C
1
), (C
2
) theo hai dy cung c di bng nhau.
2. Trong khng gian vi h ta Oxyz cho hai ng thng
1
x 1 t
d : y 2t
z 2 t

'
v
2
x t
d : y 1 3t
z 1 t

+
'
. Lp phng
trnh mt cu c ng knh l on vung gc chung ca d
1
v d
2
.
Cu VIIb (1 im) Trong cc s phc z tha mn iu kin
z 1 2i 1 + +
, tm s phc z c modun nh nht.
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Mn thi: TON
4
4x
2
+ m (C)
1. Kho st hm s vi m = 3.
2. Gi s th (C) ct trc honh ti 4 im phn bit. Tm m hnh phng gii hn bi th (C) v trc
honh c din tch phn pha trn v phn pha di trc honh bng nhau.
Cu II (2 im)
2 2
x 3x 2 2x 3x 1 x 1 + +
2. Gii phng trnh:
3 3
2
4
cos xcos3x sin xsin3x +
Cu III (1 im)
Tnh tch phn: I =
2
3
0
7sin x 5cos x
dx
(sin x cos x)
Cu IV (1 im) Cho hnh chp u S.ABCD c di cnh y bng a, mt bn to vi mt y gc 60

o
.
Mt phng (P) cha AB v i qua trng tm tam gic SAC ct SC, SD ln lt ti M, N. Tnh th tch hnh
chp S.ABMN theo a.
Cu V (1 im) Cho 4 s thc a, b, c, d tho mn a
2
+ b
2
= 1; c d = 3. Cmr:
9 6 2
F ac bd cd
4
+
+ .
Cu VIa (2 im)
1. Tm phng trnh chnh tc ca elip (E), bit tiu c l 8 v (E) qua im M( 15 ; 1).
2. Trong khng gian vi h ta Oxyz cho 2 ng thng
1
x y z
d :
1 1 2
v
2
x 1 2t
d : y t
z 1 t

'
.
Xt v tr tng i ca d
1
v d
2
. Vit phng trnh ng thng qua O, ct d
2
v vung gc vi d
1
.
Cu VIIa (1 im)
Mt hp ng 5 vin bi , 6 vin bi trng v 7 vin bi vng. Ngi ta chn ra 4 vin bi. Hi c bao nhiu
cch chn trong s bi ly ra khng c c 3 mu?
Cu VIb (2 im)
1. Trong mt phng vi h trc ta Oxy cho Hypebol (H):
2 2
x y
1
16 9
. Vit phng trnh chnh tc ca
elip (E) c tiu im trng vi tiu im ca (H) v ngoi tip hnh ch nht c s ca (H).
2. Trong khng gian Oxyz cho (P): x + 2y z + 5 = 0 v (d):
x 3
y 1 z 3
2
+
+ , im A(2; 3; 4). Gi l
ng thng nm trn (P) i qua giao im ca (d) v (P) ng thi vung gc vi d. Tm trn im M sao
cho khong cch AM ngn nht.
Cu VIIb (1 im) Tm h s ca x
3
trong khai trin
n
2
2
x
x
_
+

,
bit n tho mn
1 3 2n 1 23
2n 2n 2n
C C ... C 2
+ + + .
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Mn thi: TON
Cu I (2 im) Cho hm s
2x 1
y
x 1
+
c th (C).
1. Kho st s bin thin v v th hm s.
2. Vi im M bt k thuc th (C) tip tuyn ti M ct 2 tim cn ti Av B. Gi I l giao hai tim cn, tm
v tr ca M chu vi tam gic IAB t gi tr nh nht.
Cu II (2 im)
1. Gii phng trnh:
3sin 2x 2sin x
2
sin 2x cos x
2. Gii h phng trnh:

4 2 2
2 2
x 4x y 6y 9 0
x y x 2y 22 0
+ +
'
+ +
.
Cu III (1 im) Tnh tch phn sau:
2
2
sin x 3
0
I e sin x cos xdx
Cu IV (1 im) Cho hnh chp t gic u S.ABCD c cnh bn bng a, mt bn hp vi y gc . Tm

th tch ca hnh chp t gi tr ln nht.
Cu V (1 im) Cho 3 s dng x, y, z tho mn x + 3y + 5z 3. Chng minh rng
4 4 4
3xy 625z 4 15yz x 4 5zx 81y 4 45 5xyz + + + + +
Cu VIa (2 im)
1. Trong mt phng vi h ta Oxy cho hnh ch nht ABCD c tm I(1/2; 0). ng thng cha cnh AB
c phng trnh x 2y + 2 = 0, AB = 2AD. Tm ta cc nh A, B, C, D, bit A c honh m.
2. Trong khng gian Oxyz cho 2 ng thng (d
1
):
x 1 y 1 z 2
2 3 1
+
v (d
2
):
x 4 y 1 z 3
6 9 3

. Lp
phng trnh mt phng cha (d
1
) v (d
2
).
Cu VIIa (1 im) Tm m phng trnh 10x
2
+ 8x + 4 = m(2x + 1)
2
x 1 +
c 2 nghim phn bit
Cu VIb (2 im)
1. Trong mt phng vi h ta Oxy cho hnh vung ABCD bit M(2; 1); N(4; 2); P(2; 0); Q(1; 2) ln lt
thuc cnh AB, BC, CD, AD. Hy lp phng trnh cc cnh ca hnh vung.
2. Trong khng gian vi h ta Oxyz cho 2 ng thng () v (d) c phng trnh.
2x y 7 0
( ) :
z 4

'
v (d):
x 2 y z 2
2 2 4
+
. Vit phng trnh ng vung gc chung ca () v (d)
Cu VIIb (1 im) Gii v bin lun phng trnh
2 2 3 2
mx 1 (m x 2mx 2) x 3x 4x 2 + + + +
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Mn thi: TON
Cu I: (2 im) Cho hm s
2x 3
y
x 2
(C)
2. Cho M l im bt k trn (C). Tip tuyn ca (C) ti M ct cc ng tim cn ti A v B. Gi I l giao
im ca cc ng tim cn. Tm im M sao cho ng trn ngoi tip IAB c din tch nh nht.
Cu II (2 im)
1. Gii phng trnh:
2 2
x x x
1 sin sin x cos sin x 2cos ( )
2 2 4 2
+
2
2 2
1
log (4x 4x 1) 2 2 (x 2) log ( x)
2
+ > + +
e
2
1
ln x
I 3x ln x dx
x 1 ln x
_
+

+
,
Cu IV (1 im) Cho hnh chp S.ABC c AB = AC = a. BC = a/2. SA = a 3 ,

o
SAB SAC 30
. Tnh th
tch khi chp S.ABC.
Cu V (1 im) Cho a, b, c l ba s dng tho mn: a + b + c = 3/4. Tm gi tr nh nht ca biu thc
3 3 3
1 1 1
P
a 3b b 3c c 3a
+ +
+ + +
Cu VIa (2 im)
1. Trong mt phng Oxy, cho im A(1; 1) v B(3; 3), ng thng (D): 3x 4y + 8 = 0. Lp phng trnh
ng trn qua A, B v tip xc vi ng thng (D).
2. Trong khng gian Oxyz cho hai im A(0; 0; 3), B(2; 0; 1) v mp (P): 3x 8y + 7z + 1 = 0. Vit pt chnh
tc ng thng d nm trn mp (P) v vung gc vi AB ti giao im ca ng thng AB v (P).
Cu VIIa (1 im) Tm s nguyn dng n bit
2 3 k k 2 k 2n 1 2n 1
2n 1 2n 1 2n 1 2n 1
2C 3.2.2C ... ( 1) k(k 1)2 C ... 2n(n 1)2 C 40200
+
+ + + +
+ + + +
Cu VIb (2 im)
1. Trong mt phng Oxy cho cho hai ng thng d
1
: 2x y + 5 = 0, d
2
: 3x + 6y 7 = 0. Lp phng trnh
ng thng i qua im P(2; 1) sao cho ng thng ct hai ng thng d
1
v d
2
to ra mt tam gic
cn c nh l giao im ca hai ng thng d
1
, d
2
.
2. Trong khng gian Oxyz cho 4 im A(1; 1; 2), B(1; 3; 2), C(4; 3; 2), D(4; 1; 2) v mt phng (P) c
phng trnh x + y + z 2 = 0. Gi A l hnh chiu ca A ln mt phng Oxy. Gi (S) l mt cu i qua 4
im A, B, C, D. Xc nh ta tm v bn knh ca ng trn (C) l giao ca (P) v (S).
Cu VIIb (1 im) Gii h phng trnh
3x 1 y 2 y 3x
2
2 2 3.2
3x xy 1 x 1
+ +
+
'
+ + +
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Mn thi: TON
3
3(m + 1)x
2
+ 9x m
1. Kho st s bin thin v v th ca hm s khi m = 1.
2. Xc nh cc gi tr m hm s nghch bin trn mt khong c di bng 2.
Cu II (2 im)
1. Gii phng trnh:
2
x
x
2x 1
3 2 6
2. Gii phng trnh:

tan x tan x .sin3x sin x sin 2x
6 3
_ _
+ +

, ,
Cu III (1 im)
Tnh tch phn
( )
2
3
0
sin xdx
sin x 3 cos x
Cu IV (1 im) Tnh th tch hnh chp S.ABC bit SA = a, SB = b, SC = c, gc ASB = 60

o
, BSC = 90
o
, v
CSA = 120
o
.
Cu V (1 im) Tm gi tr nh nht ca biu thc P =
2 2 2
2 2 2
log x 1 log y 1 log z 4 + + + + + trong x, y, z
l cc s dng tho mn iu kin xyz = 8.
Cu VIa (2 im)
1. Trong mp Oxy cho hai ng thng (d
1
): x + y + 1 = 0, (d
2
): 2x y 1 = 0. Lp phng trnh ng thng
(d) i qua M(1; 1) ct (d
1
) v (d
2
) tng ng ti A v B sao cho
2MA MB 0 +
uuuur uuur r
.
2. Trong khng gian Oxyz cho mt phng (P): x + 2y 2z + 1 = 0 v hai im A(1; 7; 1), B(4; 2; 0). Lp
phng trnh ng thng (D) l hnh chiu vung gc ca ng thng AB trn (P).
Cu VIIa (1 im) K hiu x
1
v x
2
l hai nghim phc ca phng trnh 2x
2
2x + 1 = 0. Tnh gi tr cc s
phc:
2
1
1
x
v
2
2
1
x
.
Cu VIb (2 im)
1. Trong mt phng Oxy, cho hypebol (H) c phng trnh
2 2
x y
1
9 4
. Gi s (d) l mt tip tuyn thay i
v F l mt trong hai tiu im ca (H), k FM vung gc vi (d). Chng minh rng M lun nm trn mt
ng trn c nh, vit phng trnh ng trn .
2. Trong khng gian Oxyz, cho ba im A(1; 0; 0), B(0; 2; 0), C(0; 0; 3). Tm ta trc tm ca tam gic
ABC.
Cu VIIb (1 im) Ngi ta s dng 5 cun sch Ton, 6 cun Vt l, 7 cun Ho hc (cc cun sch cng
loi ging nhau) lm gii thng cho 9 hc sinh, mi hc sinh c 2 cun sch khc loi. Trong 9 hc
sinh trn c hai bn Ngc v Tho. Tm sc xut hai bn Ngc v Tho c phn thng ging nhau.
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Mn thi: TON
3
+ 2mx
2
+ (m + 3)x + 4 c th l (C
m
).
1.Kho st s bin thin v v th (C
1
) ca hm s trn khi m = 1.
2. Cho (d) l ng thng c phng trnh y = x + 4 v im K(1; 3). Tm cc gi tr ca tham s m sao cho
(d) ct (C
m
) ti ba im phn bit A(0; 4), B, C sao cho tam gic KBC c din tch bng
8 2
.
Cu II (2 im)
1. Gii phng trnh: cos2x + 5 = 2(2 cosx)(sinx cosx)
2 3
2 3
2
log (x 1) log (x 1)
0
x 3x 4
+ +
>

Cu III (1 im) Tnh tch phn I =
6 6 4
x
4
sin x cos x
dx
6 1
+
+
Cu IV (1 im)
Cho hnh chp S.ABCD c y ABCD l hnh vung cnh a, tm O. Hai mt bn SAB v SAD cng vung
gc vi mt phng y v SA = 2a. Gi H, K ln lt l hnh chiu ca A ln SB, SD. Tnh th tch khi chp
OAHK.
Cu V (1 im) Cho ba s thc dng a, b, c tha mn abc = 1. Chng minh rng:
3 3 3
4a 4b 4c
3
(1 b)(1 c) (1 c)(1 a) (1 a)(1 b)
+ +
+ + + + + +
Cu VIa (2 im)
1. Trong mt phng Oxy, cho ba im I(2; 4); B(1; 1); C(5; 5). Tm im A sao cho I l tm ng trn ni
tip ABC.
2. Trong khng gian Oxyz cho ba im A(2; 0; 1) B(1; 0; 0), C(1; 1; 1) v mt phng (P): x + y + z 2 = 0.
Vit phng trnh mt cu i qua ba im A, B, C v c tm thuc mt phng (P)
Cu VIIa (1 im) Gii phng trnh:
2 2
x 4 x 2 3x 4 x + +
Cu VIb (2 im)
1. Trong mp Oxy, cho hnh thang ABCD c AB//CD v A(10; 5), B(15; 5), D (20; 0). Tm ta C.
2. Trong khng gian Oxyz cho ng thng (d):
x t
y 1 2t
z 2 t

+
'
v mt phng (P): 2x y 2z 2 = 0. Vit

phng trnh mt cu (S) c tm I trn (d), khong cch t I n mp (P) l 2 v (S) ct mp (P) theo giao tuyn
l ng trn (C) c bn knh r = 3
Cu VIIb (1 im) Tm cc gi tr ca tham s thc m sao cho phng trnh sau c nghim thc
2 2
1 1 x 1 1 x
9 (m 2)3 2m 1 0
+ +
+ + +
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Mn thi: TON
Cu I (2 im) Cho hm s y =
x 3
x 1
+
1. Kho st s bin thin v v th (C) ca hm s cho.

2. Cho im M
o
(x
o
; y
o
) thuc th (C). Tip tuyn ca (C) ti M
o
ct cc tim cn ca (C) ti cc im A v
B. Chng minh M
o
l trung im ca on thng AB.
Cu II (2 im)
1. Gii phng trnh: 4sin
3
x + 4sin
2
x + 3sin2x + 6cosx = 0
2. Gii phng trnh:
2
x 2 7 x 2 x 1 x 8x 7 1 + + + +
2
1
I (x 2) ln xdx
Cu IV (1 im) Cho hnh lp phng ABCD.ABCD c cnh bng a v im K thuc cnh CC sao cho
CK = 2a/3. Mt phng () i qua A, K v song song BD chia khi lp phng thnh hai khi a din. Tnh th
tch ca hai khi a din .
Cu V (1 im) Cho a, b, c l ba s dng. Chng minh rng
3 3 3 2 2 2 2 2 2
2 2 2
a b c a b b c c a 9
2abc c ab a bc b ac 2
+ + + + +
+ + +
+ + +
Cu VIa. (2 im)
1. Trong mt phng Oxy, lp phng trnh chnh tc ca elip (E) c di trc ln bng 4
2
, cc nh trn
trc nh v cc tiu im ca (E) cng nm trn mt ng trn.
2. Trong khng gian Oxyz, cho A(1; 2; 0), B(0; 4; 0), C(0; 0; 3).
a) Vit phng trnh ng thng qua O v vung gc vi mt phng (ABC).
b) Vit phng trnh (P) cha OA, sao cho khong cch t B n (P) bng khong cch t C n (P).
Cu VIIa. (1 im)
Gii phng trnh: 2(log
2
x + 1)log
4
x log
2
4 = 0
Cu VIb. (2 im)
1. Trong mt phng Oxy, cho ng thng (d): 2x y 4 = 0. Lp phng trnh ng trn tip xc vi cc
trc ta v c tm trn ng thng (d).
2. Trong khng gian Oxyz, cho (P): x + y + 2z 5 = 0 v mt cu (S): (x 1)
2
+ (y 1)
2
+ (z 2)
2
= 25
a) Lp phng trnh tip din ca mt cu song song vi Ox v vung gc vi (P)
b) Lp phng trnh mt phng i qua hai A(1; 4; 4) im B(3; 5; 1) v hp vi (P) mt gc 60
o
Cu VIIb. (1 im)
T cc ch s 0, 1, 2, 3, 4, 5, 6 c th lp c bao nhiu s t nhin chn c 5 ch s khc nhau m mi s
lp c u nh hn 25000?
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Mn thi: TON
Cu I: Cho hm s
x
y
x 1
(C)
1. Kho st s bin thin v v th (C) ca hm s cho
2. Vit phng trnh tip tuyn vi th (C), bit rng khong cch t tm i xng ca (C) n tip tuyn l
ln nht.
Cu II:
1. Gii phng trnh:
1
cos3x cos2x cosx
2
+
2
x 4 x 4
x x 16 3
2
+ +
+
Cu III: Tnh tch phn:
e
1
2
I x ln xdx
x
_
+

,
.
Cu IV: Cho hnh chp lc gic u S.ABCDEF vi SA = a, AB = b. Tnh th tch ca hnh chp v
khong cch gia cc ng thng SA, BE.
Cu V: Cho x, y l cc s thc tha mn iu kin
2 2
x xy y 3. + + Chng minh rng:
2 2
(4 3 3) x xy 3y 4 3 3. +
Cu VIa:
1. Trong mt phng Oxy, cho ABC vi B(2; 7), phng trnh ng cao AA: 3x + y + 11 = 0; phng
trnh trung tuyn CM: x + 2y + 7 = 0. Vit phng trnh tng qut ca ng thng AB v AC.
2. Trong khng gian Oxyz, cho (P): 3x + 2y z + 4 = 0 v im A(4; 0; 0), B(0; 4; 0). Gi I l trung im ca
on thng AB.
a) Tm ta giao im E ca ng thng AB vi mt phng (P).
b) Xc nh ta im K sao cho KI vung gc vi mt phng (P) ng thi K cch u gc ta O v
mt phng (P).
Cu VIIa: (1 im): Gii bt phng trnh:
x x
x x
3log 3 2log 2
3
log 3 log 2
+
+
Cu VIb:
1. Vit phng trnh ng thng (d) i qua M(1; 4) v ct hai tia Ox, Oy ti hai im A, B sao cho di OA
+ OB t gi tr nh nht.
2. Trong khng gian Oxyz, cho A(1; 0; 2); B(3; 1; 0); C(0; 1; 1) v ng thng (d) l giao tuyn ca hai mt
phng (P): 3x z + 5 = 0; (Q): 4x + y 2z + 1 = 0
a) Vit phng trnh tham s ca (d) v phng trnh mt phng (P) qua A; B; C.
b) Tm giao im H ca (d) v (P). Chng minh H l trc tm ca tam gic ABC.
Cu VIIb: Cho tp A = {0; 1; 2; 3; 4; 5; 6}. C bao nhiu s t nhin c 5 ch s khc nhau chn trong A sao
cho s chia ht cho 15.
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Mn thi: TON
Cu I (2 im): Gi (C
m
) l th ca hm s y = x
3
+ (2m + 1)x
2
(m + 1)
1. Kho st s bin thin v v th khi m = 1.
2. Tm m th (C
m
) tip xc vi ng thng y = 2mx (m + 1)
Cu II (2 im)
1. Tm nghim trn (0, /2) ca phng trnh:
(1 cos x) (sin x 1)(1 cos x) (1 cos x) (sin x 1)(1 cos x) sin x 2 + + + + +
2. Gii h phng trnh:
2 2
2 2
x 2 x y 3 y 5
x 2 x y 3 y 2
+ + + + +
'
+ + +
.
Cu III (1 im)
Tnh tch phn
4
2 4
0
sin 4x
I dx
cos x tan x 1
Cu IV (1 im): Cho khi lng tr tam gic ABC.ABC c y ABC l tam gic u cnh a v nh A
cch u cc nh A, B, C. Cnh bn AA to vi y gc 60
o
. Tnh th tch ca khi lng tr theo a.
Cu V (1 im) Cho 4 s thc x, y, z, t 1. Tm gi tr nh nht ca biu thc:
4 4 4 4
1 1 1 1
P (xyzt 1)
x 1 y 1 z 1 t 1
_
+ + + +

+ + + +
,
Cu VIa (2 im)
1. Trong mt phng Oxy cho ABC c cnh AC i qua im M(0; 1). Bit AB = 2AM, pt ng phn gic
trong (AD): x y = 0, ng cao (CH): 2x + y + 3 = 0. Tm ta cc nh ca ABC.
2. Trong khng gian Oxyz cho 4 im A(3; 0; 0), B(0; 1; 4), C(1; 2; 2), D(1; 3; 1). Chng t A, B, C, D l 4
nh ca mt t din v tm trc tm ca tam gic ABC.
Cu VIIa (1 im)
Cho tp hp X = {0; 1; 2; 3; 4; 5; 6}. T cc ch s ca tp X c th lp c bao nhiu s t nhin c 5 ch
s khc nhau v phi c mt ch s 1 v 2.
Cu Vib (2 im)
1. Vit phng trnh ng thng (d) qua A(1; 2) v to vi ng thng (D):
x 3 y 5
1 2
+
mt gc 45
0
.
2. Trong khng gian Oxyz cho ng thng d l giao tuyn ca 2 mp (P): x my + z m = 0 v (Q): mx + y
mz 1 = 0, m l tham s.
a) Lp phng trnh hnh chiu ca (d) ln mt phng Oxy.
b) Chng minh rng khi m thay i, ng thng lun tip xc vi mt ng trn c nh trong mt phng
Oxy.
Cu VIIb (1 im)
Gii phng trnh sau trn tp C: (z
2
+ z)
2
+ 4(z
2
+ z) 12 = 0
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Mn thi: TON
Cu I (2 im)
Cho hm s y = x
3
+ (1 2m)x
2
+ (2 m)x + m + 2 (m l tham s) (1)
1. Kho st s bin thin v v th ca hm s (1) khi m = 2
2. Tm cc gi tr ca m th hm s (1) c im cc i, im cc tiu, ng thi honh ca im cc
tiu nh hn 1.
Cu II (2 im)
1. Gii phng trnh: cos2x + (1 + 2cosx)(sinx cosx) = 0
2. Gii h phng trnh:
2 2
2 2
(x y)(x y ) 13
(x y)(x y ) 25
+
'
+

e
1
3 2ln x
I dx
x 1 2ln x
Cu IV (1 im)
Cho lng tr ABC.ABC c A.ABC l h.chp tam gic u cnh y AB = a, cnh bn AA = b. Gi l
gc gia hai mp (ABC) v (ABC). Tnh tan v th tch ca khi chp A.BBCC
Cu V (1 im)
Cho hai s dng x, y thay i tha mn iu kin x + y 4. Tm gi tr nh nht ca biu thc
A =
2 3
2
3x 4 2 y
4x y
+ +
+
Cu Via. (2 im)
1. Trong mt phng Oxy, cho tam gic ABC c nh A(2; 1), ng cao qua nh B c phng trnh l x 3y
7 = 0 v ng trung tuyn qua nh C c phng trnh l x + y + 1 = 0. Xc nh ta cc nh B v C
ca tam gic.
2. Trong khng gian Oxyz, cho im G(1; 1; 1).
a) Vit phng trnh mt phng (P) qua G v vung gc vi ng thng OG.
b) (P) ct Ox, Oy, Oz ti A, B, C. Chng minh tam gic ABC u v G l trc tm tam gic ABC.
Cu VIIa. (1 im)
Cho hai ng thng song song d
1
v d
2
. Trn ng thng d
1
c 10 im phn bit, trn ng thng d
2
c n
im phn bit (n 2). Bit rng c 2800 tam gic c nh l cc im cho. Tm n.
Cu VIb. (2 im)
1. Trong mt phng Oxy, cho (E): 9x
2
+ 16y
2
= 144 Vit phng trnh ng thng i qua M(2; 1) v ct
elip (E) ti A v B sao cho M l trung im ca AB
2. Trong khng gian Oxyz, cho mt phng (P): 2x y + 2z + 5 = 0 v cc im A(0; 0; 4), B(2; 0; 0)
a) Vit phng trnh hnh chiu vung gc ca ng thng AB trn mt phng (P)
b) Vit phng trnh mt cu i qua O, A, B v tip xc vi mt phng (P).
Cu VIIb. (1 im)
Tm cc gi tr x trong khai trin nh thc Newton
( )
x
n
5 lg(10 3 ) (x 2)lg3
2 2

+ bit rng s hng th 6 ca khai
trin bng 21 v
1 3 2
n n n
C C 2C + .
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Mn thi: TON
Cu I (2 im)
Cho hm s y = (1/3)x
3
mx
2
+(m
2
1)x + 1 c th (C
m
)
1. Kho st s bin thin v v th (C
2
) khi m = 2.
2. Tm m, hm s (C
m
) c cc i, cc tiu v y
C
+ y
CT
> 2.
Cu II (2 im)
x 1 x x 1
15.2 1 2 1 2
+ +
+ +
2. Tm m phng trnh:
2
2 0,5
4(log x) log x m 0 + c nghim thuc (0, 1).
Cu III (2 im): Tnh tch phn: I =
( )
3
6 2
1
dx
x 1 x +
.
Cu IV (1 im): Tnh th tch ca hnh chp S.ABC, bit y ABC l mt tam gic u cnh a, mt bn
(SAB) vung gc vi y, hai mt bn cn li cng to vi y gc .
Cu V (1 im): Tm gi tr nh nht ca hm s: y =
2
cos x
sin x(2cos x sin x)
vi 0 < x /3.
Cu VIa (2 im)
1. Vit phng trnh chnh tc ca (E) c hai tiu im F
1
, F
2
bit (E) qua
3 4
M ;
5 5
_

,
v MF
1
F
2
vung ti
M.
2. Trong khng gian Oxyz cho 2 ng thng (d
1
):
x t
y 4 t
z 6 2t

+
'
v (d
2
):
x t '
y 3t ' 6
z t ' 1

'
Gi K l hnh chiu vung gc ca im I(1; 1; 1) trn (d

2
). Tm phng trnh tham s ca ng thng qua
K vung gc vi (d
1
) v ct (d
1
).
Cu VIIa (1 im): Gii phng trnh:
2
4 3
z
z z z 1 0
2
+ + + trn tp s phc.
Cu Vib (2 im)
1. Trong mt phng Oxy cho hai ng trn (C
1
): x
2
+ y
2
2x 2y 2 = 0; (C
2
): x
2
+ y
2
8x 2y + 16 = 0.
Vit phng trnh tip tuyn chung ca (C
1
) v (C
2
).
2. Trong khng gian Oxyz cho hai ng thng D
1
:
x 2 y 1 z
1 1 2

, D
2
:
x 2x 2 0
y 3
+
'
a) Chng minh rng D

1
cho D
2
. Vit phng trnh ng vung gc chung ca D
1
v D
2
b) Vit phng trnh mt cu c ng knh l on vung gc chung ca D
1
v D
2
Cu VIIb (1 im) Tnh tng
0 1 2 2012
2012 2012 2012 2012
S C 2C 3C ... 2013C + + + +
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Mn thi: TON
Cu I (2,0 im)
1. Kho st s bin thin v v th (C) ca hm s: y = x
3
3x
2
+ 2
2. Bin lun theo m s nghim ca phng trnh:
2
m
(x 2x 2)
x 1

Cu II (2,0 im)
1. Gii phng trnh:
11 5x 7 x 3x 2013
cos( ) sin( ) 2 sin( )
4 2 4 2 2 2

+ +
2. Gii h phng trnh:
2 2
2 2
2 2
30x 9x y 25y 0
30y 9y z 25z 0
30x 9z x 25x 0

'
Cu III (1,0 im) Tnh tch phn: I =

3
1
(x 4)dx
3 x 1 x 3
+
+ + +
Cu IV (1,0 im) Cho hnh chp S.ABCD c y ABCD l hnh ch nht vi AB = a, AD = 2a. Cnh SA
vung gc vi mt phng y, cnh bn SB to vi mt phng y mt gc 60
0
. Trn cnh SA ly im M sao
cho AM =
a 3
3
, mt phng (BCM) ct cnh SD ti N. Tnh th tch khi chp S.BCNM.
Cu V (1,0 im) Cho x, y, z l ba s thc tha mn: 2
x
+ 2
y
+2
z
= 1. Chng minh rng:
x y z x y z
x y z y x z z x y
4 4 4 2 2 2
2 2 2 2 2 2 4
+ + +
+ +
+ +
+ + +
Cu VI.a (2,0 im)
1.Trong mt phng Oxy, cho ng thng d c phng trnh
x 2 2t
y 3 t
+
'
+
v im A(0; 1). Tm im M thuc

d sao cho AM ngn nht.
2. Trong khng gian Oxyz cho hai ng thng d
1
:
x 2 y z 1
4 6 8
+

; d
2
:
x 7 y 2 z
6 9 12

a) Chng minh rng d

1
v d
2
song song. Vit phng trnh mt phng (P) qua d
1
v d
2
.
b) Cho im A(1; 1; 2), B(3; 4; 2). Tm im I trn ng thng d
1
sao cho IA + IB t gi tr nh nht
Cu VII.a (1,0 im) Gii phng trnh:
2 3
9 27
3 3
log (x 1) log 2 log 4 x log (x 4) + + + +
Cu VI.b (2,0 im)
1. Vi gi tr no ca m th phng trnh x
2
+ y
2
2(m + 2)x + 4my + 19m 6 = 0 l phng trnh ng trn
2. Trong khng gian Oxyz, cho ba im A(1; 2; 1); B(2; 1; 3); C(4; 7; 5) v (P): x 2y + z = 0
a) Vit phng trnh ng thng (d) qua A, song song mt phng (P) v vung gc ng thng BC
b) Tm im M trn (P) sao cho di AM + BM t gi tr nh nht.
CuVII.b (1,0 im) Cho phng trnh:
2 2
5 5
log x 2 log x 1 m 2 0 + + .
Tm cc gi tr ca tham s m phng trnh cho c t nht mt nghim thuc on
3
1;5
1
]
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THI TH I HC CAO NG
Mn thi: TON
I. PHN BT BUC DNH CHO TT C TH SINH (7,0 im)
Cu I. Cho hm s y = (x m)
3
3x (1)
1. Xc nh m hm s (1) t cc tiu ti im c honh x = 0.
2. Kho st s bin thin v v th (C) ca hm s (1) khi m = 1.
Cu II (2 im)
1. Tm tng tt c cc nghim x thuc [2; 40] ca phng trnh sinx cos2x = 0.
2. Gii h phng trnh:
x y x y 8
y x y 2
+ +
'

Cu III. Tm k h bt phng trnh sau c nghim:

( )
3
3
2
2 2
x 1 3x k 0
1 1
log x log x 1 1
2 3
<
'
+
Cu IV.
Cho hnh chp S.ABCD c y ABCD l hnh thoi cnh a, gc BAD = 60
o
, SA vung gc mt phng
(ABCD), SA = a. Gi C l trung im ca SC. Mt phng (P) i qua AC v song vi BD, ct cc cnh SB,
SD ca hnh chp ln lt ti B, D. Tnh th tch ca khi chp S.ABCD.
Cu V. Cho a, b, c l ba cnh ca mt tam gic. Chng minh bt ng thc:
( ) ( ) ( )
ab bc ca a b c
c c a a a b b b c c a a b b c
+ + + +
+ + + + + +
Cu Via.
1. Trong mt phng Oxy, vit phng trnh ng trn (C) ngoi tip tam gic ABC bit A(1; 4), B(7; 4),
C(2; 5)
2. Trong khng gian Oxyz, cho hai ng thng
1
x 1 t
( ) : y 1 t
z 2
+

'
, ( )
2
x 3 y 1 z
:
1 2 1

a) Vit phng trnh mt phng cha

1
v song song vi
2
.
b) Xc nh im A trn
1
v im B trn
2
sao cho on AB c di nh nht.
Cu VIIa. Tm s phc z tha mn iu kin
z 5
v phn thc ca z bng hai ln phn o ca n.
Cu Vib.
1. Trong mt phng Oxy, vit phng trnh ng thng (D) qua A(2; 0) v to vi ng thng (d): x + 3y
3 = 0 mt gc 45
o
2. Cho mt phng (P): 2x y + 2z 3 = 0 v mt cu (S): (x 1)
2
+ (y + 1)
2
+ (z 2)
2
= 25
a) Chng t rng mt phng (P) v mt cu (S) ct nhau. Tm bn knh ca ng trn giao tuyn
b) Lp phng trnh cc tip din ca mt cu song song vi mt phng (P).
Cu VIIb. Tnh tng
2 3 25
25 25 25
S 1.2.C 2.3.C ... 24.25.C + + + .
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Mn thi: TON
4
2mx
2
+ m 1 (1), vi
m
l tham s thc.
1. Kho st s bin thin v v th hm s (1) khi m = 1.
2. Xc nh
m
hm s (1) c ba im cc tr, ng thi cc im cc tr ca th to thnh mt tam gic
c bn knh ng trn ngoi tip bng 1.
Cu II (2 im)
1. Gii phng trnh:
2 3
2
2
cos x cos x 1
cos 2x tan x
cos x
+
.
2. Gii h phng trnh:
2 2
2 2
x y xy 1 4y
y(x y) 2x 7y 2
+ + +
'
+ + +

e 3
2
2
1
log x
I dx
x 1 3ln x
.
Cu IV. (1 im) Cho hnh hp ng ABCD.ABCD c cc cnh AB = AD = a, AA =
a 3
2
v gc BAD =
60
o
. Gi M v N ln lt l trung im ca cc cnh AD v AB. Chng minh AC vung gc vi mt
phng (BDMN). Tnh th tch khi chp A.BDMN.
Cu V. (1 im) Cho a, b, c l cc s thc khng m tha mn a + b + c = 1. Chng minh rng:
7
ab bc ca 2abc
27
+ + .
Cu VIa. (2 im)
1. Trong mt phng Oxy, cho tam gic ABC bit A(5; 2). Phng trnh ng trung trc cnh BC, ng
trung tuyn CC ln lt l x + y 6 = 0 v 2x y + 3 = 0. Tm ta cc nh ca tam gic ABC.
2. Trong khng gian Oxyz, hy xc nh ta tm v bn knh ng trn ngoi tip tam gic ABC, bit A(
1; 0; 1), B(1; 2; 1), C(1; 2; 3).
Cu VIIa. (1 im)
Cho z
1
, z
2
l cc nghim phc ca phng trnh 2z
2
4z + 11 = 0. Tnh gi tr ca biu thc
2 2
1 2
2
1 2
z z
(z z )
+
+
.
Cu VIb. (2 im)
1. Trong mt phng Oxy cho hai ng thng (d): x + 3y + 8 = 0, (d) : 3x 4y + 10 = 0 v im A(2; 1).
Vit phng trnh ng trn c tm thuc (d), i qua A v tip xc vi (d).
2. Trong khng gian Oxyz, Cho ba im A(0; 1; 2), B(2; 2; 1), C(2; 0; 1). Vit phng trnh mt phng
(ABC) v tm im M thuc mt phng 2x + 2y + z 3 = 0 sao cho MA = MB = MC.
Cu VIIb. (1 im) Gii h phng trnh:
2
1 x 2 y
1 x 2 y
2log ( xy 2x y 2) log (x 2x 1) 6
log (y 5) log (x 4) 1
+
+
+ + + +
'
+ +
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Mn thi: TON
I. PHN CHUNG CHO TT C CC TH SINH (7 im)
3
+ 3x
2
+ mx 2 (1), m l tham s thc.
1. Kho st s bin thin v v th hm s khi m = 0.
2. Tm cc gi tr ca m hm s (1) nghch bin trn khong (0; 2).
Cu II. (2 im)
1. Gii phng trnh:
2
2
tan x tan x 2
sin x
tan x 1 2 4
+ _
+

+ ,
2. Gii h phng trnh:
1 2
2
(1 4 ).5 1 3
1
3 1 2
+ +
+ +
'

x y x y x y
x y y y
x
Cu III. (1 im) Tnh tch phn:
4
0
sin x dx
4
sin 2x 2(sin x cos x) 2
_

,
+ + +
.
Cu IV. (1 im)
Cho hnh chp S.ABC c y ABC l tam gic vung cn ti nh B, AB = a, SA = 2a v SA vung gc mt
phng y. Mt phng qua A vung gc vi SC ct SB, SC ln lt ti H, K. Tnh theo a th tch khi t din
SAHK.
Cu V. (1 im) Tm cc gi tr ca tham s m phng trnh sau c ng mt nghim thc
4 2
x 2x 4 x 1 m + + +
Cu VI.a (2 im)
1. Cho ng trn (C): (x 3) + (y +1) = 4 v im M(1; 3) Vit phng trnh tip tuyn (d) ca (C), bit (d)
i qua M.
2. Trong khng gian Oxyz, cho M(1; 2; 3). Lp phng trnh mt phng i qua M ct ba tia Ox ti A, Oy ti B,
Oz ti C sao cho th tch t din OABC nh nht.
Cu VII.a (1 im) Gii bt phng trnh:
2x 1 2x 1 x
3 2 5.6 0
+ +
+ .
B.Theo chng trnh Nng cao:
Cu VI.b (2 im)
1. Chng minh rng trong cc tip tuyn ca (P): y
2
= 4x k t cc im A(0; 1); B(2; 3) c hai tip tuyn
vung gc vi nhau
2. Trong khng gian vi h ta Oxyz, cho hai ng thng
1
x 4 y 1 z 5
d :
3 1 2
+

v
2
x 2 t
d : y 3 3t
z t
+
+
'

a. Chng minh rng d
1
v d
2
cho nhau, tnh khong cch gia d
1
v d
2
.
b. Vit phng trnh mt cu c bn knh nh nht tip xc vi c hai ng thng d
1
v d
2
.
Cu VII.b (1 im) Gii phng trnh:
7 3
log x log (2 x) +
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Mn thi: TON
4
(2m + 1)x
2
+ 2m.
1. Kho st s bin thin v v th (C) ca hm s khi m = 2.
2. Tm tt c cc gi tr ca m th hm s ct trc Ox ti 4 im phn bit cch u nhau.
Cu II (2 im)
2 2
1 8 21 1
2cos x cos x 3 sin 2(x ) 3cos(x ) sin x
3 3 2 3
+ + + + + + .
2. Gii h phng trnh:
2 2
2 2 2
x xy y 3(x y)
x xy y 7(x y)
+
'
+ +
Cu III (1 im)
Tnh din tch hnh phng gii hn bi cc ng sau: y = 0, x = 1,
( )
x
2
xe
y
x 1
+
.
Cu IV (1 im)
Cho hnh chp S.ABCD c y ABCD l hnh thang AB = a, BC = a, gc BAD = 90
o
, cnh SA =
a 2
v SA
vung gc vi y, tam gic SCD vung ti C. Gi H l hnh chiu ca A trn SB, tnh th tch ca t din
SBCD v khong cch t im H n mt phng (SCD).
Cu V (1 im) Vi mi s thc x, y, z > 1 v tha iu kin
1 1 1
2
x y z
+ +
. Tm GTLN ca biu thc A = (x
1)(y 1)(z 1)
Cu VIa (2 im)
1. Trong mt phng Oxy, cho ABC vi A(1; 1); B(2; 0); C(2; 2). Vit phng trnh ng thng cch u
cc nh ca ABC.
2. Trong khng gian Oxyz, cho 2 im A(4; 0; 0), B(0; 0; 4) v mp (P): 2x y + 2z 4 = 0
a. Chng minh rng ng thng AB song song vi mt phng (P), vit phng trnh mt phng trung trc
ca on AB.
b. Tm im C trn mt phng (P) sao cho tam gic ABC u.
Cu VIIa (1 im): Tm phn thc ca s phc z = (1 + i)
n
, trong n l s t nhin v tha mn
( ) ( )
4 5
log n 3 log n 6 4 + +
.
Cu VIb (2 im)
1. Trong mt phng Oxy, cho (H):
2 2
x y
1
4 5
v ng thng (d): x y + m = 0. CMR (d) lun ct (H) ti
hai im M, N thuc hai nhnh khc nhau ca (H).
2. Trong khng gian Oxyz, cho cc im A(1; 3; 5), B(4; 3; 2), C(0; 2; 1). Tm ta tm ng trn ngoi
tip tam gic ABC.
Cu VIIb (1 im) Cho s phc z 1 3i . Hy vit s z
n
dng lng gic bit rng l s t nhin v tha
mn:
2
3 3
log (n 2n 6) log 5 2 2
n 2n 6 4 (n 2n 6)
+
+ + + .
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Mn thi: TON
Cu I (2,0 im) Cho hm s
2x 1
y
x 1
+
+
(C)
1. Kho st s bin thin v v th (C) ca hm s cho
2. Tm trn th (C) nhng im c tng khong cch n hai tim cn ca (C) nh nht.
Cu II (2,0 im)
1. Gii h phng trnh:
2 2
3 3
2y x 1
2x y 2y x

'

.
2.Gii phng trnh sau: ( )
6 6
8 sin x cos x 3 3sin 4x 3 3 cos 2x 9sin 2x 11 + + +
.
Cu III (1,0 im) Tnh tch phn: I =
2 1
x
x
1
2
1
(x 1 )e dx
x
+
+
.
Cu IV (1,0 im) Cho t din ABCD c AC
2
= AD
2
= 2a
2
, BC = BD = a, khong cch t B n (ACD) l
a 3
3
. Bit th ca t din ABCD bng
3
a 15
27
. Tnh gc gia hai mt phng (ACD) v (BCD).
Cu V (1,0 im) Vi mi s thc x, y tha iu kin ( )
2 2
2 x y xy 1 + +
. Tm gi tr ln nht v gi tr nh
nht ca biu thc
4 4
x y
P
2xy 1
+
+
.
II. PHN T CHN (3,0 im). Tt c th sinh ch c lm mt trong hai phn: A hoc B.
A.Theo chng trnh Chun
Cu Via (2,0 im)
1. Trong mp Oxy cho ng trn (C): x
2
+ y
2
2x + 6y 15 = 0. Vit phng trnh ng thng () vung
gc vi ng thng (d): 4x 3y + 2 = 0 v ct ng trn (C) ti A; B sao cho AB = 6.
2. Trong khng gian Oxyz cho hai ng thng d
1
:
x 2 y z 1
4 6 8
+

v d
2
:
x 7 y 2 z
6 9 12

. Xt v tr
tng i ca d
1
v d
2
. Cho hai im A(1; 1; 2) v B(3; 4; 2), Tm ta im I trn ng thng d
1
sao
cho IA + IB t gi tr nh nht.
Cu VIIa (1,0 im) Gii phng trnh trn tp hp C: (z
2
+ i)(z
2

z
) = 0
Cu Vib (2,0 im)
1. Trong mt phng Oxy cho elip (E):
2 2
x y
1
4 3
+ v ng thng : 3x + 4y =12. T im M bt k trn
k ti (E) cc tip tuyn MA, MB. Chng minh rng ng thng AB lun i qua mt im c nh.
2. Trong khng gian Oxyz, cho (d):
x 3 y 2 z 1
2 1 1
+ +

v mt phng (P): x + y + z + 2 = 0. Lp phng

trnh ng thng (D) nm trong (P), vung gc (d) v khong cch t giao im ca (d) v (P) n ng
thng (D) l
42
.
Cu VIIb (1,0 im) Gii h phng trnh:
2 2 2
2 2 2
x log y ylog 3 log x
xlog 72 log x 2y log y
+ +
'
+ +
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Mn thi: TON ( 113)
3
+ (1 2m)x
2
+ (2 m)x + m + 2 (1) m l tham s.
1. Kho st s bin thin v v th (C) ca hm s (1) vi m = 2.
2. Tm tham s m th ca hm s (1) c tip tuyn to vi ng thng d: x + y + 7 = 0 gc sao cho
1
cos
26

.
Cu II.
2
2
2x
log 4 5
4 x
_

,
.
2. Gii phng trnh: ( ) 3sin 2x 2cos x 1 2 cos3x cos 2x 3cos x + + +
Cu III. Tnh tch phn:
( )
4
2
0
x 1
I dx
1 1 2x
+
+ +
.
Cu IV. Cho hnh chp S.ABC c y ABC l tam gic vung cn nh A, BC = 2a. Gi I l trung im ca
BC, hnh chiu vung gc H ca S ln mt y (ABC) tha mn
IA 2IH
uur uur
, gc gia SC v mt y (ABC)
bng 60
o
. Hy tnh th tch khi chp S.ABC v khong cch t trung im K ca SB ti (SAH).
Cu V. Cho x, y, z l ba s thc dng thay i v tha mn x
2
+ y
2
+ z
2
xyz. Hy tm gi tr ln nht ca
biu thc:
2 2 2
x y z
P
x yz y zx z xy
+ +
+ + +
.
Cu Via.
1. Trong mt phng Oxy, cho tam gic ABC bit A(3; 0), ng cao t nh B c phng trnh x + y + 1 = 0,
trung tuyn t nh C c phng trnh 2x y 2 = 0. Vit phng trnh ng trn ngoi tip tam gic ABC.
2. Trong khng gian Oxyz, cho cc im A(1; 1; 0), B(0; 0; 2) v C(1; 1; 1). Hy vit phng trnh mt
phng (P) qua hai im A v B, ng thi khong cch t C ti mt phng (P) bng 3 .
Cu VIIa. Cho khai trin
10 2 2 2 14
0 1 2 14
(1 2x) (x x 1) a a x a x ... a x + + + + + + + . Hy tm gi tr ca a
6
.
Cu Vib.
1. Trong mt phng Oxy, cho tam gic ABC bit A(1; 1), B(2; 1), din tch bng 5,5 v trng tm G thuc
ng thng d: 3x + y 4 = 0. Tm ta nh C.
2. Trong khng gian Oxyz, cho mt phng (P): x + y z + 1 = 0, ng thng d:
x 2 y 1 z 1
1 1 3

. Gi I l
giao im ca d v (P). Vit phng trnh ca ng thng d nm trong (P), vung gc vi d v cch
I mt khong bng
3 2
.
Cu VIIb (1 im) Gii phng trnh trn tp hp C:
3
z i
1
i z
+ _
,
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Mn thi: TON
Cu I (2 im) Cho hm s y = 2x
3
3(2m + 1)x
2
+ 6m(m + 1)x +1 c th (C
m
).
1. Kho st s bin thin v v th ca hm s khi m = 0.
2. Tm m hm s ng bin trn khong (2; +)
Cu II (2 im)
1. Gii phng trnh: 2cos3x(2cos2x + 1) = 1
2. Gii phng trnh:
2 2
3
(3x 1) 2x 1 5x x 3
2
+ +
3ln2
3 x 2
0
dx
I
( e 2)
Cu IV (1 im) Cho hnh lng tr ABC.ABC c y l tam gic u cnh a, hnh chiu vung gc ca A
ln mt phng (ABC) trng vi tm O ca tam gic ABC. Tnh th tch khi lng tr ABC.ABC bit
khong cch gia AA v BC l
a 3
4
.
Cu V (1 im) Cho x, y tho mn x
2
xy + y
2
= 1. Tm gi tr ln nht, nh nht ca biu thc
4 4
2 2
x y 1
P
x y 1
+ +
+ +
Cu VIa (2 im)
1. Cho hnh tam gic ABC c din tch bng 2. Bit A(1; 0), B(0; 2) v trung im I ca AC nm trn ng
thng y = x. Tm ta nh C.
2. Trong khng gian Oxyz, cho cc im A(1; 0; 0); B(0; 2; 0); C(0; 0; 2) tm ta im O i xng vi
O qua (ABC).
Cu VIIa (1 im) Gii phng trnh trn tp s phc:
2
(z z)(z 3)(z 2) 10 + +
Cu VIb (2 im)
1. Trong mp Oxy, cho im A(1; 0), B(1; 2) v ng thng (d): x y 1 = 0. Lp phng trnh ng trn
i qua 2 im A, B v tip xc vi ng thng (d).
2. Trong khng gian Oxyz, cho hai ng thng
1
x 4 y 1 z 5
d :
3 1 2
+

v
2
x 2 y 3 z
d :
1 3 1
+
. Vit
phng trnh mt cu c bn knh nh nht tip xc vi c hai ng thng d
1
v d
2
Cu VIIb (1 im) Gii bt phng trnh:
2 2
x(3log x 2) 9log x 2 >
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Mn thi: TON
Cu I: (2,0 im)
1. Kho st s bin thin v v th (C) ca hm s
3 2
1
y x 2x 3x.
3
+
2. Vit phng trnh tip tuyn ca th (C), bit tip tuyn ny i qua gc ta O.
Cu II: (2,0 im)
1. Gii phng trnh
2 sin 2x 3sin x cos x 2
4
_
+ + +

,
.
2. Gii h phng trnh
2 2
2
3
4xy 4(x y ) 7
(x y)
1
2x 3
x y
+ + +
'
.
Cu III: (1,0 im) Tm cc gi tr ca tham s m phng trnh
2
m x 2x 2 x 2 + +
c 2 nghim phn
bit.
Cu IV: (1,0 im) Cho hnh chp t gic u S.ABCD c tt c cc cnh u bng a. Tnh theo a th tch
khi chp S.ABCD v tnh bn knh mt cu nt tip hnh chp .
Cu V: (1,0 im) Vi mi s thc dng a; b; c tha mn iu kin a + b + c = 1. Tm gi tr nh nht ca
biu thc:
( ) ( ) ( )
3 3 3
2 2 2
a b c
P
1 a 1 b 1 c
+ +

Cu VIa: (2,0 im)
1. Trong mt phng Oxy, cho ng trn (C): (x 1)
2
+ (y + 1)
2
= 25 v M(7; 3). Lp phng trnh ng
thng (d) i qua M v ct (C) ti hai im A, B sao cho MA = 3MB.
2. Trong khng gian Oxyz, cho im I(1; 2; 3). Vit phng trnh mt cu tm I v tip xc vi trc Oy.
Cu VIIa: (1,0 im)
1. Gii phng trnh 2.27
x
+ 18
x
= 4.12
x
+ 3.8
x
.
2. Tm nguyn hm ca hm s
2
tan x
f (x)
1 cos x
+
.
Cu VIb: (2,0 im)
1. Trong mt phng Oxy, cho ng trn (C): x
2
+ y
2
+ 2x = 0. Vit phng trnh tip tuyn ca (C), bit gc
gia tip tuyn ny v trc tung bng 30
o
.
2. Cho hnh hp ch nht ABCD.A
1
B
1
C
1
D
1
c cc cnh AA
1
= a, AB = AD = 2a. Gi M, N, K ln lt l
trung im cc cnh AB,AD, AA
1
.
a) Tnh theo a khong cch t C
1
n mt phng (MNK).
b) Tnh theo a th tch ca t din C
1
MNK
Cu VIIb: (1,0 im)
1. Gii bt phng trnh
3
4 log x
x 243
+
> .
2. Tm m hm s
2
mx 1
y
x

c 2 im cc tr A, B v on AB ngn nht
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1. Vit phng trnh tng qut ca mt phng i qua im A v vung gc vi ng thng d.

2. Gii phng trnh: cosx = 8sin

Cu IV. Cho hnh nn c nh S, y l ng trn tm O, SA v SB l hai ng sinh, bit SO = 3, khong

PHN RING (3 im)

B. Theo chng trnh nng cao.

v gii bt phng trnh

Cu III. Tnh din tch ca min phng gii hn bi cc ng

2. Tm cc gi tr ca tham s thc m sao cho phng trnh sau c nghim.

Cu IV (1,0 im) Cho lng tr ng ABC.ABC c y l tam gic u. Mt phng ABC to vi y mt

2) Cho Elip c phng trnh chnh tc

2. Gii phng trnh:

v mt phng (P): x y z = 0. Tm ta hai im

2. Tnh cc gc ca tam gic ABC bit: 2A = 3B;

Cu IV. Cho lng tr tam gic ABC.A

. Lp phng trnh mt phng (P) i

Cu 3: Cho s thc b ln2. Tnh J =

. Tm cc im thuc (C) cch u 2 tim cn.

2. Gii phng trnh:

2. Gii phng trnh sau trn tp s phc C:

a. Chng minh rng (d

Cu 4: Cho hnh chp tam gic u S.ABC di cnh bn bng 1. Cc mt bn hp vi mt phng y mt

v hai im A(1; 2; 1), B(7; 2; 3). Tm trn (d) nhng

Cu III. (1.0 im)

Cu IV. (1.0 im)

Vit phng trnh chnh tc ca ng thng i qua im M, ct v vung gc vi ng thng d.

Cu III. (1,0 im) Tnh tch phn

Cu VI. (1,0 im)

2. Gii phng trnh: 2 2 sin(x ) cos x 1

Cu III (1 im) Tnh tch phn

Cu IV (1 im) Cho hnh chp S.ABC c gc ((SBC), (ACB)) = 60

Cu III (1 im) Tnh tch phn:

Cu IV (1 im) Cho hnh chp u S.ABCD c di cnh y bng a, mt bn to vi mt y gc 60

2. Gii h phng trnh:

Cu IV (1 im) Cho hnh chp t gic u S.ABCD c cnh bn bng a, mt bn hp vi y gc . Tm

Cu IV (1 im) Cho hnh chp S.ABC c AB = AC = a. BC = a/2. SA = a 3 ,

2. Gii phng trnh:

Cu IV (1 im) Tnh th tch hnh chp S.ABC bit SA = a, SB = b, SC = c, gc ASB = 60

v mt phng (P): 2x y 2z 2 = 0. Vit

1. Kho st s bin thin v v th (C) ca hm s cho.

Cu III (1 im) Tnh tch phn:

Gi K l hnh chiu vung gc ca im I(1; 1; 1) trn (d

a) Chng minh rng D

Cu III (1,0 im) Tnh tch phn: I =

v im A(0; 1). Tm im M thuc

a) Chng minh rng d

Cu III. Tm k h bt phng trnh sau c nghim:

a) Vit phng trnh mt phng cha

Cu III (1 im) Tnh tch phn:

v mt phng (P): x + y + z + 2 = 0. Lp phng

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