BỘ ĐỀ THI THỬ ĐẠI HỌC 2009 CHỌN LỌC (toanhoccapba.wordpress.com)
Transcript of BỘ ĐỀ THI THỬ ĐẠI HỌC 2009 CHỌN LỌC (toanhoccapba.wordpress.com)
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s 1Cu1: (2,5 im)
Cho hm s: y = -x3 + 3mx2 + 3(1 - m2)x + m3 - m21) Kho st s bin thin v v th ca hm s trn khi m = 1.2) Tm k phng trnh: -x3 + 3x2 + k3 - 3k2 = 0 c 3 nghim phn bit.
3) Vit phng trnh ng thng i qua 2 im cc tr ca th hm s trn.Cu2: (1,75 im)
Cho phng trnh: 01212323 =++ mxlogxlog (2)
1) Gii phng trnh (2) khi m = 2.
2) Tm m phng trnh (2) c t nht 1 nghim thuc on
331; .
Cu3: (2 im)
1) Tm nghim (0; 2) ca pt : 32221
335 +=
+++ xcos
xsinxsinxcos
xsin
2) Tnh din tch hnh phng gii hn bi cc ng: y = 342 + xx , y = x + 3
Cu4: (2 im)1) Cho hnh chp tam gic u S.ABC nh S c di cnh y bng a. Gi M
v N ln lt l trung im ca cc cnh SB v SC. Tnh theo a din tch AMN bitrng mt phng (AMN) vung gc mt phng (SBC).
2) Trong khng gian Oxyz cho 2 ng thng: 1:
=++=+
0422
042
zyx
zyx
v 2:
+=+=+=
tz
ty
tx
21
2
1
a) Vit phng trnh mt phng (P) cha ng thng 1 v song song vi ngthng 2.
b) Cho im M(2; 1; 4). Tm to im H thuc ng thng 2 sao cho onthng MH c di nh nht.Cu5: (1,75 im)
1) Trong mt phng vi h to cc vung gc Oxy xt ABC vung ti
A, phng trnh ng thng BC l: 033 = yx , cc nh A v B thuc trchonh v bn knh ng trn ni tip bng 2. Tm to trng tm G ca ABC
2 Khai trin nh thc:
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nxnn
nxxnn
xnx
n
nx
n
nxx
CC...CC
+
++
+
=
+
3
1
32
1
13
1
2
1
12
1
032
1
22222222
Bit rng trong khai trin 13 5 nn CC = v s hng th t bng 20n, tm n v x
s 2Cu1: (2 im)
Cu Cho hm s: y = mx4 + (m2 - 9)x2 + 10 (1)
1) Kho st s bin thin v v th ca hm s (1) khi m = 1.
2) Tm m hm s (1) c ba im cc tr.
Cu2: (3 im)
1) Gii phng trnh: sin23x - cos24x = sin25x - cos26x
2) Gii bt phng trnh: logx(log3(9x - 72)) 1
3) Gii h phng trnh:
++=+
=
2
3
yxyx
yxyx
Cu3: (1,25 im)
Tnh din tch hnh phng gii hn bi cc ng: y =x
yvx 2
2444
2
=
Cu4: (2,5 im)1) Trong mt phng vi h to cc vung gc Oxy cho hnh ch nht
ABCD c tm I
0
2
1; , phng trnh ng thng AB l x - 2y + 2 = 0 v AB = 2AD.
Tm to cc nh A, B, C, D bit rng nh A c honh m
2) Cho hnh lp phng ABCD.A1B1C1D1 c cnh bng a
a) Tnh theo a khong cch gia hai ng thng A1B v B1D.
b) Gi M, N, P ln lt l cc trung im ca cc cnh BB1, CD1, A1D1. Tnh gcgia hai ng thng MP v C1N.
Cu5: (1,25 im)
Cho a gic u A1A2...A2n (n 2, n Z) ni tip ng trn (O). Bit rng s
tam gic c cc nh l 3 im trong 2n im A1, A2, ... ,A2n nhiu gp 20 ln s hnh
ch nht c cc nh l 4 im trong 2n im A1, A2, ... ,A2n . Tm n.
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s 3Cu1: (3 im)
Cho hm s: y =( )
1
122
xmxm
(1) (m l tham s)
1) Kho st s bin thin v v th (C) ca hm s (1) ng vi m = -1.
2) Tnh din tch hnh phng gii hn bi ng cong (C) v hai trc to .
3) Tm m th ca hm s (1) tip xc vi ng thng y = x.
Cu2: (2 im)
1) Gii bt phng trnh: (x2 - 3x) 0232 2 xx .
2) Gii h phng trnh:
=+
+
=+
y
yy
x
xx
x
22
24
452
1
23
Cu3: (1 im)
Tm x [0;14] nghim ng phng trnh: cos3x - 4cos2x + 3cosx - 4 = 0 .Cu4: (2 im)
1) Cho hnh t din ABCD c cnh AD vung gc vi mt phng (ABC); AC =
AD = 4 cm ; AB = 3 cm; BC = 5 cm. Tnh khong cch t im A ti mt phng
(BCD).
2) Trong khng gian vi h to cc vung gc Oxyz, cho mt phng
(P): 2x - y + 2 = 0 v ng thng dm:
( ) ( )
( )
=++++
=+++02412
01112
mzmmx
mymxm
Xc nh m ng thng dm song song vi mt phng (P) .
Cu5: (2 im)
1) Tm s nguyn dng n sao cho: 243242 210 =++++ nnn
nnn C...CCC .
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2) Trong mt phng vi h to cc vung gc Oxy cho Elp (E) c
phng trnh: 1916
22
=+ yx . Xt im M chuyn ng trn tia Ox v im N chuyn
ng trn tia Oy sao cho ng thng MN lun tip xc vi (E). Xc nh to ca
M, N on MN c di nh nht. Tnh gi tr nh nht .
s 4
Cu1: (2 im)
Cho hm s: y =1
32
+
xx
1) Kho st s bin thin v v th hm s.
2) Tm trn ng thng y = 4 cc im m t k c ng 2 tip tuyn
n th hm s.Cu2: (2 im)
1) Gii h phng trnh:
=++
=++
0
123
yxyx
yxyx
2) Gii bt phng trnh: ( ) 012
1 2 >++ xxlnxln
Cu3: (2 im)
1) Gii phng trnh: cosx+ cos2x + cos3x + cos4x + cos5x = -2
1
2) Chng minh rng ABC tho mn iu kin
224
22
2
7 Bcos
Acos
CsinCcosBcosAcos ++=+ th ABC u
Cu4: (2 im)
1) Trn mt phng to cho A(1, 0); B(0, 2); O(0, 0) v ng trn (C) c
phng trnh: (x - 1)2 +2
21
y = 1. Vit phng trnh ng thng i qua cc giao
im ca ng thng (C) v ng trn ngoi tip OAB.
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2) Cho hnh chp S.ABC c y ABC l tam gic vung cn vi AB = AC = a,
SA = a, SA vung gc vi y. M l mt im trn cnh SB, N trn cnh SC sao cho
MN song song vi BC v AN vung gc vi CM. Tm t sMBMS
.
Cu5: (2 im)
1) Tnh din tch phn mt phng gii hn bi cc ng cong: y = x3 - 2 v
(y + 2)2 = x.
2) Vi cc ch s 1, 2, 3, 4, 5, 6 c th lp c bao nhiu s c 3 ch s khc
nhau, bit rng cc s ny chia ht cho 3.
s 5
Cu1: (2 im)
Cho hm s: y = x + 1 +1
1x
.
1) Kho st s bin thin v v th (C) hm s.
2) T mt im trn ng thng x = 1 vit phng trnh tip tuyn n th (C).
Cu2: (2 im)
1) Gii phng trnh: 1635223132 2 +++=+++ xxxxx
2) Tm cc gi tr x, y nguyn tho mn: ( ) yyxxlogy
3732
282
2
2
+++
+
Cu3: (2 im)
1) Gii phng trnh: (cos2x - 1)(sin2x + cosx + sinx) = sin22x
2) ABC c AD l phn gic trong ca gc A (D BC) v sinBsinC 2
2 Asin .
Hy chng minh AD2 BD.CD .
Cu4: (2 im)
1) Trn mt phng to vi h to cc vung gc Oxy, cho elip cphng trnh: 4x2 + 3y2 - 12 = 0. Tm im trn elip sao cho tip tuyn ca elip ti
im cng vi cc trc to to thnh tam gic c din tch nh nht.
2) Trong khng gian vi h trc to cc vung gc Oxyz, cho hai mt
phng (P): x - y + z + 5 = 0 v (Q): 2x + y + 2z + 1 = 0. Vit phng trnh mt cu c
tm thuc mt phng (P) v tip xc vi mt phng (Q) ti M(1; - 1; -1).
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Cu5: (2 im)
1) Tnh din tch hnh phng gii hn bi cc ng: y = 2 -4
2xv x + 2y = 0
2) a thc P(x) = (1 + x + x2)10 c vit li di dng: P(x) = a0 + a1x + ... +
a20x20. Tm h s a4 ca x4.
s 6
Cu1: (2 im)
Cho hm s: y =1
2
++
xmxmx (1) (m l tham s)
1) Kho st s bin thin v v th ca hm s (1) khi m = -1.
2) Tm m th hm s (1) ct trc honh ti hai im phn bit v hai im
c honh dng.
Cu2: (2 im)
1) Gii phng trnh: cotgx - 1 =tgx
xcos+1
2+ sin2x -
2
1sin2x
2) Gii h phng trnh:
+=
=
12
11
3xy
yy
xx
Cu3: (3 im)
1) Cho hnh lp phng ABCD.A'B'C'D'. Tnh s o ca gc phng nh din
[B, A'C, D].
2) Trong khng gian vi h to cc Oxyz cho hnh hp ch nht
ABCD.A'B'C'D' c A trng vi gc ca h to , B(a; 0; 0), D(0; a; 0), A'(0; 0; b)(a > 0, b > 0). Gi M l trung im cnh CC'.
a) Tnh th tch khi t din BDA'M theo a v b.
b) Xc nh t sba
hai mt phng (A'BD) v (MBD) vung gc vi nhau.
Cu4: (2 im)
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1) Tm h s ca s hng cha x8 trong khai trin nh thc Niutn ca:n
xx
+ 53
1, bit rng: ( )373
14 += +
++ nCC
nn
nn (n N*, x > 0)
2) Tnh tch phn: I = +
32
52
4xx
dx
Cu5: (1 im)
Cho x, y, z l ba s dng v x + y + z 1. Chng minh rng:
821112
2
2
2
2
2 +++++z
zy
yx
x
s 7Cu1: (2 im)
Cho hm s: y = x3 - 3x2 + m (1)1) Tm m th hm s (1) c hai im phn bit i xng vi nhau qua gc
to .2) Kho st s bin thin v v th ca hm s (1) khi m = 2 .
Cu2: (2 im)
1) Gii phng trnh: cotgx - tgx + 4sin2x =xsin2
2
2) Gii h phng trnh:
+=
+
=
2
2
2
2
23
2
3
y
xx
x
y
y
Cu3: (3 im)
1) Trong mt phng vi h ta cc vung gc Oxy cho ABC c: AB =
AC, = 900. Bit M(1; -1) l trung im cnh BC v G
0
3
2; l trng tm ABC.
Tm to cc nh A, B, C .2) Cho hnh lng tr ng ABCD.A'B'C'D' c y ABCD l mt hnh thoi cnh a,
gc = 600 . gi M l trung im cnh AA' v N l trung im cnh CC'. Chngminh rng bn im B', M, D, N cng thuc mt mt phng. Hy tnh di cnhAA' theo a t gic B'MDN l hnh vung.
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3) Trong khng gian vi h to cc Oxyz cho hai im A(2; 0; 0) B(0; 0; 8)
v im C sao cho ( )060 ;;AC = . Tnh khong cch t trung im I ca BC n ngthng OA.Cu4: (2 im)
1) Tm gi tr ln nht v nh nht ca hm s: y = x +2
4 x
2) Tnh tch phn: I =
+4
0
2
21
21dx
xsinxsin
Cu5: (1 im)Cho n l s nguyn dng. Tnh tng:
nn
n
nnn Cn
...CCC1
12
3
12
2
121
23
12
0
+++++
+
( knC l s t hp chp k ca n phn t) s 8
Cu1: (2 im)
1) Kho st s bin thin v v th ca hm s: y =2
422
+
xxx
(1)
2) Tm m ng thng dm: y = mx + 2 - 2m ct th ca hm s (1) ti hai
im phn bit.
Cu2: (2 im)1) Gii phng trnh: 0
242
222 =
xcosxtgxsin
2) Gii phng trnh: 32222
2 = + xxxx Cu3: (3 im)
1) Trong mt phng vi h ta trc cc vung gc Oxy cho ng trn:
(C): (x - 1)2 + (y - 2)2 = 4 v ng thng d: x - y - 1 = 0
Vit phng trnh ng trn (C') i xng vi ng trn (C) qua ng thng d.
Tm ta cc giao im ca (C) v (C').2) Trong khng gian vi h to cc vung gc Oxyz cho ng thng:
dk:
=++=++01
023
zykx
zkyx
Tm k ng thng dk vung gc vi mt phng (P): x - y - 2z + 5 = 0.
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3) Cho hai mt phng (P) v (Q) vung gc vi nhau, c giao tuyn l ng
thng . Trn ly hai im A, B vi AB = a. Trong mt phng (P) ly im C, trong
mt phng (Q) ly im D sao cho AC, BD cng vung gc vi v AC = BD = AB.Tnh bn knh mt cu ngoi tip t din ABCD v tnh khong cch t A n mt
phng (BCD) theo a.
Cu4: (2 im)
1) Tm gi tr ln nht v gi tr nh nht ca hm s: y =1
1
2 +
+
x
x
trn on [-1; 2]
2) Tnh tch phn: I = 2
0
2 dxxx
Cu5: (1 im)
Vi n l s nguyn dng, gi a3n - 3 l h s ca x3n - 3 trong khai trin thnh athc ca (x2 + 1)n(x + 2)n. Tm n a3n - 3 = 26n.
s 9
Cu1: (2 im)
Cho hm s: y =( )12
332
+
xxx
(1)
1) Kho st s bin thin v v th ca hm s (1).
2) Tm m ng thng y = m ct th hm s (1) ti hai im A, B sao cho
AB = 1.
Cu2: (2 im)
1) Gii bt phng trnh:( )
3
73
3
1622
>+
x
xx
x
x
2) Gii h phng trnh:
( )
=+
=
25
11
22
4
4
1
yxy
logxylog
Cu3: (3 im)
1) Trong mt phng vi h ta cac Oxy cho im A(0; 2) v B ( )13 ; .
Tm to trc tm v to tm ng trn ngoi tip OAB.
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2) Trong khng gian vi h to cc Oxyz cho hnh chp S.ABCD c y
ABCD l hnh thoi, AC ct BD ti gc to O. Bit A(2; 0; 0) B(0; 1; 0)
S(0; 0; 2 2 ). Gi M l trung im ca cnh SC.
a) Tnh gc v khong cch gia hai ng thng SA v BM.
b) Gi s mt phng (ABM) ct SD ti N. Tnh th tch hnh chp S.ABMN.Cu4: (2 im)
1) Tnh tch phn: I = +
2
1 11dx
x
x
2) Tm h s ca x8 trong khai trin thnh a thc ca: ( )[ ]82 11 xx + Cu5: (1 im)
Cho ABC khng t tho mn iu kin: cos2A + 2 2 cosB + 2 2 cosC = 3
Tnh cc gc ca ABC.
s 10Cu1: (2 im)
Cho hm s: y = xxx 323
1 23 + (1) c th (C)
1) Kho st s bin thin v v th ca hm s (1).
2) Vit phng trnh tip tuyn ca (C) ti im un v chng minh rng ltip tuyn ca (C) c h s gc nh nht.
Cu2: (2 im)
1) Gii phng trnh: 5sinx - 2 = 3(1 - sinx)tg2x
2) Tm gi tr ln nht v gi tr nh nht ca hm s: y =x
xln2trn on
[ ]31 e; .Cu3: (3 im)
1) Trong mt phng vi h ta cc Oxy cho im A(1; 1), B(4; -3). Tmim C thuc ng thng y = x - 2y - 1 = 0 sao cho khong cch t C n ngthng AB bng 6.
2) Cho hnh chp t gic u S.ABCD c cnh y bng a, gc gia cnh bn
v mt y bng (00 < < 900). Tnh tang ca gc gia hai mt phng (SAB) v(ABCD) theo a v .
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3) Trong khng gian vi h to cc Oxyz cho im A(-4; -2; 4) v ng
thng d:
+==
+=
tz
ty
tx
41
1
23
(t R). Vit phng trnh ng thng i qua im A, ct v
vung gc vi ng thng d.Cu4: (2 im)
1) Tnh tch phn I = +e
xdxlnx
xln
1
31
2) Trong mt mn hc, thy gio c 30 Cu hi khc nhau gm 5 Cu hi kh,
10 Cu hi trung bnh, 15 Cu hi d. T 30 Cu hi c th lp c bao nhiu kim tra, mi gm 5 Cu hi khc nhau, sao cho trong mi nht thit phi c 3 loi Cu hi (kh, d, trung bnh) v s Cu hi d khng t hn 2?
Cu5: (1 im)Xc nh m phng trnh sau c nghim:
224221112211 xxxxxm ++=
++
s 11
Cu1: (2 im)
Cho hm s y = x3 - 3mx2 + 9x + 1 (1) (m l tham s)
1) Kho st s bin thin v v th ca hm s (1) khi m = 2.2) Tm m im un ca th hm s (1) thuc ng thng y = x + 1.
Cu2: (2 im)
1) Gii phng trnh: ( )( ) xsinxsinxcosxsinxcos =+ 2212
2) Tm m h phng trnh sau:
=+
=+
myyxx
yx
31
1c nghim.
Cu3: (3 im)
1) Trong mt phng vi h ta cc Oxy cho ABC c cc nh A(-1; 0);
B(4; 0); C(0; m) vi m 0. Tm to trng tm G ca ABC theo m. Xc nh m
GAB vung ti G.
2) Trong khng gian vi h to cc Oxyz cho hnh lng tr ng
ABC.A1B1C1. Bit A(a; 0; 0); B(-a; 0; 0); C(0; 1; 0); B1(-a; 0; b) a > 0, b > 0.
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a) Tnh khong cch gia hai ng thng B1C v AC1 theo a, b.
b) Cho a, b thay i nhng lun tho mn a + b = 4. Tm a, b khong cch
gia 2 ng thng B1C v AC1 ln nht.
3) Trong khng gian vi h to cc Oxyz cho 3 im A(2; 0; 1) B(1; 0; 0)
C(1; 1; 1) v mt phng (P): x + y + x - 2 = 0. Vit phng trnh mt cu i qua 3im A, B, C v c tm thuc mt phng (P).
Cu4: (2 im)
1) Tnh tch phn I = ( ) 3
2
2 dxxxln
2) Tm cc s hng khng cha x trong khai trin nh thc Newtn ca
7
4
3 1
+ xx vi x > 0
Cu5: (1 im)
Chng minh rng phng trnh sau c ng 1 nghim: x5 - x2 - 2x - 1 = 0
s 12Cu1: (2 im)
Gi (Cm) l th ca hm s: y = mx +1
x(*) (m l tham s)
1. Kho st s bin thin v v th ca hm s (*) khi m = 14
2. Tm m hm s (*) c cc tr v khong cch t im cc tiu ca (Cm)n tim cn xin ca (Cm) bng
1
2
Cu2: (2 im)
1. Gii bt phng trnh: 5 1 1 2 4 x x x > 2. Gii phng trnh: cos23xcos2x - cos2x = 0
Cu3: (3 im)
1. Trong mt phng vi h to Oxy cho hai ng thngd1: x - y = 0 v d2: 2x + y - 1 = 0
Tm to cc nh ca hnh vung ABCD bit rng nh A thuc d1, nh Cthuc d2 v cc nh B, D thuc trc honh.
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2. Trong khng gian vi h to Oxyz cho ng thng d:1 3 3
1 2 1
x y z + = =
v mt phng (P): 2x + y - 2z + 9 = 0.
a. Tm to im I thuc d sao cho khong cch t I n mt phng(P) bng 2
b. Tm to giao im A ca ng thng d v mt phng (P). Vitphng trnh tham s ca ng thng nm trong mt phng (P),bit i qua A v vung gc vi d.
Cu4: (2 im)
1. Tnh tch phn I = 20
sin 2 sin
1 3cos
x xdx
x
++
2. Tm s nguyn dng n sao cho:( )1 2 2 3 3 4 2 12 1 2 1 2 1 2 1 2 12.2 3.2 4.2 ... 2 1 2 2005n nn n n n nC C C C n C 2 ++ + + + + + + + + =
Cu5: (1 im)
Cho x, y, z l cc s dng tho mn:1 1 1
4 x y z
+ + = . Chng minh rng:
1 1 11
2 2 2x y z x y z x y z+ +
+ + + + + +
s 13
Cu1: (2 im)
Gi (Cm) l th hm s y =( )2 1 1
1
x m x m
x
+ + + ++
(*) m l tham s
1. Kho st s bin thin v v th ca hm s (*) khi m = 1.2. Chng minh rng vi m bt k, th (Cm) lun lun c im cc i, cc
tiu v khong cch gia hai im bng 20
Cu2: (2 im)
1. Gii h phng trnh: ( )2 39 31 2 1
3log 9 log 3
x y
x y
+ = =
2. Gii phng trnh: 1 + sinx + cosx + sin2x + cos2x = 0Cu3: (3 im)
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1. Trong mt phng vi h to Oxy cho A(2; 0) v B(6; 4). Vit phngtrnh ng trn (C) tip xc vi trc honh ti hai im v khong cch t
tm ca (C) n im B bng 5.
2. Trong khng gian vi h to Oxyz cho hnh lng tr ng ABC.A1B1C1vi A(0; -3; 0) B(4; 0; 0) C(0; 3; 0) B1(4; 0; 4)
a. Tm to cc nh A1, C1. Vit phng trnh mt cu c tm l A vtip xc vi mt phng (BCC1B1).
b. Gi M l trung im ca A1B1. Vit phng trnh mt phng P) i quahai im A, M v song song vi BC1. mt phng (P) ct ng thng
A1C1 ti im N. Tnh di on MN
Cu4: (2 im)
1. Tnh tch phn: I = 20
sin 2 cos
1 cos
x xdx
x
+
2. Mt i thanh nin tnh nguyn c 15 ngi, gm 12 nam v 3 n. Hi cbao nhiu cch phn cng i thanh nin tnh nguyn v gip 3 tnh
min ni, sao cho mi tnh c 4 nam v 1 n?
Cu5: (2 im)
Chng minh rng vi mi x thuc R ta c:
12 15 203 4 5
5 4 3
x x x
x x x + + + +
Khi no ng thc xy ra?
s 14Cu1: (2 im)
Gi (Cm) l th hm s: y =3 21 1
3 2 3
mx x + (*) (m l tham s)
1. Kho st s bin thin v v th ca hm s (*) khi m = 22. Gi M l im thuc (Cm) c honh bng -1. Tm m tip tuyn ca
(Cm) ti im M song song vi ng thng 5x - y = 0Cu2: (2 im)
Gii cc phng trnh sau:
1. 2 2 2 1 1 4 x x x+ + + + = 2. 4 4 3cos sin cos sin 3 0
4 4 2 x x x x
+ + =
Cu3: (3 im)
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1. Trong mt phng vi h to Oxy cho im C(2; 0) v Elip (E):2 2
14 1
x y+ = . Tm to cc im A, B thuc (E), bit rng A, B i xng
vi nhau qua trc honh va ABC l tam gic u.2. Trong khng gian vi h to Oxyz cho hai ng thng:
d1:1 2 1
3 1 2
x y z + += =
v d2:2 0
3 12 0
x y z
x y
+ = + =
a. Chng minh rng: d1 v d2 song song vi nhau. Vit phng trnh mtphng (P) cha c hai ng thng d1 v d2
b. mt phng to Oxz ct hai ng thng d1, d2 ln lt ti cc imA, B. Tnh din tch OAB (O l gc to )
Cu4: (2 im)
1. Tnh tch phn: I = ( )2
sin
0
cos cosxe x xdx
+
2. Tnh gi tr ca biu thc M =( )
4 3
13
1 !
n nA A
n
+ ++
bit rng
2 2 2 2
1 2 3 42 2 149n n n nC C C C + + + ++ + + =
Cu5: (1 im)Cho cc s nguyn dng x, y, z tho mn xyz = 1. Chng minh rng:
3 3 3 3 3 31 1 13 3
x y y z z x
xy yz zx
+ + + + + ++ +
Khi no ng thc xy ra?
s 15Phn chung c tt c cc th sinh
Cu1: (2 im)1. Kho st s bin thin v v th ca hm s: y = 2x3 - 9x2 + 12x - 42. Tm m phng trnh sau c 6 nghim phn bit: 3 22 9 12 x x x m + =
Cu2: (2 im)
1. Gii phng trnh: ( )6 62 sin sin .cos 02 2sin
cos x x x x
x
+ =
2. Gii h phng trnh: 31 1 4
xy xy
x y
=
+ + + =
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Cu3: ( 2 im) Trong khng gian vi h to Oxyz. Cho hnh lp phngABCD.ABCD vi A(0; 0; 0) B(1; 0; 0) D(0; 1; 0) A(0; 0; 1). Gi M v N ln ltl trung im ca AB v CD.
1. Tnh khong cch gia hai ng thng AC v MN.2. Vit phng trnh mt phng cha AC v to vi mt phng Oxy mt gc
bit cos = 16
Cu4: (2 im)
1. Tnh tch phn: I = 22 2
0
sin2
cos 4sin
xdx
x x
+
2. Cho hai s thc x 0, y 0 thay i v iu kin: (x + y)xy = x2 + y2 - xy.Tm GTLN ca biu thc A =
3 3
1 1
x y+
Phn T chn: Th sinh chn Cu 5.a hc Cu 5.bCu5a:Theo chng trnh khng phn ban: (2 im)
1. Trong mt phng vi h to Oxy cho cc ng thng:d1: x + y + 3 = 0 d2: x - y - 4 = 0 d3: x - 2y = 0.Tm to im M nm trn ng thng d3 sao cho khong cch t M n
ng thng d1 bng hai ln khong cch t M n ng thng d2
2. Tm h s ca s hng cha x26 trong khai trin nh thc: 74
1n
xx
+
, bit
rng: 1 2 02 1 2 1 2 1... 2 1n
n n nC C C2
+ + ++ + + = Cu5b:Theo chng trnh phn ban: (2 im)
1. Gii phng trnh: 3.8x + 4.12x - 18x - 2.27x = 02. Cho hnh lng tr c cc y l hai hnh trn tm O v O, bn knh bng
chiu cao v bng a. Trn ng trn y tm O ly im A, trn ng trn y tmO ly im B sao cho AB = 2a. Tnh th tch ca khi t din OOAB.
s 16Phn chung c tt c cc th sinh
Cu1: (2 im)
Cho hm s: y =
21
2
x x
x
+ +
1. Kho st s bin thin v v th (C) ca hm s.2. Vit phng trnh tip tuyn ca th (C), bit tip tuyn vung gc vi
tim cn xin ca (C).Cu2: (2 im)
1. Gii phng trnh: cotx + sinx 1 tan .tan 42
xx
+ =
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2. Tm m phng trnh sau c hai nghim thc phn bit:2
2 2 1 x mx x+ + = Cu3: (2 im)
Trong khng gian vi h to Oxyz cho im A(0; 1; 2) v hai ng thng :
d1:1 1
2 1 1
x y z += = d2:
1
1 2
2
x t
y t
z t
= +
= = +
1. Vit phng trnh mt phng (P) qua A, ng thi song song vi d1 v d2.2. Tm to cc im M d1, N d2 sao cho ba im A, M, N thng hng
Cu4: (2 im)
1. Tnh tch phn: I = ln5ln3
2 3x x
dx
e e+
2. Cho x, y l cc s thc thay i. Tm GTNN ca biu thc:A = ( ) ( )
2 22 2
1 1 2 x y x y y + + + + + Phn T chn: Th sinh chn Cu 5.a hc Cu 5.b
Cu5a:Theo chng trnh khng phn ban: (2 im)1. Trong mt phng vi h to Oxy cho ng trn (C): x2 + y2 -2x - 6y + 6
= 0 v im M(-3; 1). Gi T1 v T2 l cc tip im ca cc tip tuyn k t M n(C). Vit phng trnh ng thng T1T2
2. Cho tp hp A gm n phn t (n 4). Bit rng s tp con gm 4 phn t caA bng 20 ln s tp con gm 2 phn t ca A. Tm k {1, 2,..., n} sao cho s tpcon gm k phn t ca A l ln nht.Cu5b:Theo chng trnh phn ban: (2 im)
1. Gii bt phng trnh: ( ) ( )25 5 5log 4 144 4log 2 1 log 2 1x x+ < + +
2. Cho hnh chp S.ABCD c y ABCD l hnh ch nht vi AB = a, AD =
a 2 , SA = a v SA vung gc vi mt phng (ABCD). Gi M v N ln lt l trungim ca AD v SC; I l giao im ca BM v AC. Chng minh rng: mt phng(SAC) vung gc vi mt phng (SMB). Tnh th tch ca khi t din ANIB
s 17Phn chung c tt c cc th sinh
Cu1: (2 im)
Cho hm s y = x3 - 3x + 21. Kho st s bin thin v v th (C) ca hm s cho.2. Gi d l ng thng i qua im A(3; 2) v c h s gc l m. Tm m
ng thng d ct th (C) ti ba im phn bit.Cu2: (2 im)
1. Gii phng trnh: cos3x + cos2x - cosx - 1 = 0
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2. Gii phng trnh: 22 1 3 1 0 x x x + + = (x R)Cu3: (2 im)
Trong khng gian vi h to Oxyz, cho im A(1; 2; 3) v hai ng thng
d1:2 2 3
2 1 1
x y z + = =
d2:1 1 1
1 2 1
x y z += =
1. Tm to im A i xng vi im A qua ng thng d12. Vit phng trnh ng thng i qua A vung gc vi d1 v ct d2
Cu4: (2 im)
1. Tnh tch phn: I = ( )1 20
2x
x e dx
2. Chng minh rng: vi mi a > 0, h phng trnh sau c nghim duy nht:( ) ( )ln 1 ln 1x ye e x y
y x a
= + +
=
Phn T chn: Th sinh chn Cu 5.a hc Cu 5.b
Cu5a:Theo chng trnh khng phn ban: (2 im)1. Trong mt phng vi h to Oxy cho ng trn (C): x2 + y2 - 2x - 2y + 1
= 0 v ng thng d: x - y + 3 = 0. Tm to im M nm trn d sao cho ngtrn tm M, c bn knh gp i bn knh ng trn (C) tip xc ngoi vi ngtrn (C)
2. i thanh nin xung kch ca mt trng ph thng c 12 hc sinh, gm 5hc sinh lp A, 4 hc sinh lp B v 3 hc sinh lp C. Cn chn 4 hc sinh i lm
nhim v, sao cho 4 hc sinh ny thuc khng qu 2 trong 3 lp trn. Hi c baonhiu cch chn nh vy?Cu5b:Theo chng trnh phn ban: (2 im)
1. Gii phng trnh:2 2 22 4.2 2 4 0 x x x x x+ + =
2. Cho hnh chp S.ABC c y ABC l tam gic u cnh a, SA = 2a v SAvung gc vi mt phng (ABC). Gi M v N ln lt l hnh chiu vung gc ca Atrn cc ng thng SB v SC. Tnh th tch ca khi chp A.BCNM
s 18
Phn chung c tt c cc th sinhCu1: (2 im)
Cho hm s: y =( )2 22 1 4
2
x m x m m
x
+ + + ++
(1) m l tham s
1. Kho st s bin thin v v th ca hm s (1) khi m = -1.2. Tm m hm s (1) c cc i v cc tiu, ng thi cc im cc tr ca
th cng vi gc to to thnh mt tam gic vung ti O
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Cu2: (2 im)
1. Gii phng trnh: ( ) ( )2 21 sin cos 1 cos sin 1 sin 2 x x x x x+ + + = + 2. Tm m phng trnh sau c nghim thc: 243 1 1 2 1 x m x x + + =
Cu3: (2 im)Trong khng gian vi h to Oxyz cho hai ng thng
d1:1 2
2 1 1
x y z += =
v d2:
1 2
1
3
x t
y t
z
= + = + =
1. Chng minh rng: d1 v d2 cho nhau.2. Vit phng trnh ng thng d vung gc vi mt phng (P): 7x + y - 4z =
0 v ct hai ng thng d1, d2Cu4: (2 im)
1. Tnh din tch hnh phng gii hn bi cc ng: y = (e + 1)x, y = (1 + ex)x2. Cho x, y, z l cc s thc dng thay i v tho mn iu kin: xyz = 1.
Tm GTNN ca biu thc: P =( ) ( ) ( )2 2 2
2 2 2
x y z y z x z x y
y y z z z z x x x x y y
+ + ++ +
+ + +
Phn T chn: Th sinh chn Cu 5.a hc Cu 5.b
Cu5a:Theo chng trnh khng phn ban: (2 im)
1. Trong mt phng vi h to Oxy cho ABC c A(0; 2) B(-2 -2) vC(4; -2). Gi H l chn ng cao k t B; M v N ln lt l trung im ca cc cnhAB v BC. Vit phng trnh ng trn i qua cc im H, M, N
2. Chng minh rng:
21 3 5 2 1
2 2 2 2
1 1 1 1 2 1
...2 4 6 2 2 1
nn
n n n nC C C C n n
+ + + + = + Cu5b:Theo chng trnh phn ban: (2 im)
1. Gii bt phng trnh: ( ) ( )3 13
2log 4 3 log 2 3 2x x + +
2. Cho hnh chp S.ABCD c y l hnh vung cnh a, mt bn SAD l tamgic u v nm trong mt phng vung gc vi y. Gi M, N, P ln lt l trungim ca cc cnh SB, BC, CD. Chng minh AM vung gc vi BP v tnh th tchca khi t din CMNP.
s 19Phn chung c tt c cc th sinh
Cu1: (2 im)Cho hm s: y = -x3 + 3x2 + 3(m2 -1)x - 3m2 - 1 (1) m l tham s1. Kho st s bin thin v v th ca hm s (1) khi m = 12. Tm m hm s (1) c cc i, cc tiu v cc im cc tr ca th hm
s (1) cch u gc to O.Cu2: (2 im)
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1. Gii phng trnh: 2sin22x + sin7x - 1 = sinx2. Chng minh rng vi mi gi tr dng ca tham s m, phng trnh sau c
hai nghim thc phn bit: x2 + 2x - 8 = ( )2m x
Cu3: (2 im)Trong khng gian vi h to Oxyz cho mt cu (S): x2 + y2 + z2 - 2x + 4y +
2z - 3 = 0 v mt phng (P): 2x - y + 2z - 14 = 01. Vit phng trnh mt phng (Q) cha trc Ox v ct (S) theo mt ng
trn c bn knh bng 3.2. Tm to im M thuc mt cu (S) sao cho khong cch t M n mt
phng (P) ln nhtCu4: (2 im)
1. Cho hnh phng H gii hn bi cc ng: y = xlnx, y = 0, x = e. Tnh thtch ca khi trn xoay to thnh khi quay hnh H quanh trc Ox.
2. Cho x, y, z l ba s thc dng thay i. Tm gi tr nh nht ca biu thc:P =
1 1 1
2 2 2
x y z x y z
yz zx xy
+ + + + +
Phn T chn: Th sinh chn Cu 5.a hc Cu 5.b
Cu5a:Theo chng trnh khng phn ban: (2 im)1. Tm h s ca s hng cha x10 trong khai trin nh thc ca (2 + x)n bit
( )0 1 1 2 2 3 33 3 3 3 ... 1 2048nn n n n n
n n n n nC C C C C + + + =
2. Trong mt phng vi h to Oxy cho im A(2; 2) v cc ng thng:d1: x + y - 2 = 0 d2: x + y - 8 = 0
Tm to cc im B v C ln lt thuc d1 v d2 sao cho ABC vung cn tiA.Cu5b:Theo chng trnh phn ban: (2 im)
1. Gii phng trnh: ( ) ( )2 1 2 1 2 2 0x x
+ =
2. Cho hnh chp t gic u S.ABCD c y l hnh vung cnh a. Gi E lim i xng ca D qua trung im ca SA, M l trung im ca AE, N l trungim ca BC. Chng minh MN vung gc vi BD v tnh theo a khong cch gia haing thng MN v AC.
s 20Phn chung c tt c cc th sinh
Cu1: (2 im) Cho hm s: y =2
1
x
x +
1. Kho st s bin thin v v th (C) ca hm s cho.2. Tm to im M thuc (C), bit tip tuyn ca (C) ti M ct hai trc Ox,
Oy ti A, B v tam gic OAB c din tch bng1
4
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Cu2: (2 im)
1. Gii phng trnh: 2sin cos 3 cos 22 2
x xx
+ + =
2. Tm gi tr ca tham s m h phng trnh sau c nghim thc:
3 3
3 3
1 15
1 115 10
x y
x y
x y mx y
+ + + = + + + =
Cu3: (2 im)Trong khng gian vi h to Oxyz cho hai im A(1; 4; 2 B(-1 2; 4) v
ng thng :1 2
1 1 2
x y z += =
1. Vit phng trnh ng thng d i qua trng tm G ca tam gic OAB vvung gc vi mt phng (OAB).
2. Tm to im M thuc ng thng sao cho MA2 + MB2- nh nhtCu4: (2 im)
1. Tnh tch phn: I = 3 21
ln
e
x xdx
2. Cho a b > 0. Chng minh rng: 1 12 22 2
b a
a b
a b
+ +
Phn T chn: Th sinh chn Cu 5.a hc Cu 5.bCu5a:Theo chng trnh khng phn ban: (2 im)
1. Tm h s ca x5 trong khai trin thnh a thc ca: x(1 - 2x)5 + x2(1 + 3x)102. Trong mt phng vi h to Oxy cho ng trn (C): (x - 1)2 + (y + 2)2 =
9 v ng thng d: 3x - 4y + m = 0.
Tm m trn d c duy nht mt im P m t c th k c hai tip tuynPA, PB ti (C) (A, B l cc tip im) sao cho PAB uCu5b:Theo chng trnh phn ban: (2 im)
1. Gii phng trnh: ( )2 2 1log 4 15.2 27 2log 04.2 3
x x
x+ + + =
2. Cho hnh chp S.ABCD c y l hnh thang, ABC = BAD = 900 , BA =BC = a, AD = 2a. cnh bn SA vung gc vi y v SA = a 2 . Gi H lhnh chiu vung gc ca A trn SB. Chng minh tam gic SCD vung vtnh theo a khong cch t H n mt phng (SCD)
s 21Cu1: (2 im)
Cho hm s: y = x4 - mx2 + m - 1 (1) (m l tham s)
1) Kho st s bin thin v v th ca hm s (1) khi m = 8.2) Xc nh m sao cho th ca hm s (1) ct trc honh ti bn im phn bit.
Cu2: (2 im)
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1) Gii bt phng trnh: ( ) ( )xxx 2.32log44log 122
1
2
1 ++
2) Xc nh m phng trnh: ( ) 02sin24coscossin4 44 =+++ mxxxx c t
nht mt nghim thuc on
2
;0
Cu3: (2 im)
1) Cho hnh chp S.ABC c y ABC l tam gic u cnh a v cnh bn SA vunggc vi mt phng y (ABC). Tnh khong cch t im A ti mt phng (SBC) theo
a, bit rng SA =2
6a
2) Tnh tch phn: I = +
1
02
3
1x
dxx
Cu4: (2 im)Trong mt phng vi h to cc vung gc Oxy, cho hai ng trn:
(C1): x2 + y2 - 10x = 0, (C2): x
2 + y2 + 4x - 2y - 20 = 01) Vit phng trnh ng trn i qua cc giao im ca (C1), (C2) v c tm nm
trn ng thng x + 6y - 6 = 0.2) Vit phng trnh tip tuyn chung ca cc ng trn (C1) v (C2).
Cu5: (2 im)
1) Gii phng trnh: 16212244 2 +=++ xxxx
2) i tuyn hc sinh gii ca mt trng gm 18 em, trong c 7 hc sinh khi12, 6 hc sinh khi 11 v 5 hc sinh khi 10. Hi c bao nhiu cch c 8 hc sinhtrong i i d tri h sao cho mi khi c t nht mt em c chn.
Cu6: ( Tham kho)
Gi x, y, z l khong cch t im M thuc min trong ca ABC c 3 gc nhn n
cc cnh BC, CA, AB. Chng minh rng:R
cbazyx
2
222 ++++ ; a, b, c l
ba cnh ca , R l bn knh ng trn ngoi tip. Du "=" xy ra khi no?
s 22
Cu1: (2 im)
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1) Tm s n nguyn dng tho mn bt phng trnh: nCA nnn 9223 + , trong
knA v
knC ln lt l s chnh hp v s t hp chp k ca n phn t.
2) Gii phng trnh: ( ) ( ) ( )xxx 4log1log4
13log
2
12
842 =++
Cu2: (2,5 im)
Cho hm s: y =2
22
+
x
mxx(1) (m l tham s)
1) Xc nh m hm s (1) nghch bin trn on [-1; 0].
2) Kho st s bin thin v v th ca hm s (1) khi m = 1.
3) Tm a phng trnh sau c nghim:
( ) 01232922
1111 =+++ ++ aa tt
Cu3: (1,5 im)
1) Gii phng trnh:x
xgx
xx
2sin8
12cot
2
1
2sin5
cossin44
=+
2) Xt ABC c di cc cnh AB = c; BC = a; CA = b. Tnh din tch ABC, bit
rng: bsinC(b.cosC + c.cosB) = 20
Cu4: (3 im)
1) Cho t din OABC c ba cnh OA; OB v OC i mt vung gc. Gi ; ; ln
lt l cc gc gia mt phng (ABC) vi cc mt phng (OBC); (OCA) v (OAB).
Chng minh rng: 3coscoscos ++ .
2) Trong khng gian vi h to cc Oxyz cho mt phng (P): x- y + z + 3 = 0
v hai im A(-1; -3; -2), B(-5; 7; 12).
a) Tm to im A' l im i xng vi im A qua mt phng (P).
b) Gi s M l mt im chy trn mt phng (P), tm gi tr nh nht ca biu thc:
MA + MB.
Cu5: (1,0 im)
Tnh tch phn: I =
( )
+
3ln
03
1x
x
e
dxe
s 23
Cu1: (3,0 im)
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Cho hm s: y =3
122
3
1 23 + mxmxx (1) (m l tham s)
1) Cho m =2
1
a) Kho st s bin thin v v th (C) ca hm s (1)
b) Vit phng trnh tip tuyn ca th (C), bit rng tip tuyn song song ving thng d: y = 4x + 2.
2) Tm m thuc khong
6
5;0 sao cho hnh phng gii hn bi th ca hm s (1)
v cc ng x = 0, x = 2, y = 0 c din tch bng 4.Cu2: (2 im)
1) Gii h phng trnh:
=
=+
0loglog
034
24 yx
yx
2) Gii phng trnh:x
xxxtg4
24
cos
3sin2sin21 =+
Cu3: (2 im)1) Cho hnh chp S.ABCD c y ABCD l hnh vung cnh a, SA vung gc vi
mt phng (ABCD) v SA = a. Gi E l trung im ca cnh CD. Tnh theo a khongcch t im S n ng thng BE.2) Trong khng gian vi h to cc Oxyz cho ng thng
:
=+++
=+++
02
012
zyx
zyxv mt phng (P): 4x - 2y + z - 1 = 0
Vit phng trnh hnh chiu vung gc ca ng thng trn mt phng (P).Cu4: (2 im)
1) Tm gii hn: L =x
xx
x
3
0
11lim
++
2) Trong mt phng vi h ta cac Oxy cho hai ng trn:(C1): x
2 + y2 - 4y - 5 = 0 v (C2): x2 + y2 - 6x + 8y + 16 = 0
Vit phng trnh cc tip tuyn chung hai ng trn (C1) v (C2)
Cu5: (1 im)Gi s x, y l hai s dng thay i tho mn iu kin x + y =
4
5. Tm gi tr nh
nht ca biu thc: S =yx 4
14 +
s 24
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Cu1: (2 im)
1) Gii bt phng trnh: 12312 +++ xxx
2) Gii phng trnh: tgx + cosx - cos2x = sinx(1 + tgxtg2
x)
Cu2: (2 im)Cho hm s: y = (x - m)3 - 3x (m l tham s)
1) Xc nh m hm s cho t cc tiu ti im c honh x = 0.
2) Kho st s bin thin v v th ca hm s cho khi m = 1.
3) Tm k h bt phng trnh sau c nghim:( )
+
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Cu1: (2 im)
Cho hm s: y =x
mxx
+
1
2
(1) (m l tham s)
1) Kho st s bin thin v v th ca hm s (1) khi m = 0.2) Tm m hm s (1) c cc i v cc tiu. Vi gi tr no ca m th khong cch
gia hai im cc tr ca th hm s (1) bng 10.Cu2: (2 im)
1) Gii phng trnh: 0log3log16 2327 3 = xx xx
2) Cho phng trnh: axx
xx =+++
3cos2sin
1cossin2(2) (a l tham s)
a) Gii phng trnh (2) khi a =3
1.
b) Tm a phng trnh (2) c nghim.Cu3: (3 im)1) Trong mt phng vi h ta cac Oxy cho ng thng d: x - y + 1 = 0 v
ng trn (C): x2 + y2 + 2x - 4y = 0. Tm to im M thuc ng thng d m qua ta k c hai ng thng tip xc vi ng trn (C) ti A v B sao cho gcAMB bng 600.2) Trong khng gian vi h to cc Oxyz cho ng thng
d:
=+=+
0422
0122
zyx
zyxv mt cu (S): x2 + y2 + z2 + 4x - 6y + m = 0.
Tm m ng thng d ct mt cu (S) ti hai im M, N sao cho khong cch giahai im bng 9.3) Tnh th tch khi t din ABCD, bit AB = a; AC = b; AD = c v cc gc BAC;
CAD; DAB u bng 600Cu4: (2 im)
1) Tnh tch phn: I = 2
0
56 3cossincos1
xdxxx
2) Tm gii hn:x
xx
x cos1
1213lim
23 2
0 ++
Cu5: (1 im)Gi s a, b, c, d l bn s nguyn thay i tho mn 1 a < b < c < d 50. Chng
minh bt ng thc:b
bb
d
c
b
a
50
502 +++ v tm gi tr nh nht ca biu thc:
S =d
c
d
a +
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s 26
Cu1: (2 im)
1) Kho st s bin thin v v th ca hm s: y = xxx 323
1 23 +
2) Tnh din tch hnh phng gii hn bi th hm s (1) v trc honh.Cu2: (2 im)
1) Gii phng trnh: xx
sincos8
12
=
2) Gii h phng trnh:( )( )
=+
=+
3532log
3532log
23
23
xyyy
yxxx
y
x
Cu3: (2 im)
1) Cho hnh t din u ABCD, cnh a = 6 2 cm. Hy xc nh v tnh di on
vung gc chung ca hai ng thng AD v BC.
2) Trong mt phng vi h ta cac Oxy cho elip (E): 149
22
=+ yx v ng
thng dm: mx - y - 1 = 0.
a) Chng minh rng vi mi gi tr ca m, ng thng dm lun ct elp (E) ti hai
im phn bit.
b) Vit phng trnh tip tuyn ca (E), bit rng tip tuyn i qua im N(1; -3)
Cu4: (1 im)
Gi a1, a2, ..., a11 l h s trong khai trin sau:
( ) ( ) 119
210
11110
...21 axaxaxxx ++++=++
Hy tnh h s a5
Cu5: (2 im)
1) Tm gii hn: L = ( )2
6
1 1
56
lim +
x
xx
x
2) Cho ABC c din tch bng2
3. Gi a, b, c ln lt l di ca cc cnh BC,
CA, AB v ha, hb, hc tng ng l di cc ng cao k t cc nh A, B, C ca tam
gic. Chng minh rng: 3111111
++
++
cba hhhcba
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s 27
Cu1: (2 im)
1) Kho st s bin thin v v th ca hm s y =( )12
3422
x
xx
2) Tm m phng trnh: 2x2 - 4x - 3 + 2m 1x = 0 c hai nghim phn bit.
Cu2: (2 im)
1) Gii phng trnh: ( ) 0623 =++ xcosxsintgxtgx
2) Gii h phng trnh:
=+
=
322yx
xy ylogxylog
Cu3: (3 im)
1) Trong mt phng vi h ta cc Oxy cho parabol (P) c phng trnh y2 = x
v im I(0; 2). Tm to hai im M, N thuc (P) sao cho INIM 4= .
2) Trong khng gian vi h to cc Oxyz cho t din ABCD vi A(2; 3; 2),
B(6; -1; -2), C(-1; -4; 3), D(1; 6; -5). Tnh gc gia hai ng thng AB v CD. Tm
to im M thuc ng thng CD sao cho ABM c chu vi nh nht.
3) Cho lng tr ng ABC. A'B'C' c y ABC l tam gic cn vi AB = AC = a v
gc BAC = 1200
, cnh bn BB' = a. Gi I l trung im CC'. Chng minh rng AB'Ivung A. Tnh cosin ca gc gia hai mt phng (ABC) v (AB'I).
Cu4: (2 im)
1) C bao nhiu s t nhin chia ht cho 5 m mi s c 4 ch s khc nhau?
2) Tnh tch phn: I = +
4
02cos1
dxx
x
Cu5: (1 im)Tm gi tr ln nht v gi tr nh nht ca hm s: y = sin5x + 3 cosx
]
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s 28
Cu1: (2 im)
Cho hm s: y =( )
( )mx
mmxmx
+
+++++
2
41222
(1) (m l tham s)
1) Tm m hm s (1) c cc tr v tnh khong cch gia hai im cc tr ca
th hm s (1).
2) Kho st s bin thin v v th ca hm s (1) khi m = 0
Cu2: (2 im)
1) Gii phng trnh: cos2x + cosx(2tg2x - 1) = 2
2) Gii bt phng trnh: 11 21212.15 ++ ++ xxx
Cu3: (3 im)1) Cho t din ABCD vi AB = AC = a, BC = b. Hai mt phng (BCD) v (ABC)
vung gc vi nhau v gc BDC = 900. Xc nh tm v tnh bn knh mt cu ngoi
tip t din ABCD thao a v b.
2) Trong khng gian vi h to cc Oxyz cho hai ng thng:
d1:12
1
1
zyx =+= v d2:
=+=+
012
013
yx
zx
a) Chng minh rng d1, d2 cho nhau v vung gc vi nhau.
b) Vit phng trnh tng qut ca ng thng d ct c hai ng thng d1, d2 v
song song vi ng thng :2
3
4
7
1
4
== zyx
Cu4: (2 im)
1) 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?
2) Tnh tch phn: I = 1
0
231 dxxx
Cu5: (1 im)
Tnh cc gc ca ABC bit rng:( )
=
8
332
2sin
2sin
2sin
4
CBA
bcapp
trong BC = a, CA = b, AB = c, p =2
cba ++
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s 29
Cu1: (2 im)
Cho hm s: y = (x - 1)(x2
+ mx + m) (1) (m l tham s)1) Tm m th hm s (1) ct trc honh ti ba im phn bit.
2) Kho st s bin thin v v th ca hm s (1) khi m = 4.
Cu2: (2 im)
1) Gii phng trnh: 032943 26 =++ xcosxcosxcos
2) Tm m phng trnh: ( ) 042
12
2 =+ mxlogxlog c nghim thuc
khong (0; 1).
Cu3: (3 im)
1) Trong mt phng vi h ta cc Oxy cho ng thng d: x - 7y + 10 = 0.
Vit phng trnh ng trn c tm thuc ng thng : 2x + y = 0 v tip xc vi
ng thng d ti im A(4; 2).
2) Cho hnh lp phng ABCD.A'B'C'D'. Tm im M thuc cnh AA' sao cho
mt phng (BD'M) ct hnh lp phng theo mt thit din c din tch nh nht.
3) Trong khng gian vi h to cc Oxyz cho t din OABC vi A(0; 0;
3a ), B(0; 0; 0), C(0; a 3 ; 0) (a > 0). Gi M l trung im ca BC. Tnh khong
cch gia hai ng thng AB v OM.
Cu4: (2 im)
1) Tm gi tr ln nht v gi tr nh nht ca hm s: y = x6 + ( )3214 x trnon [-1; 1].
2) Tnh tch phn: I =
5
2
2
1
ln
lnx
x
e
dxe
Cu5: (1 im)
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T cc ch s 1, 2, 3, 4, 5, 6 c th lp c bao nhiu s t nhin, mi s c 6
ch s v tho mn iu kin: Su ch s ca mi s l khc nhau v trong mi s
tng ca ba ch s u nh hn tng ca ba ch s cui mt n v?
s 30
Cu1: (2 im)
Cho hm s: y =1
12
x
x(1)
1) Kho st s bin thin v v th ca hm s (C) ca hm s (1).2) Gi I l giao im ca hai ng tim cn ca (C). Tm im M thuc (C) sao cho
tip tuyn ca (C) ti M vung gc vi ng thng IM.
Cu2: (2 im)
1) Gii phng trnh:( )
11cos2
42sin2cos32
2
=
x
xx
2) Gii bt phng trnh: ( ) 06log1log2log 24
1
2
1 ++ xx
Cu3: (3 im)
1) Trong mt phng vi h ta cc Oxy cho elip (E): 114
22
=+ yx , M(-2; 3),
N(5; n). Vit phng trnh cc ng thng d1, d2 qua M v tip xc vi (E). Tm n trong s cc tip tuyn ca (E) i qua N v c mt tip tuyn song song vi d1 hoc d2
2) Cho hnh chp u S.ABC, y ABC c cnh bng a, mt bn to vi y mt gc
bng (00 < < 900). Tnh th tch khi chp S.ABC v khong cch t nh A nmt phng (SBC).3) Trong khng gian vi h to cc Oxyz cho hai im I(0; 0; 1), K(3; 0; 0).
Vit phng trnh mt phng i qua hai im I, K v to vi vi mt phng xOy mtgc bng 300
Cu4: (2 im)1) T mt t gm 7 hc sinh n v 5 hc sinh nam cn chn ra 6 em trong s hc
sinh n phi nh hn 4. Hi c bao nhiu cch chn nh vy?
2) Cho hm s f(x) =( )
xbxe
x
a ++ 31
. Tm a v b bit rng
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f'(0) = -22 v ( ) 51
0
= dxxf
Cu5: (1 im)
Chng minh rng:2
2cos2
xxxex ++ x R
s 31
Cu1: (2 im)
Cho hm s: y =3
6522
++++
xmxx
(1) (m l tham s)
1) Kho st s bin thin v v th ca hm s (1) khi m = 1.
2) Tm m hm s (1) ng bin trn khong (1; + ).Cu2: (2 im)
1) Gii phng trnh:( )
( )xsinxcosxsin
xcosxcos +=+
12
12
2) Cho hm s: f(x) = 2xlogx (x > 0, x 1)
Tnh f'(x) v gii bt phng trnh f'(x) 0
Cu3: (3 im)
1) Trong mt phng vi h ta cc Oxy cho ABC c nh A(1; 0) v hai
ng thng ln lt cha cc ng cao v t B v C c phng trnh tng ng l:
x - 2y + 1 = 0 v 3x + y - 1 = 0 Tnh din tch ABC.
2) Trong khng gian vi h to cc Oxyz cho mt phng
(P): 2x + 2y + z - m2 - 3m = 0 (m l tham s)
v mt cu (S): ( ) ( ) ( ) 9111 222 =+++ zyx
Tm m mt phng (P) tip xc vi mt cu (S). Vi m tm c, hy xc nh to
tip im ca mt phng (P) v mt cu (S).
3) Cho hnh chp S.ABC c y ABC l tam gic vung ti B, AB = a, BC = 2a, cnh
SA vung gc vi y v SA = 2a. Gi M l trung im ca SC. Chng minh rng
AMB cn ti M v tnh din tch AMB theo a.Cu4: (2 im)
1) T 9 ch s 0, 1, 2, 3, 4, 5, 6, 7, 8 c th lp c bao nhiu s t nhin chn m
mi s gm 7 ch s khc nhau?
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2) Tnh tch phn: I = 1
0
32
dxex x
Cu5: (1 im)
Tm cc gc A, B, C ca ABC biu thc: Q = CsinBsinAsin 222 + t gi tr
nh nht.
s 32
Cu1: (2 im)
1) Kho st s bin thin v v th ca hm s (C) ca hm s: y = 2x3 - 3x2 - 1
2) Gi dk l ng thng i qua im M(0 ; -1) v c h s gc bng k. Tm k
ng thng dk ct (C) ti ba im phn bit.
Cu2: (2 im)
1) Gii phng trnh:xsinxcos
tgxgxcot2
42+=
2) Gii phng trnh: ( ) xlog x = 1455 Cu3: (3 im)
1) Trong khng gian vi h to cc Oxyz cho hai im A(2; 1; 1), B(0; -1; 3)
v ng thng d:
=+
=
083
01123
zy
yx
a) Vit phng trnh mt phng (P) i qua trung im I ca AB v vung gc vi
AB. Gi K l giao im ca ng thng d v mt phng (P), chng minh rng d
vung gc vi IK.
b) Vit phng trnh tng qut ca hnh chiu vung gc ca d trn mt phng c
phng trnh: x + y - z + 1 = 0.
2) Cho t din ABCD c AD vung gc vi mt phng (ABC) v ABC vung ti A,
AD = a, AC = b, AB = c. Tnh din tch ca BCD theo a, b, c v chng minh rng:
2S ( )cbaabc ++
Cu4: (2 im)
1) Tm s t nhin n tho mn: 1002 333222 =++ nnnnnnnn CCCCCC
trong knC l s t hp chp k ca n phn t.
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2) Tnh tch phn: I = +e
xdxlnx
x
1
21
Cu5: (1 im)
Xc nh dng ca ABC, bit rng: ( ) ( ) BsinAsincBsinbpAsinap =+ 22
trong BC = a, CA = b, AB = c, p =2
cba ++
s 33
Cu1: (2,5 im)
1) Cho hm s: y =1
12
+
xmxx
(*)
a) Kho st s bin thin v v th (C) ca hm s khi m = 1.
b) Tm nhng im trn (C) c to l nhng s nguyn.
c) Xc nh m ng thng y = m ct th ca hm s (*) ti hai im phn bit
A, B sao cho OA vung gc vi OB.
Cu2: (1 im)
Cho ng trn (C): x2 + y2 = 9 v im A(1; 2). Hy lp phng trnh ca ng
thng cha dy cung ca (C) i qua A sao cho di dy cung ngn nht.
Cu3: (3,5 im)
1) Cho h phng trnh:
+=+=+
12
3
mymx
myx
a) Gii v bin lun h phng trnh cho.
b) Trong trng hp h c nghim duy nht, hy tm nhng gi tr ca m sao
cho nghim (x0; y0) tho mn iu kin
>>
0
0
0
0
y
x
2) Gii cc phng trnh v bt phng trnh sau:
a) sin(cosx) = 1b) 11252 5
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b) 6 ng trn phn bit.
2) T kt qu ca 1) hy suy ra s giao im ti a ca tp hp cc ng ni trn.
Cu5: (2 im)
Cho hnh chp t gic u S.ABCD c cc cnh bn bng a v mt cho SAC l tam
gic u.
1) Tm tm v bn knh ca mt cu ngoi tip hnh chp.
2) Qua A dng mt phng () vung gc vi SC. Tnh din tch thit din to bi
mt phng () v hnh chp.
s 34
Cu1: (2 im)
Cho hm s: y =12
1
xx
1) Kho st s bin thin v v th ca hm s.
2) Tm cc im trn th hm s c to l cc s nguyn.
Cu2: (2 im)
1) Gii phng trnh: xsinxcostgxxtg 33
12 =
2) Gii bt phng trnh: ( ) ( ) ( ) 04221 33
1
3
1
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Cu5: (2 im)
1) Cho a gic li c n cnh. Xc nh n a gic c s ng cho gp i s
cnh.
2) Tnh tch phn: I =( ) ++
1
0
2
11dx
xx
x
s 35
Cu1: (3,5 im)
Cho hm s: y = 14
2
+
x
xx
(1)
1) Kho st s bin thin v v th (C) ca hm s (1).
2) Tm m ng thng (d) c phng trnh y = mx ct (C) ti hai im phn bit.
3) Tnh din tch hnh phng c gii hn bi (C); tim cn xin v cc ng
thng x = 2; x = 4.
Cu2: (1 im)
Gii phng trnh: ( ) ( ) 021223 =++++ xcosxsinxsinxcosxsin
Cu3: (2 im)
Cho phng trnh: 04 22 =+ xx (2)
1) Gii phng trnh (2) khi m = 2.
2) Xc nh m phng trnh (2) c nghim.
Cu4: (1 im)
Cho cc ch s: 0, 1, 2, 3, 4. C bao nhiu s t nhin chn gm 5 ch s khc nhau
lp t cc ch s trn?Cu5: ( 2,5 im)
Cho elip (E) c hai tiu im l F1( 03; ); ( )032 ;F v mt ng chun c
phng trnh: x =3
4.
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1) Vit phng trnh chnh tc ca (E).
2) M l im thuc (E). Tnh gi tr ca biu thc:
P = MF.MFOMMFMF 2122
22
1 3 +
3) Vit phng trnh ng thng (d) song song vi trc honh v ct (E) ti hai
im A, B sao cho OA OB.
s 36
Cu1: (2,5 im)
Cho hm s: y =x
xx 232 +
1) Kho st s bin thin v v th (C) ca hm s.
2) Tm trn ng thng x = 1 nhng im M sao cho t M k c hai tip tuyn
ti (C) v hai tip tuyn vung gc vi nhau.
Cu2: (1,5 im) Gii cc phng trnh:
1) ( ) ( ) 24224 =+ xloglogxloglog
2)5
5
3
3 xsinxsin =
Cu3: (2 im)
Gii cc bt phng trnh:
1) ( ) ( ) 06140252 1 + xxx
Cu4: (2 im) Cho In = ( ) 1
0
221 dxxx
nv J n = ( )
1
0
21 dxxx
n
vi n nguyn dng.
1) Tnh Jn v chng minh bt ng thc:( )12
1
+
nIn
2) Tnh In + 1 theo In v tmn
n
x II
lim 1+
Cu5: (2 im)
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1) Trong mt phng (P) cho ng thng (D) c nh, A l mt im c nh nm
trn (P) v khng thuc ng thng (D); mt gc vung xAy quay quanh A, hai tia
Ax v Ay ln lt ct (D) ti B v C. Trn ng thng (L) qua A v vung gc vi
(P) ly im S c nh khc A. t SA = h v d l khong cch t im A n (D).
Tm gi tr nh nht ca th tch t din SABC khi xAy quay quanh A.
2) Trong mt phng vi h ta cc Oxy cho ABC. im M(-1; 1) l trungim ca cnh BC; hai cnh AB v AC theo th t nm trn hai ng thng c
phng trnh l: x + y - 2 = 0; 2x + 6y + 3 = 0.
Xc nh to ba nh A, B, C.
s 37
Cu1: (3 im)
Cho hm s: y = x3 - 3mx + 2 c th l (Cm) (m l tham s)
1) Kho st s bin thin v v th (C1) ca hm s khi m = 1.
2) Tnh din tch hnh phng gii hn bi (C1) v trc honh.
3) Xc nh m (Cm) tng ng ch c mt im chung vi trc honh.
Cu2: (1 im)
1) Chng minh rng vi mi s nguyn dng n ta u c:
nnnnn
nnnnn C...CCCC...CCC
22
42
22
02
122
52
32
12 ++++=++++
2) T cc ch s 1, 2, 3, 4, 5 c th lp c bao nhiu s gm 3 ch s khc nhau
nh hn 245.
Cu3: (1,5 im)
1) Gii h phng trnh:( )( )( )( )
=++
=
15
3
22
22
yxyx
yxyx
2) Gii phng trnh: xx +=+ 173
Cu4: (1,5 im)Cho phng trnh: ( ) 01122 =++ mxcosmxcos (m l tham s)
1) Gii phng trnh vi m = 1.
2) Xc nh m phng trnh c nghim trong khong
;
2.
Cu5: (3 im)
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1) Cho khi chp t gic u S.ABCD c cc cnh bn v cnh y u bng a. Gi
M, N v P ln lt l trung im ca cc cnh AD, BC v SC. Mt phng (MNP) ct
SD ti Q. Chng minh rng MNPQ l hnh thang cn v tnh din tch ca n.
2) Trong khng gian vi h to cc Oxyz cho hai ng thng:
(D1):
== =
tz
ty tx1
v (D2):
===
'tz
'ty 'tx 12
(t, t' R)
a) Chng minh (D1), (D2) cho nhau v tnh khong cch gia hai ng thng y.
b) Tm hai im A, B ln lt trn (D1), (D2) sao cho AB l on vung gc chung
ca (D1) v (D2).
s 38
Cu1: (3 im)
Cho hm s: y =1
12
+
xmxx
1) Kho st s bin thin v v th ca hm s khi m = 1.
2) Xc nh m hm s ng bin trn cc khong (- ; 1) v (1; + )3) Vi gi tr no ca m th tim cn xin ca th hm s to vi cc trc to
mt tam gic c din tch bng 4 (n v din tch).
Cu2: (2 im)
Cho phng trnh: ( ) ( ) mtgxtgx
=++ 223223
1) Gii phng trnh khi m = 6.
2) Xc nh m phng trnh c ng hai nghim phn bit nm trong khong
22; .
Cu3: (2 im)
1) Gii bt phng trnh: ( )4
3
16
1313
4
14
xx loglog
2) Tnh tch phn: I =
2
0
32 xdxsinxsinxsin
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Cu4: (2 im)
Trong mt phng vi h ta cc Oxy cho ABC v im M(-1; 1) l trung
im ca AB. Hai cnh AC v BC theo th t nm trn hai ng:
2x + y - 2 = 0 v x + 3y - 3 = 0
1) Xc nh ta ba nh A, B, C ca tam gic v vit phng trnh ng cao CH.2) Tnh din tch ABC.
Cu5: (1 im)
Gi s x, y l cc nghim ca h phng trnh:
+=+
=+
32
12
222 aayx
ayx
Xc nh a tch P = x.y t gi tr nh nht.
s 39
Cu1: (2 im)
Cho hm s: y =2
52
+
xxx
1) Kho st s bin thin v v th ca hm s cho.
2) Bin lun theo m s nghim ca phng trnh:2
52
+
x
xx= m
Cu2: (2 im)
1) Gii phng trnh: 01 =++ xcosxsin
2) Gii bt phng trnh: ( ) xlogxlog x 22
22 + 4
Cu3: (1 im)
Gii h phng trnh:( )
++=+
=
2
7
22
33
yxyx
yxyx
Cu4: (1,5 im)
Tnh cc tch phn sau: I1 = ( )
+2
0
442 dxxcosxsinxcos I2 =
2
0
5 xdxcos
Cu5: (3,5 im)
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1) Trong mt phng vi h ta cc Oxy cho ng trn (S) c phng trnh:
x2 + y2 - 2x - 6y + 6 = 0 v im M(2 ; 4)
a) Chng minh rng im M nm trong ng trn.
b) Vit phng trnh ng thng i qua im M, ct ng trn ti hai im A v B
sao cho M l trung im ca AB.c) Vit phng trnh ng trn i xng vi ng trn cho qua ng thng
AB.
2) Cho hnh chp t gic S.ABCD c di tt c cc cnh u bng a. Chng minh
rng:
a) y ABCD l hnh vung.
b) Chng minh rng nm im S, A, B, C, D cng nm trn mt mt cu. Tm tm v
bn knh ca mt cu .
s 40
Cu1: (2 im)
Cho hm s: y =( )
( )1
1322
++
mxmxmx
1) Kho st s bin thin v v th ca hm s khi m = 2.
2) Tm tt c cc gi tr ca m hm s cho ng bin trong khong (0; + ).Cu2: (2 im)
1) Tnh tch phn: I = ( )
2
0
33 dxxsinxcos
2) T 5 ch s 0, 1, 2, 5, 9 c th lp c bao nhiu s l, mi s gm 4 ch s
khc nhau.
Cu3: (3 im)1) Gii phng trnh: ( ) 442 =+ xsinxcosxsin
2) Gii h phng trnh:
+=
+=
432
432
22
22
yxy
xyx
3) Cho bt phng trnh: ( ) ( ) 114 2525
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Tm m bt phng trnh nghim ng vi mi x thuc khong (2 ; 3)
Cu4: (3 im)
Trong khng gian vi h to cc Oxyz cho hai ng thng (1) v (2) c
phng trnh: 1
:
=+
=+
0104
0238
zy
yx
2
:
=++
=
022
032
zy
zx
1) Chng minh (1) v (2) cho nhau.
2) Vit phng trnh ng thng () song song vi trc Oz v ct cc ng thng
(1) v (2).
s 41
Cu1: (2,5 im)
Cho hm s: y = x3 - mx2 + 1 (Cm)
1) Khi m = 3
a) Kho st s bin thin v v th ca hm s.
b) Tm trn th hm s tt c cc cp im i xng nhau qua gc to .
2) Xc nh m ng cong (Cm) tip xc vi ng thng (D) c phng trnh
y = 5. Khi tm giao im cn li ca ng thng (D) vi ng cong (Cm).
Cu2: (1,5 im)
1) Gii bt phng trnh: ( ) ( ) 13
3
1
310310
++
+ xx
xx
0
2) Gii phng trnh: ( ) 01641 323 =++ xlogxxlogx
Cu3: (2 im)
1) Gii phng trnh: ( )( ) 45252 =++++ xxxx
2) Gii phng trnh:xcos
xcosxcos1
7822 =+
Cu4: (2 im)
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1) Trong khng gian vi h to cc Oxyz cho im A(-1; 2; 5), B(11; -16; 10).
Tm trn mt phng Oxy im M sao cho tng cc khong cch t M n A v B l b
nht.
2) Tnh tch phn: I = +
3
2 48
7
21
dxxx
x
Cu5: (2 im)
Trn tia Ox, Oy, Oz i mt vung gc ln lt ly cc im khc O l M, N v S
vi OM = m, ON = n v OS = a.
Cho a khng i, m v n thay i sao cho m + n = a.
1) a) Tnh th tch hnh chp S.OMN
b) Xc nh v tr ca cc im M v N sao cho th tch trn t gi tr ln nht.
2) Chng minh:
s 42
Cu1: (2 im)
1) Kho st s bin thin v v th (C) ca hm s: y =2
1
+
x
x
2) Tm cc im trn th (C) ca hm s c to l nhng s nguyn.
3) Tm cc im trn th (C) sao cho tng khong cch t im n hai tim
cn l nh nht.
Cu2: (2 im)
1) Gii phng trnh: 012315 = xxx
2) Gii h phng trnh:( )
( )
=+
=+
223
223
xylog
yxlog
y
x
Cu3: (1 im)
Gii phng trnh lng gic: 022 3 =+ xcosxcosxsin
Cu4: (2 im)
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Cho D l min gii hn bi cc ng y = tg2x; y = 0; x = 0 v x =4
.
1) Tnh din tch min D.
2) Cho D quay quanh Ox, tnh th tch vt th trn xoay c to thnh.
Cu5: (1,5 im)Trong khng gian vi h to cc Oxyz cho ba im A(1; 4; 0), B(0; 2; 1),
C(1; 0; -4).
1) Vit phng trnh tng qut ca mt phng () i qua im C v vung gc vi
ng thng AB.
2) Tm to im C' i xng vi im C qua ng thng AB.
Cu6: (1,5 im)
1) Gii phng trnh: xxCCC xxx 14966 2321 =++ (x 3, x N)
2) Chng minh rng: 1919201720
520
320
120 2=+++++ CC...CCC
s 43
Cu1: (2,5 im)
1) Kho st s bin thin v v th ca hm s y = 1
2
xx .
2) Bin lun theo tham s m s nghim ca phng trnh: mxx = 1
2
Cu2: (2,5 im)
1) Chng minh rng nu x, y l hai s thc tho mn h thc:
x + y = 1 th x4 + y48
1
2) Gii phng trnh: 12822324222
212 ++>++ + x.x..xx xxx
Cu3: (2,5 im)
1) Gii phng trnh: 0239624
22
=+xcos
xcosxsinxsin
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2) Cc gc ca ABC tho mn iu kin:
( )CcosBcosAcosCsinBsinAsin 222222 3 ++=++ Chng minh rng ABC l tam gic u.
Cu4: (2,5 im)
1) Tnh tch phn: e
xdxlnx1
22
2) Cho hnh lp phng ABCD.A'B'C'D' vi cc cnh bng a. Gi s M, N ln
lt l trung im ca BC, DD'. Tnh khong cch gia hai ng thng BD v MN
theo a.
s 44
Cu1: (3 im)
Cho hm s: y = x3 - 3mx2 + 3(2m - 1)x + 1 (1)1) Kho st s bin thin v v th ca hm s (1) khi m = 2.
2) Xc nh m sao cho hm s (1) ng bin trn tp xc nh.
3) Xc nh m sao cho hm s (1) c mt cc i v mt cc tiu. Tnh to ca
im cc tiu.
Cu2: (2 im)
1) Gii phng trnh: 23sin2sinsin 222 =++ xxx
2) Tm m phng trnh: ( )33 2422
122 =+ xlogmxlogxlog
c nghim thuc khong [32; + ).
Cu3: (2 im)
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1) Gii h phng trnh:
=+
=+
015132
932
22
22
yxyx
yxyx
2) Tnh tch phn: e
dxx
xln
13
Cu4: (1,5 im)
Cho hnh chp S.ABC c y ABC l tam gic u cnh a v SA vung gc vi mt
phng (ABC). t SA = h.
1) Tnh khong cch t A n mt phng (SBC) theo a v h.
2) Gi O l tm ng trn ngoi tip tam gic ABC v H l trc tm tam gic SBC.
Chng minh: OH (SBC).
Cu5: (1,5 im)Trong khng gian vi h to cc Oxyz cho ng thng d v mt phng (P):
d:
==+032
03
zy
zx(P): x + y + z - 3 = 0
1) Vit phng trnh mt phng (Q) cha ng thng d v qua im M(1; 0; -2).
2) Vit phng trnh hnh chiu vung gc ca ng thng d trn mt phng (P).
s 45
Cu1: (3 im)
Cho hm s: y =1
12
xxx
(C)
1) Kho st s bin thin v v th ca hm s (C).
2) Lp phng trnh tip tuyn vi (C) ti im c honh x = 0.
3) Tm h s gc ca ng thng ni im cc i, cc tiu ca th (C).
Cu2: (2,5 im)
1) Gii phng trnh: xxx .4269 =+ .
2) Tnh: ++
2
02
3
12
3
xx
dxx
Cu3: (2,5 im)
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1) Gii h phng trnh:
=+
=+
26
2
33 yx
yx
2) Tnh gc C ca ABC nu: ( )( ) 211 =++ gBcotgAcot
Cu4: (2 im)
Trong khng gian vi h to cc Oxyz :
1) Cho 2 ng thng:
(1):
==
0
0
y
x(2):
==+
0
01
z
yx
Chng minh (1) v (2) cho nhau.
2) Cho 2 im A(1 ; 1 ; -1), B(3 ; 1 ; 1) v mt phng (P) c phng trnh:
x + y + z - 2 = 0Tm trn mt phng (P) cc im M sao cho MAB l tam gic u.
s 46Cu1: (2,5 im)
Cho hm s: y = x3 - (2m + 1)x2 - 9x (1)
1) Vi m = 1;
a) Kho st s bin thin v v th (C) ca hm s (1).
b) Cho im A(-2; -2), tm to im B i xng vi im A qua tm i xng
ca th (C).
2) Tm m th ca hm s (1) ct trc honh ti ba im phn bit c cc honh
lp thnh mt cp s cng.Cu2: (2 im)
1) Gii phng trnh: 03sin2cos4cossin =+ xxxx
2) Cho ABC cnh a, b, c tho mn h thc: 2b = a + c.
Chng minh rng: 32
cot2
cot =CgAg .
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Cu3: (2 im)
1) Gii bt phng trnh: ( ) ( )12lg2
13lg
22 +> xxx
2) Tm a h phng trnh sau c nghim duy nht:( )
( )
=+
=+
1
1
2
2
xayxy
yaxxy
Cu4: (1,5 im)
1) Tnh tch phn: I = +++2
05cos3sin4
1sin3cos4
dxxx
xx
2) Tnh tng: P = 51054
1043
1032
1021
10110 33333 CCCCCC ++
1010
10910
9810
8710
7610
633333 CCCCC +++
Cu5: (2 im)1) Trong khng gian vi h to cc Oxyz cho mt phng (P) v mt cu (S) ln
lt c phng trnh: (P): y - 2z + 1 = 0 (S): x2 + y2 + z2 - 2z = 0.
Chng minh rng mt phng (P) v mt cu (S) ct nhau. Xc nh tm v bn knh
ca ng trn giao tuyn.
2) Cho hnh chp u S.ABC nh S, chiu cao l h, y l tam gic u cnh a. Qua
cnh AB dng mt phng vung gc vi SC. Tnh din tch thit din to thnh theo a
v h.
s 47
Cu1: (2,5 im)
Cho hm s: y =1
2222
+++
xmxmx
(m l tham s)
1) Kho st s bin thin v v th ca hm s khi m = 0.
2) Tm m trn th c hai im i xng nhau qua gc to .
Cu2: (2 im)
1) Gii phng trnh: 09328322
122 =+ +++ xxxx .
2) Cho ABC. Chng minh rng nuCsin
BsintgCtgB
2
2
= th tam gic l tam gic vung
hoc cn.
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Cu3: (2 im)
1) Tnh tch phn: 9
1
3 1 dxxx
2) Gii h phng trnh: ( )
+=+
+=+
yxyx
yyxx
322
22
Cu4: (2,5 im)
1) Cho hnh chp tam gic u S.ABC c gc gia mt bn v mt y l v SA =
a. Tnh th tch hnh chp cho.
2) Trong khng gian vi h to cc Oxyz vi h to vung gc Oxyz, cho
hai ng thng: 1:3
3
2
2
1
1 == zyx 2:
=+
=+
0532
02
zyx
zyx
Tnh khong cch gia hai ng thng cho.
Cu5: ( 1 im)
Chng minh rng: P1 + 2P2 + 3P3 + ... + nPn = Pn + 1 - 1
Trong n l s t nhin nguyn dng v Pn l s hon v ca n phn t.
s 48
Cu1: (3 im)
Cho hm s: y = x3 + 3x2 + 1 (1)
1) Kho st s bin thin v v th ca hm s (1).
2) ng thng (d) i qua im A(-3 ; 1) c h gc l k. Xc nh k (d) ct th
hm s (1) ti ba im phn bit.Cu2: (2,5 im)
1) Gii phng trnh: 0221 =++++ xcosxsinxcosxsin
2) Gii h phng trnh:( )( )
=++
=++
095
1832
2
2
yxx
yxxx
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Cu3: (2 im)
1) Gii bt phng trnh: ( )382
4 1+ xlogxlog 1
2) Tm gii hn:xcos
xxlimx
++ 1
121323 2
0
Cu4: (1,5 im)
Trong mt phng vi h ta cc Oxy cho hai im A(1; 2), B(3; 4). Tm trn
tia Ox mt im P sao cho AP + PB l nh nht.
Cu5: (1 im)
Tnh tch phn: I = +
+2
03 23
1dx
x
x
s 49
Cu1: (2,5 im)
Cho hm s: y = ( ) ( ) 4313
1 23 +++ xmxmx (1) (m l tham s)
1) Kho st s bin thin v v th ca hm s khi m = 0.
2) Xc nh m hm s (1) ng bin trong khong: 0 < x < 3
Cu2: (2 im)
1) Gii phng trnh: 0322212 333 =+++++ xxx (1)
2) Cho phng trnh: ( ) 061232 2 =++ mxcosxsinmxsin
a) Gii phng trnh vi m = 1.
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b) Vi gi tr no ca m th phng trnh (1) c nghim.
Cu3: (1 im)
Gii h bt phng trnh:
>+
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Xc nh n du "=" xy ra?
Cu3: (2 im)
1) Cho phng trnh: xsinmxcosxsin 266 =+ a) Gii phng trnh khi m = 1.
b) Tm m phng trnh c nghim.
2) Chng minh rng ABC u khi v ch khi
++=
=
acbacb
a
Ccosba333
2
2
Cu4: (2,5 im)
1) Trong mt phng vi h ta cac Oxy cho im A(8; 6). Lp phng trnh
ng thng qua A v to vi hai trc to mt tam gic c din tch bng 12.
2) Trong khng gian vi h to cc Oxyz Cho A(1; 2; 2), B(-1; 2; -1),
C(1; 6; -1), D(-1; 6; 2)a) Chng minh rng ABCD l hnh t din v tnh khong cch gia hai ng
thng AB v CD.
b) Vit phng trnh mt cu ngoi tip t din ABCD.
Cu5: (1,5 im)
Cho hai hm s f(x), g(x) xc nh, lin tc v cng nhn gi tr trn on [0; 1].
Chng minh rng: ( ) ( ) ( ) ( )
1
0
1
0
21
0
dxxgdxxfdxxgxf
s 51
Cu1: (2 im)
Cho hm s: y =( )( )
mmxmxxm
+++ 421 2
(Cm) (m l tham s, m 0, -4
1)
1) Kho st s bin thin v v th ca hm s (C2) vi m = 2.
2) Tm m hm s (Cm) c cc i, cc tiu v gi tr cc i, cc tiu cng du.
Cu2: (2 im)
1) Gii h phng trnh:
++=
++=
22
22
3
3
yxy
xyx
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2) Gii phng trnh: tg2x + cotgx = 8cos2x
Cu3: (2,5 im)
1) 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 .
2) Trong khng gian vi h to cc Oxyz cho hai ng thng:
(D1):
=+=+
0104
0238
zy
zx(D2):
=++=
022
032
zy
zx
a) Vit phng trnh cc mt phng (P) v (Q) song song vi nhau v ln lt i qua
(D1) v (D2).
b) Vit phng trnh ng thng (D) song song vi trc Oz v ct c hai ng
thng (D1), (D2)
Cu4: (2 im)
1) Tnh tng: S = ( ) nnn
nnnn nC....CCCC 14324321 +++
Vi n l s t nhin bt k ln hn 2, knC l s t hp chp k ca n phn t.
2) Tnh tch phn: I = +
2
1 12xx
dx
Cu5: (1,5 im)
Cho ba s bt k x, y, z. Chng minh rng:
222222 zyzyzxzxyxyx +++++++
s 52
Cu1: (2 im)
Cho hm s: y =1
1
+
xx
(1) c th (C)
1) Kho st s bin thin v v th ca hm s (1).2) Chng minh rng ng thng d: y = 2x + m lun ct (C) ti hai im A, B
thuc hai nhnh khc nhau. Xc nh m on AB c di ngn nht.
Cu2: (2,5 im)
Cho phng trnh: 03232322
224 =+ . xx (1)
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1) Gii phng trnh (1) khi m = 0.
2) Xc nh m phng trnh (1) c nghim.
Cu3: (2,5 im)
Gii cc phng trnh v bt phng trnh sau:
1) xtgxsinxcos
xcosxsin 28
1322
66=
+
2) ( ) ( )2431243 2329 ++>+++ xxlogxxlog Cu4: (1,5 im)
Trong khng gian vi h to cc Oxyz Cho A(1; 1; 1), B(1; 2; 0) v mt
cu (S): x2 + y2 + z2 - 6x - 4y - 4z + 13 = 0. Vit phng trnh mt phng cha ng
thng AB v tip xc vi (S).Cu5: (1,5 im)
Tnh tng: S = nnnnn Cn
...CCC1
1
3
1
2
1 211
+++++
Bit rng n l s nguyn dng tho mn iu kin: 7921 =++ nnnn
nn CCC
knC l s t hp chp k ca n phn t.
s 53
Cu1: (2 im)
Cho hm s: y = -x3 + 3x2 - 2
1) Kho st s bin thin v v th (C) ca hm s.
2) Tm t phng trnh: 023 223 =+ tlogxx c 6 nghim phn bit.
Cu2: (3 im)
1) Trong mt phng vi h ta cc Oxy cho ng trn
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(C): ( ) ( ) 413 22 =+ yx . Vit phng trnh tip tuyn ca (C) bit rng tip
tuyn ny i qua im M0(6; 3)
2) Trong khng gian vi h to cc Oxyz cho hnh hp ABCD.A'B'C'D'
Vi A(2; 0; 2), B(4; 2; 4), D(2; -2; 2) v C'(8; 10; -10).
a) Tm to cc nh cn li ca hnh hp ABCD.A'B'C'D'.
b) Tnh th tch ca hnh hp ni trn.
Cu3: (2 im)
1) Gii phng trnh: 21 +=++ xxx
2) Gii h phng trnh:
=
=+
22
1
22 yyx
x
ysinxsin
Cu4: (2 im)
1) Chng minh rng: knkn
kn
kn CCCCCCC =++
22
22
12
122
02
n k + 2 ; n v k l cc s nguyn dng, knC l s t hp chp k ca n phn t.
2) Tnh din tch hnh phng gii hn bi parabol: y = -x2 - 4x; ng thng x = -1;
ng thng x = -3 v trc Ox
Cu5: (1 im)
Cho 2 s nguyn dng m, n l s l
Tnh theo m, n tch phn: I =
2
0
xdxcosxsin mn
s 54Cu1: (2 im)
1) Kho st s bin thin v v th ca hm s: y = xxx 323
23 +
2) Da v th (C) Cu trn, hy bin lun theo tham s m s nghim ca
phng trnh: meee xx
x=+ 32
3
23
Cu2: (3 im)
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1) Trong mt phng vi h ta cc Oxy cho elp (E) c phng trnh:
12
2
2
2
=+b
y
a
x(a > 0, b > 0)
a) Tm a, b bit Elip (E) c mt tiu im l F1(2; 0) v hnh ch nht c s ca (E)
c din tch l 12 5 (vdt).b) Tm phng trnh ng trn (C) c tm l gc to . Bit rng (C) ct (E) vatm c Cu trn ti 4 im lp thnh hnh vung.
2) Trong khng gian vi h to cc Oxyz tm theo a, b, c (a, b, c 0) to cc nh ca hnh hp ABCD.A'B'C'D'. Bit A(a; 0; 0); B(0; b; 0) C(0; 0; c) vD'(a; b; c).Cu3: (2 im)
1) Gii v bin lun phng trnh sau theo tham s m:
( ) 012333
= mlogxlogxlog
2) Gii phng trnh: ( ) 032332 =++++ xcosxcosxcosxsinxsinxsin Cu4: (2 im)
1) Cho f(x) l hm lin tc trn on [0; 1]. Chng minh rng:
( ) ( )
=2
0
2
0
dxxcosfdxxsinf
2) Tnh cc tch phn:
I =
+
2
020032003
2003
xcosxsin
xdxsinJ =
+
2
020032003
2003
xcosxsin
xdxcos
Cu5: (1 im)
Gii bt phng trnh: ( ) nnn
nnn C.C.C.!n 32
3 720knC l t hp chp k ca n phn t.
s 55
Cu1: (2 im)1) Kho st s bin thin v v th ca hm s: y = x4 - 10x2 + 9
2) Tm tt c cc gi tr ca tham s m phng trnh: x - 3mx + 2 = 0 c
nghim duy nht.
Cu2: (2 im)
1) Tm tt c cc ng tim cn xin ca th hm s: y = 2x + 21 x+
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2) Tnh th tch ca vt th trn xoay c to ra khi cho hnh phng gii hn
bi cc ng: y = ex ; y =e1
; y = e v trc tung quay xung quanh Oy.
Cu3: (2 im)
1) Cho a thc: P(x) = ( )
2005
1516 x , khai trin a thc di dng:P(x) = 20052005
2210 xa...xaxaa ++++
Tnh tng: S = 2005210 a...aaa ++++
2) Gii h phng trnh:( )
=+=
5
115223
22 logyxlog
yx
Cu4: (2 im)
1) Cho ABC c di cc cnh BC, CA, AB theo th t lp thnh cp s
cng. Tnh gi tr ca biu thc: P =22CgcotAgcot
2) Trong mt phng vi h to cc vung gc Oxy cho hypebol (H):
1916
22
= yx . Lp phng trnh ca elp (E), bit rng (E) c cc tiu im l cc tiu
im ca (H) v (E) ngoi tip hnh ch nht c s ca (H)
Cu5: (2 im)
1) Trong khng gian vi h to cc Oxyz cho ABC c im B(2; 3; -4),
ng cao CH c phng trnh:52
2
5
1
== zyx v ng phn gic trong gc A l
AI c phng trnh:2
1
1
3
7
5 +== zyx . Lp phng trnh chnh tc ca cnh AC.
2) CMR: trong mi hnh nn ta lun c:2
6
V
3
3
2
S
(V l th tch hnh nn, S l din tch xung quanh ca hnh nn)
s 56Cu1: (2 im)
Cho hm s: y =( )
1
112
+++
xmxmx
(1) (m l tham s)
1) Kho st s bin thin v v th ca hm s (1) khi m = 1.
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2) Chng minh rng hm s (1) lun c gi tr cc i (yC) v gi tr cc tiu
(yCT) vi m. Tm cc gi tr ca m (yC)2 = 2yCT
Cu2: (2 im)
1) Gii phng trnh: 3cosx ( ) 1221 2 = xsinxsinxcosxsin
2) Gii h bt phng trnh:
+
045
02
24
2
xx
xx
Cu3: (2 im)
1) Tnh tch phn: I = +3
0
231 dxxx
2) Tm s nguyn dng n tho mn ng thc: nCA nn 16223 =+
Cu4: (3 im)
1) Cho t din ABCD c di cnh AB = x (x > 0), tt c cc cnh cn li c
di bng 1. Tnh d di on vung gc chung ca hai cnh AB v CD. Tm iu
kin i vi x Cu ton c ngha.
2) Trong khng gian vi h to cc Oxyz cho t din OABC c O l gc
ta , A Ox, B Oy, C Oz v mt phng (ABC) c phng trnh:
6x + 3y + 2z - 6 = 0.a) Tnh th tch khi t din OABC.
b) Xc nh to tm v tnh bn knh ca mt cu ngoi tip khi t din OABC.
Cu5: (1 im)
Cho x, y l hai s thc dng khc 1.
Chng minh rng nu: ( ) ( )yloglogxloglog xyyx = th x = y.
s 57
Cu1: (2 im)
Cho hm s: y =2
52
xx
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1) Kho st s bin thin v v th ca hm s.
2) Vit phng trnh tip tuyn ca th hm s, bit tip tuyn i qua im
A(-2; 0).
Cu2: (3 im)
1) Gii phng trnh: xsinxsin 243 = +
2) Gii bt phng trnh: ( ) ( )1111
2 +>+ xlogxlog xx
3) Gii h phng trnh:
=
=+
72
3432
22
22
yx
xyyx
Cu3: (2 im)
1) Tnh tch phn: ++
2
02
3
12 dxxx
x
2) Tm h s ln nht ca a thc trong khai trin nh thc Niutn ca:15
3
2
3
1
+ x
Cu4: (3 im)
1) Cho hnh lp phng ABCD.A'B'C'D'. Chng minh rng cc im gia ca 6
cnh khng xut pht t hai u ng cho AC' l nhng nh ca mt lc gic
phng u.
2) Trong mt phng vi h ta cc Oxy cho hai ng thng:x + y - 1 = 0 v 3x - y + 5 = 0
Hy tm din tch hnh bnh hnh c hai cnh nm trn hai ng thng cho, mt
nh l giao im ca hai ng v giao im ca hai ng cho l I(3; 3).
3) Trong khng gian vi h to cc Oxyz cho hai ng thng:
d1:
=+=+053
0523
zy
yxv d2:
25
2
1
2
=+= zyx
Chng minh rng hai ng thng cho nhau v tm phng trnh ng vunggc chung ca chng.
s 58
Cu1: (4 im)
Cho hm s: y =mx
mx
+ 13(1)
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1) Xc nh m hm s (1) nghch bin trong khong (1; + )
2) Kho st s bin thin v v th ca hm s (1) khi m = 1, gi th ca
hm s ny l (C).
3) Tm hai im A, B thuc (C) sao cho A v B i xng vi nhau qua ng
thng (d): x + 3y - 4 = 0.Cu2: (2 im)
Cho phng trnh: x2 - 2ax + 2 - a = 0 (1)
1) Xc nh a phng trnh (1) c hai nghim x1, x2 sao cho: -2 < x1 < 3 < x2
2) Xc nh a phng trnh (1) c hai nghim x1, x1 sao cho:22
21 xx + t gi
tr nh nht.
Cu3: (1 im)
Cho ABC c 3 gc tho mn iu kin sau: sinA + cosA + sinB - cosB + sinC
- cosC = 1. Chng minh rng: ABC l tam gic vung.
Cu4: (3 im)
Cho ABC c A(-1; 5) v phng trnh ng thng BC: x - 2y - 5 = 0 (xB 0). Dng hnh hp ch nht nhn O, A, B, C
lm bn nh v gi D l nh i din vi nh O ca hnh hp .
1) Tnh khong cch t im C n mt phng (ABD).
2) Tnh to hnh chiu vung gc ca C xung mt phng (ABD). Tm iukin i vi a, b, c hnh chiu nm trn mt phng (xOy)
Cu5: (2 im)
1) Tnh tch phn: +
1
0 1xe
dx
2) Tnh h nguyn hm ca: f(x) = x(1 - x)20
s 76
Cu1: (2 im)
1) Kho st s bin thin v v th ca hm s: y = x3 - x2 - x + 1
2) Bin lun theo tham s m s nghim ca phng trnh: ( ) mxx =+ 112
Cu2: (2 im)
Gii cc phng trnh:
1) sin4x + cos2x + 4cos6x = 0
2) xlogx
logx
logxlogxlog xx 244244
22
222 =+++
Cu3: (1 im)
Tm tt c cc gi tr ca tham s m phng trnh sau c nghim:( )( ) mxxxx =+++ 2222
Cu4: (1,5 im)
Cho t din SABC vi gc tam din nh S l vung. Gi H l trc tm ca
ABC. Chng minh rng:
1) SH (ABC).
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2)2222
1111
SCSBSASH++=
Cu5: (2 im)
Cho n N
1) Tnh tch phn: ( ) +1
0
21 dxxx n
2) Chng minh rng:1
12
1
1
4
1
3
1
2
11
1321
+=
++++++
+
nC
n...CCC
nnnnnn
Cu6: (1,5 im)
1) Tnh tch phn: I = ( ) +1
0
321 dxxx
n(n N)
2) Lp phng trnh ng thng i qua im M(1; 0) sao cho ng thng
cng vi hai ng thng: (d1): 2x - y + 1 = 0 (d2): x + 2y - 2 = 0 to ra mt tam gic
cn c nh l giao im ca hai ng thng d1, d2.
s 77
Cu1: (2 im)
Cho hm s: y = x3 + 3mx2 + 3(m2 - 1)x + m3 - 3m1) Kho st s bin thin v v th ca hm s ng vi m = 0.
2) Chng minh rng vi mi m hm s cho lun lun c cc i v cc tiu;
ng thi chng minh rng khi m thay i cc im cc i v cc tiu ca th
hm s lun lun chy trn hai ng thng c nh.
Cu2: (2 im)
1) Gii phng trnh lng gic:
sinx + sin2x + sin3x + sin4x = cosx + cos2x + cos3x + cos4x2) Chng minh rng trong ABC ta c:
+++=++
2222222
1111 Cgcot
Bgcot
Agcot
Ctg
Btg
Atg
CsinBsinAsin
Cu3: (2 im)
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1) Gii h phng trnh:
=+
=+
13
5
4224
22
yyxx
yx
2) Vi nhng gi tr no ca m th phng trnh: 15
1 2434
2
+=
+mm
xx
c bn nghim phn bit.
Cu4: (2 im)
Cho gc tam din ba mt vung Oxyz. Trn Ox, Oy, Oz ln lt ly cc im
A, B, C.
1) Tnh din tch ABC theo OA = a
2) Gi s A, B, C thay i nhng lun c: OA + OB + AB + BC + CA = k
khng i. Hy xc nh gi tr ln nht ca th tch t din OABC.Cu5: (2 im)
1) Tm h nguyn hm ca hm s: f(x) = tg4x
2) Tm h nguyn hm ca hm s: f(x) =xx
x
3
42
.
s 78Cu1: (2 im)
Cho hm s: y = f(x) = x4 + 2mx2 + m (m l tham s)1) Kho st s bin thin v v th ca hm s khi m = -1.
2) Tm tt c cc gi tr ca m hm s f(x) > 0 vi x. Vi nhng gi tr cam tm c trn, CMR hm s: F(x) = f(x) + f'(x) + f"(x) + f"'(x) + f(4)(x) > 0 xCu2: (2 im)
1) Gii phng trnh lng gic:( )
1
2
2
1
=
+ gxcotxsinxcos
xgcottgx
2) Hai gc A, B ca ABC tho mn iu kin: 122
=+ BtgAtg . Chng minh
rng: 124
3
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Trong khng gian vi h to cc Oxyz cho ng thng (d):
==+=
tz
ty
tx
3
2
21
v mt phng (P): 2x - y - 2z + 1 = 01) Tm to cc im thuc ng thng (d) sao cho khong cch t mi
im n mt phng (P) bng 12) Gi K l im i xng ca I(2; -1; 3) qua ng thng (d). Hy xc nh
to im K.Cu4: (2 im)
1) Gii bt phng trnh: ( )32
1265
3
1
3
12
3 +>++ xlogxlogxxlog
2) Vi a > 1 th phng trnh sau v nghim:
112222
++=++ aaxcosxxsinxCu5: (2,5 im)
1) Tnh din tch ca hnh phng gii hn bi parabol (P) c phng trnh:y = x2 - 4x + 5 v hai tip tuyn ca (P) k ti hai im A(1; 2) v B(4; 5)
2) Tnh tch phn: I = ( )
+2
0
442 dxxcosxsinxcos J =
0
dxxsinxcos
3) Vit khai trin Newton ca biu thc (3x - 1)16. T chng minh rng:
161616216141161501616 2333 =++ C...CCC s 79
Cu1: (2 im)Cho hm s: y = -x4 + 2(m + 1)x2 - 2m - 11) Xc nh tham s m th hm s ct trc honh ti bn im lp thnh
mt cp s cng.2) Gi (C) l th khi m = 0. Tm tt c cc im thuc trc tung sao cho t
c th k c ba tip tuyn vi th (C).Cu2: (2 im)
1) Gii phng trnh: x2 + 11 =+x
2) Gii v bin lun phng trnh: m.cotg2x =xsinxcos
xsinxcos66
22
+
theo tham s m
Cu3: (1,5 im)1) Cho hai hm s: f(x) = 4cosx + 3sinx; g(x) = cosx + 2sinx
a) Tm cc s A, B tho mn: g(x) = A.f(x) + B.f'(x)
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b) Tnh tch phn:( )
( )
4
0
dxxfxg
2) Tm th tch vt th to bi elp:( )
1164
422
+ yx quay quanh trc Oy
Cu4: (2,5 im)1) Cho hnh hp ch nht ABCD.A1B1C1D1; H v K l cc hnh chiu vung
gc ca A v C1 xung mt phng (B1CD1). Chng minh: 12KCAH = 2) Cho hai ng trn: tm A(1; 0) bn knh rA = 4 v tm B(-1; 0) bn knh rB
= 2. Tm tp hp tm I(x, y) ca cc ng trn tip xc c 2 ng trn trn. Tp hp l ng g?
3) Vit phng trnh ng thng d vung gc vi mt phng (P): x + y + z = 1
v ct c hai ng thng d1:
11
1
2
1 zyx =
+= d2:
=++
=+
0122
042
zyx
xyx
Cu5: (2 im)1) Cho ba hp ging nhau, mi hp ng 7 bt ch khc nhau v mu sc.
Hp I c 3 bt mu , 2 bt mu xanh, 2 bt mu en;Hp II c 2 bt mu , 2 bt mu xanh, 3 bt mu en;Hp III c 5 bt mu , 1 bt mu xanh, 1 bt mu en;
Ly ngu nhin mt hp v rt h ho t hp ra 2 bt.a) Tnh tt c s cc kh nng xy ra v s kh nng 2 bt c cng mub) Tnh s kh nng 2 bt khng c mu en
2) C bao nhiu s t nhin khc nhau, nh hn 10.000 c to thnh t 5 chs: 0, 1, 2, 3, 4
s 80
Cu1: (2,5 im)
Trong mt phng vi h ta cc Oxy cho (C) l th ca hm s
y = x +x1
v (d) l ng thng c phng trnh y = ax + b
1) Tm iu kin ca a v b (d) tip xc vi (C).
2) Gi s (d) tip xc vi (C) ti I. Gi M v N theo th t l giao im ca (d)
vi trc tung v vi ng phn gic ca gc phn t th nht. Chng minh:
a) I l trung im ca on MN.
b) Tam gic OMN c din tch khng ph thuc vo a v b.
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Cu2: (1,5 im)
Tm k h phng trnh:
==+
kyx
yx 122 c nghim duy nht.
Cu3: (1,5 im)
1) Chng minh rng: 11 22 ++++ aaaa 2 a R
2) Gii h phng trnh:
=+
=
1023
122
xyyx
xyyx
Cu4: (3 im)
1) Tm h nguyn hm ca hm s: f(x) = (sin4x + cos4x)(sin6x + cos6x)
2) Trong mt phng vi h ta cc Oxy cho hai ng thng:
(1): 4x - 3y - 12 = 0 (2): 4x + 3y - 12 = 0a) Tm to cc nh ca tam gic c ba cnh ln lt nm trn cc
ng thng (1), (2) v trc tung.
b) Xc nh tm v bn knh ng tr