Đề thi thử đại học môn Toán năm 2015 (14)

2

y
CT

1
2

cos
2
x cos
2
10x 1 3 cos11x.sin 9x

z 1 i

w =1+ z + z
3
+ z
4

4
x
9
x
= 5(ln x 1)(3
x
2
x
)

+ + =
+ +
+
− − = + − −
⎧
⎨
⎪
⎪
⎪
⎪
⎪
⎩
⎪
⎪
⎪
⎪
⎪
∈

I x −x − x dx
∫
⊥

AB = a,BC = a 3

6

3

P x y + z =

+ =

x − y =

(
nx
5
1
x
2
)
n

n C
n
n

min , ,

P
x
y z
y
z x
3
2
.
z

x y
z
z
2
xy yz zx

2

y
CT

1
2

y ' = 2x
3
− 2m
2
x; y' = 0 ⇔
x = 0
x
2
= m
2
⎡
⎣
⎢
⎢

m = 0 ⇒ y' = 2x

3
; y ' = 0 ⇔ x = 0

x = 0; y
CT
= 1 ≠
1
2

m ≠ 0

±m; y
CT
= y(±m) = 1−
m
4
2

1−
m
4
2
=
1
2
⇔ m
4
= 1 ⇔ m = −1;m = 1

m = −1;m = 1

cos
2
x + cos
2
10x = 1+ 3 cos11x.sin 9x

z = 1+ i

w =1+ z + z
3
+ z
4

1+ cos2x +1+ cos20x
2
= 1+ 3 cos11x.sin9x
⇔
cos2x + cos20x
2
= 3 cos11x.sin9x
⇔ cos11x.cos9x = 3 cos11x.sin9x
⇔
cos11x = 0
tan9x =
1
3
⎡
⎣
⎢

⎢
⎢
⎢
⇔
x =
π
22
+ k
π
11
x =
π
54
+ k
π
9
⎡
⎣
⎢
⎢
⎢
⎢
⎢
⎢
,k ∈ !

x =
π
22
+ k

π
11
;x =
π
54
+ k
π
9
,k ∈ !

w =1+ (1+ i ) + (1+ i )
3
+ (1+ i)
4
= −4+ 3i ⇒ w = 5

4
x
9
x
= 5(ln x 1)(3
x
2
x
)

x > 0

(2
x

−3
x
)(2
x
+ 3
x
) = 5(ln x −1)(3
x
− 2
x
)
⇔ (2
x
−3
x
)(2
x
+ 3
x
+ 5ln x −5) = 0
⇔
2
x
= 3
x
2
x
+ 3
x
+ 5ln x −5 = 0

⎡
⎣
⎢
⎢
⎢
⇔
x = 0(l )
x = 1(t / m)
⎡
⎣
⎢
⎢

x = 1

(x + y)(
1
xy
+ 3) =
6(x
2
+ y
2
) + 4
2(x
2
+ y
2
)
4− x

2
− y
2
= 2 2xy + 2− x
2
− y
2
⎧
⎨
⎪
⎪
⎪
⎪
⎪
⎩
⎪
⎪
⎪
⎪
⎪
(x, y ∈ !)

xy > 0,x
2
+ y
2
≤ 2

3(x + y − 2(x
2

+ y
2
)) +
1
x
+
1
y
−
4
2(x
2
+ y
2
)
= 0
⇔ 3(x + y − 2(x
2
+ y
2
)) +
2(x
2
+ y
2
)(x + y)−4xy
xy 2(x
2
+ y
2

)
= 0
( 2(x
2
+ y
2
) − x − y) −3+
2(x
2
+ y
2
)(x + y)−4xy
xy 2(x
2
+ y
2
)( 2(x
2
+ y
2
) − x − y)
⎛
⎝
⎜
⎜
⎜
⎜
⎜
⎞
⎠

⎟
⎟
⎟
⎟
⎟
⎟
= 0
⇔ ( 2(x
2
+ y
2
) − x − y) A−3
( )
= 0 (*)

A =
( 2(x
2
+ y
2
)(x + y)−4xy)( 2(x
2
+ y
2
) + x + y)
xy 2(x
2
+ y
2
)(x − y)

2
=
( 2(x
2
+ y
2
)(x + y)−4xy)( 2(x
2
+ y
2
) + x + y)
xy 2(x
2
+ y
2
)(x − y)
2
=
2(x + y) + 2(x
2
+ y
2
)
xy 2(x
2
+ y
2
)

xy ≤

x
2
+ y
2
2
≤1

2(x + y) + 2(x
2
+ y
2
)
xy 2(x
2
+ y
2
)
≥
3(x + y)
xy 2(x
2
+ y
2
)
≥
6
2xy(x
2
+ y
2

)
≥
6
2.1.2
= 3

(*) ⇔
x = y = 1
x = y
⎡
⎣
⎢
⎢

x = y =1

x = y

4− 2x
2
= 2 2x + 2− 2x
2
⇔ ( 2x −1)( 2x + 3)+ ( 2− 2x
2
−1) = 0
⇔ ( 2x −1)( 2x + 3−
1+ 2x
2− 2x
2
+1

) = 0
⇔ ( 2x −1)( 2x 2− 2x
2
+ 2+ 2 2− 2x
2
) = 0
⇔ 2x −1= 0 ⇔ x =
1
2

(x; y) = (
1
2
;
1
2
)

2(x
2
+ y
2
) − x − y

(x − y)
2

(x + y)(1+
7
20xy

) =
10(x
2
+ y
2
) +7
5 2(x
2
+ y
2
)
2− x
2
− y
2
= 2 2xy −1
⎧
⎨
⎪
⎪
⎪
⎪
⎪
⎩
⎪
⎪
⎪
⎪
⎪

(x; y) = (
1
2
;
1
2
)

(2− x )(2− y) +
x
2
+ y
2
2
= 2
2− x
2
− y
2
+ 2 2xy = 4− x
2
− y
2
⎧
⎨
⎪
⎪
⎪
⎪
⎩

⎪
⎪
⎪
⎪

(x; y) = (
1
2
;
1
2
)
HD:

(2− x )(2− y) = 2−
x
2
+ y
2
2
⎛
⎝
⎜
⎜
⎜
⎜
⎜
⎞
⎠
⎟

⎟
⎟
⎟
⎟
⎟
2
⇔ (x − y)
2
2
2(x
2
+ y
2
) + x + y
−
1
2
⎡
⎣
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
= 0 ⇔ x = y

2

2(x
2
+ y
2
) + x + y
−
1
2
≥
2
2+ 2
−
1
2
= 0

I = (x +1) −x
2
−2x + 3 dx
0
1
∫

t = 3− x
2
− 2x ⇒ t
2
= 3− x
2
− 2x ⇒ 2tdt = (−2x − 2)dx = −2(x +1)dx

I = t.(−t dt )
3
0
∫
= t
2
dt
0
3
∫
=
t
3
3
3
0
= 3
⊥

AB = a,BC = a 3

6

3

OC =
AC
2
=

a
2
+ 3a
2
2
= a

SO = SC
2
−OC
2
= 6a
2
− a
2
= a 5

V
S .ABCD
=
1
3
SO.S
ABCD
=
1
3
.a 5.a.a 3 =
a
3

15
3

IH
SO
=
CH
CO
=
CI
CS
=
1
3
,IH =
1
3
SO =
a 5
3

d (SB;AI ) = d(SB;(AIM )) = d (B;(AIM )) = 2d (C ;(AIM )) = 2.
CA
HA
d (H ;(AIM )) =
12
5
d (H ;(AIM ))

HK ⊥ (AIM ) ⇒ HK = d (H ;(AIM ))

S
AMC
=
1
2
S
ABC
=
1
2
.
1
2
.a.a 3 =
a
2
3
4
,AM = a
2
+
4a
2
3
=
a 21
3

d (C ;AM ) =

2S
AMC
AM
=
2.
a
2
3
4
a 21
3
=
3a 7
14

HE =
5
6
d (C ; AM ) =
5
6
.
3a 7
14
=
5a 7
28

1
HK

2
=
1
IH
2
+
1
HE
2
=
9
5a
2
+
112
25a
2
=
157
25a
2
⇒ HK =
5a
157

d (SB;AI ) =
12
5
HK =
12a

157

(P ) : x − 2y + 2z −1= 0

d (I ;(P )) =
1− 2.2+ 2.(−1)−1
1
2
+ (−2)
2
+ 2
2
= 2

(S ) : (x −1)
2
+ (y− 2)
2
+ (z +1)
2
= 4

H (
1
2
;1;
m −1
2
)

1
2
− 2.1+ 2.
m −1
2
−1 = 0 ⇔ m −
7
2
= 0 ↔ m =
7
2
⇒ M (0;0;−
7
2
)
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">?@A8B6%CDEF%GFF%GCG%% )H89%IJ%83K=%L%3MN%OA83%83P8%QR%STA%3MN%U3V%%%
03A%?AW?6%X<?3@A8IOYZ8!
6!
0[R%\%]^_C%SA`=aY%Trong!mặt!phẳng!với!trục!toạ!độ!Oxy!cho!hình!chữ!nhật!ABCD!có!tâm!I.!Gọi!
M!là!trung!điểm!cạnh!CD,!H!là!hình!chiếu!vuông!góc!của!D!trên!AM!và!N!là!trung!điểm!AH.!
Tìm!toạ!độ!các!đỉnh!B,D!biết!phương!trình!đường!tròn!ngoại!tiếp!tam!giác!IMN!là!

(C ) : (x −
5
2
)
2
+ ( y−
9

2
)
2
=
5
2
!và!đỉnh!D!có!hoành!độ!nguyên!nằm!trên!đường!thẳng!

x − 2y = 0
.!
!
Đường!tròn!(C)!có!tâm

J (
5
2
;
9
2
)
,!bán!kính!

R =
10
2
.!
!!+)!Ta!có!tứ!giác!IMDN!nội!tiếp!đường!tròn:!
Chứng)minh.)
Gọi!E!là!trung!điểm!đoạn!HD,!ta!có!

NE = IM =
1
2
AD
NE / /IM
⎧
⎨
⎪
⎪
⎪
⎩
⎪
⎪
⎪
!nên!
tứ!giác!IMEN!là!hình!bình!hành.!
Do!đó!E!là!trực!tâm!tam!giác!MND,!và!

EM ⊥ ND ⇒ IN ⊥ ND
.!!!
Hay!góc!

IND
!
= IMD
!
= 90
0
,!do!vậy!IMDN!nội!tiếp!đường!tròn.!!
+)!Do!D!thuộc!(C)!nên!toạ!độ!D!thoả!mãn!hệ:

(x −
5
2
)
2
+ (y−
9
2
)
2
=
5
2
y − 2x = 0
⎧
⎨
⎪
⎪
⎪
⎩
⎪
⎪
⎪
⇔
x = 3, y = 6
x =
8
5
, y =

16
5
⎡
⎣
⎢
⎢
⎢
⎢
⇒ D(3;6)
.!
+)!Do!J!là!trung!điểm!của!ID!nên!I(2;3),!và!I!là!trung!điểm!của!BD!nên!B(1;0).!
!W?%@RP86!Vậy!

B(1;0),D(3;6)
.!!
! !
03b%JY!Nếu!cho!thêm!giả!thiết!điểm!

H (
63
17
;
82
17
)
!ta!tìm!được:

A(−1;2),B(1;0),C (5;4),D(3;6)
.!!
0[R%D%]C_c%SA`=aY!Tìm!số!hạng!chứa!

x
10
!trong!khai!triển!

(
nx
5
−
1
x
2
)
n
!biết!n!là!số!tự!nhiên!thoả!
mãn!

n
2
+C
n
2
= 145 (n ≥ 2)
.!!
+)!Ta!có:!
!

n
2
+

n(n −1)
2
= 145 ⇔ 3n
2
−n −290 = 0 ⇔
n =10(t / m)
n = −
29
3
(l )
⎡
⎣
⎢
⎢
⎢
⎢
.!
+)!Vì!vậy!

(
nx
5
−
1
x
2
)
n
= (2x −
1

x
2
)
10
= C
10
k
(2x )
10−k
(−
1
x
2
)
k
k=0
10
∑
= C
10
k
.(−1)
k
.2
10−k
.x
10−3k
k=0
10
∑

.!
Chọn!

10−3k = 10 ⇔ k = 0
suy!ra!số!hạng!cần!tìm!là!

C
10
0
.2
10
= 1024
.!!!!!
!
!"#$%&'('%)*%+",+% /0%&'1%2%+3456%)789%+3:83%;<=%
">?@A8B6%CDEF%GFF%GCG%% )H89%IJ%83K=%L%3MN%OA83%83P8%QR%STA%3MN%U3V%%%
03A%?AW?6%X<?3@A8IOYZ8!
7!
0[R%^C%]^_C%SA`=aY!Cho!x,y,z!là!các!số!thực!dương!thoả!mãn!

z = min x, y, z
{ }
.!Tìm!giá!trị!nhỏ!
nhất!của!biểu!thức!

P =
x
y + z
+
y

z + x
+
3
2
.
z
x + y
−
z
z
2
+ xy + yz + zx
.!
Đặt!

a =
x
z
,b =
y
z
(a,b ≥1)
!khi!đó:!
!

P =
a
b +1
+
b

a +1
+
3
2(a + b)
⎡
⎣
⎢
⎢
⎤
⎦
⎥
⎥
−
1
(a +1)(b +1)
.!
+)!Ta!có:!

(a +1)(b +1) ≥ 2,∀a,b ≥1 ⇒−
1
(a +1)(b +1)
≥−
1
2
,!và!!
!

a
b +1
+

b
a +1
+
3
2(a + b)
= (a + b +1)(
1
b +1
+
1
a +1
) +
3
2(a + b)
− 2
≥
4(a + b +1)
a + b + 2
+
3
2(a + b)
− 2
.!
Đặt!

t = a + b ≥ 2
,!ta!có:!!!

a
b +1

+
b
a +1
+
3
2(a + b)
≥ f (t) =
4(t +1)
t + 2
+
3
2t
− 2 ≥ f (2) =
7
4
.!
Do!đó!

P ≥ f (t) −
1
2
≥
5
4
.!!Dấu!bằng!đạt!tại!

x = y = z
.!Vậy!giá!trị!nhỏ!nhất!của!P!bằng!5/4.!!!
0dN3%G6!Sử!dụng!bất!đẳng!thức!AM!–GM!ta!có:!!

z
z
2
+ xy + yz + zx
=
z
(z + x)(z + y)
≤
1
2
z
z + x
+
z
z + y
⎛
⎝
⎜
⎜
⎜
⎜
⎞
⎠
⎟
⎟
⎟
⎟
.!
Suy!ra!!

x
y + z
+
y
z + x
−
z
z
2
+ xy + yz + zx
≥
x
y + z
+
y
z + x
−
1
2
z
z + x
+
z
z + y
⎛
⎝
⎜
⎜
⎜
⎜

⎞
⎠
⎟
⎟
⎟
⎟
=
1
2
2 y − z
z + x
+
2x − z
z + y
⎛
⎝
⎜
⎜
⎜
⎜
⎞
⎠
⎟
⎟
⎟
⎟
.!
Sử!dụng!bất!đẳng!thức!Cauchy!–!Shcwarz!ta!có:!
!

2y − z
z + x
+
2x − z
z + y
=
(2y − z)
2
(2y − z)(z + x )
+
(2x − z )
2
(2x − z )(z + y)
≥
(2y − z + 2x − z)
2
(2y − z)(z + x )+ (2x − z )(z + y)
=
4(x + y − z )
2
4xy + z(x + y) −2z
2
≥
4(x + y − z )
2
(x + y)
2
+ z(x + y)− 2z
2
=

4(x + y − z )
x + y + 2z
.!
Vì!vậy!

P ≥
2(x + y − z)
x + y + 2z
+
3
2
.
z
x + y
.!Đặt!

t =
x + y
z
≥ 2
ta!có!

P ≥ f (t ) =
2(t −1)
t + 2
+
3
2t
.!
Xét!hàm!số!f(t)!ta!có:

f '(t) =
6
(t + 2)
2
−
3
2t
2
=
9t
2
−12t −12
t
2
(t + 2)
2
=
3(t − 2)(3t + 2)
t
2
(t + 2)
2
≥ 0,∀t ≥ 2
.!!
!
!
!"#$%&'('%)*%+",+% /0%&'1%2%+3456%)789%+3:83%;<=%
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03A%?AW?6%X<?3@A8IOYZ8!

8!
!
Vì!vậy!

P ≥ f (t) ≥ f (2) =
5
4
.!Dấu!bằng!đạt!tại!

x = y = z
.!Vậy!giá!trị!nhỏ!nhất!của!P!bằng!5/4.!
e:A%?PU%?Qf89%?g%h%%
e:A%Oi%C^6%Cho!x,y,z!là!các!số!thực!dương!thoả!mãn!

x, y, z ≤1; x + y ≥ z +1
.!Tìm!giá!trị!nhỏ!nhất!
của!biểu!thức!

P =
x
y + z
+
y
z + x
−
z
z
2
+ xy + yz + zx
.!Đ/s:!

P
min
=
1
2
.!!
e:A%Oi%CG6!Cho!x,y,z!là!các!số!thực!dương!thoả!mãn!

z = min x, y, z
{ }
.!Tìm!giá!trị!nhỏ!nhất!của!
biểu!thức!

P =
x
y + z
+
y
z + x
+
5
4
.
z
x + y
−
z
z
2

+ xy + yz + zx
.!Đ/s:!

P
min
=
9
8
.!!!!!
!
!
!