Khoá giải đề THPT Quốc Gia Môn Toán – Thầy Đặng Thành Nam – Mathlinks.vn!
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Bi=)\)j.c*)>%d6kF!#$%!$&'!()!

y = x
3
− (m + 2)x
2
+ (2m +1)x + 2 (1)
*!
"* +$,%!( !(/!0123!.$143!5&!56!78!.$9!$&'!()!:";!5<1!

m = −2
*!!
=* !>?'!'!7@!:";!7A.!B/B!7A1!.A1!71@'!

x
1
C!7A.!B/B!.1@D!.A1!71@'!

x
2
(E%!B$%!


⎜
⎞
⎠
⎟
⎟
⎟
⎟
= 1+ sin x + tan x
*!!!
Bi=)7)j\c*)>%d6kF!>L3$!.LB$!G$M3!

I = 2x −
1
x
⎛
⎝
⎜
⎜
⎜
⎜
⎞
⎠
⎟
⎟
⎟
⎟
ln
2
x dx
1

x
2
−
2
x
3
⎛
⎝
⎜
⎜
⎜
⎜
⎞
⎠
⎟
⎟
⎟
⎟
n
*!!!!
Bi=)^)j\c*)>%d6kF)#$%!$?3$!$XG!B$Y!3$Z.![\#]*[^\^#^]^!BP!.$@!.LB$!0_3J!"*!F`1!aCbCc!Rd3!
RHe.!R&!.KD3J!71@'!B-B!7%A3!.$f3J![[^C#]C[^]*!>L3$!.$@!.LB$!T$)1!.V!g1Q3!\abc!5&!J1-!.K9!
R<3!3$h.!BiE!T$%,3J!B-B$!J1YE!$E1!7Hj3J!.$f3J!#a!5&![^b*!!
Bi=)-)j\c*)>%d6kF!>K%3J!T$U3J!J1E3!5<1!$Q!.KkB!.%A!7X!lWmO!B$%!0E!71@'![:"nono;C!\:onp=no;C!
#:"np=nq;*!r12.!G$HI3J!.K?3$!'s.!G$f3J!71!tDE!0E!71@'![C\C#*!>?'!71@'!]!.K43!.1E!lO!(E%!B$%!
.V!g1Q3![\#]!BP!.$@!.LB$!0_3J!=*!!
Bi=),)j\c*)>%d6kF)>K%3J!'s.!G$f3J!5<1!.KkB!.%A!7X!lWm!B$%!$?3$!B$Y!3$Z.![\#]!BP!

AC = 2BC
G$HI3J!.K?3$!7Hj3J!B$u%![#!R&!

x + y + 2xy
2
x
2
+ y
2
−3xy + 5y = 4 x ( 5y −1 − x )
⎧
⎨
⎪
⎪
⎪
⎪
⎩
⎪
⎪
⎪
⎪
(x, y ∈ !)
*!!!
Bi=)+)j\c*)>%d6kF!#$%!WCmCO!R&!B-B!()!.$/B!gHI3J!.$%,!'S3!

x
2
+ 8y
2
+ 8z
2
= 20
*!>?'!J1-!.K9!R<3!

+ (2m +1)x + 2 (1)
*!
"* +$,%!( !(/!0123!.$143!5&!56!78!.$9!$&'!()!:";!5<1!

m = −2
*!!
=* !>?'!'!7@!:";!7A.!B/B!7A1!.A1!71@'!

x
1
C!7A.!B/B!.1@D!.A1!71@'!

x
2
(E%!B$%!

x
1
2
− x
2
= −1
*!!!!
"* v`B!(13$!./!J1,1*!
=* >E!BPw!

y ' = 3x
2
− 2(m + 2)x + 2m +1; y ' = 0 ⇔
x = 1

z4D!BdD!.K{!.$&3$w!

1
2
−
2m +1
3
= −1 ⇔ m =
5
2
(t / m)
*!
y!b2D!

m <1 ⇒
2m +1
3
<1 ⇒ x
1
=
2m +1
3
,x
2
= 1
*!
z4D!BdD!.K{!.$&3$w!

2m +1
3

Bi=).)j\c*)>%d6kF)
E; F1,1!G$HI3J!.K?3$!

log
18
(3
x
+ 2) =
1− x
2+ log
3
2
*!
0; F1,1!G$HI3J!.K?3$!

sin 2x.sin x −
3π
2
⎛
⎝
⎜
⎜
⎜
⎜
⎞
⎠
⎟
⎟
⎟
⎟

+ 2 = 3
1−x
=
3
3
x
⇔ 3
2x
+ 2.3
x
−3 = 0 ⇔
3
x
= 1
3
x
= −3(l )
⎡
⎣
⎢
⎢
⎢
⇔ x = 0
*!
rZm!G$HI3J!.K?3$!BP!3J$1Q'!gDm!3$h.!

x = 0
*!!
0; x1|D!T1Q3w!


sin2x cos x −sin x
( )
− 2
( )
= 0
⇔
sin x + cos x = 0 (1)
sin2x cos x −sin x
( )
− 2 = 0 (2)
⎡
⎣
⎢
⎢
⎢
*!
y;!>E!BP!

(1) ⇔ sin x = −cos x ⇔ tan x = −1 ⇔ x = −
π
4
+ kπ
*!
y;!>E!BP!

VT
(2)
≤ sin 2x . sin x − cos x − 2≤ 2 −2 < 0
:5U!3J$1Q';*!
5Zm!G$HI3J!.K?3$!BP!3J$1Q'!R&!

>E!BP!

I = 2 x ln
2
x
1
2
∫
K
! "### $###
−
ln
2
x
x
dx
1
2
∫
= 2K − ln
2
xd(ln x )
1
2
∫
= 2K −
ln
3
x
3

⎧
⎨
⎪
⎪
⎪
⎪
⎪
⎩
⎪
⎪
⎪
⎪
⎪
*!
}Dm!KE!

K =
x
2
ln
2
x
2
2
1
− x ln x dx
1
2
∫
M

⎪
⎩
⎪
⎪
⎪
⎪
⎪
⇒ M =
x
2
ln x
2
2
1
−
1
2
x dx
1
2
∫
= 2ln 2−
1
4
x
2
2
1
= 2ln 2−
3

2
3
+ 4 ln
2
2−4ln 2+
3
2
*!!!
Bi=)l)j\c*)>%d6kF!!
B; >?'!B-B!()!.$/B!EC0CB!(E%!B$%!$E1!G$HI3J!.K?3$!

az
2
+ bz + c = 0;cz
2
+ bz + a +16−16i = 0
:N3!
O;!BP!!3J$1Q'!B$D3J!R&!

1+ 2i
*!
g; #$%!3!R&!()!./!3$143!.$%,!'S3!

C
n
2
+ A
n
2
= n

BP!3J$1Q'!:"y=1;!T$1!!!
Khoá giải đề THPT Quốc Gia Môn Toán – Thầy Đặng Thành Nam – Mathlinks.vn!
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•!
!

a(1+ 2i )
2
+ b(1+ 2i) + c = 0
⇔ −3a + b + c + (4a + 2b)i = 0
⇔
−3a + b + c = 0
4a + 2b = 0
⎧
⎨
⎪
⎪
⎩
⎪
⎪
(1)
*!
>HI3J!./!G$HI3J!.K?3$!

cz
2
+ bz + a +16 −16i = 0
!BP!3J$1Q'!:"y=1;!T$1!!


−3n −70 = 0 ⇔
n =10(t / m)
n = −7(l )
⎡
⎣
⎢
⎢
*!
+$1!7P!

x
2
−
2
x
3
⎛
⎝
⎜
⎜
⎜
⎜
⎞
⎠
⎟
⎟
⎟
⎟
10
= C

M (0;0;
c
2
),N (
a
2
;b;0),P(0;
b
2
;
c
2
)
*!
>$~%!J1,!.$12.!.E!BPw!

abc = 1
*!!
y;!>E!BP!

V
BMNP
=
1
6
BM
! "!!
,BN
! "!!
⎡

! "!!
,A' N
! "!!!
⎡
⎣
⎢
⎤
⎦
⎥
=
abc
4b
2
c
2
+ 4a
2
b
2
+ 9a
2
c
2
=
1
4
a
2
+
4

9
b
2
3
= 3 144
3
⇒ d (CM ; A' N ) ≤
1
3 144
3
*!
]hD!0_3J!7A.!.A1!

a
2
=
c
2
=
b
3
=
1
12
3
*!rZm!J1-!.K9!R<3!3$h.!BiE!T$%,3J!B-B$!J1YE!#a!5&![^b!
Khoá giải đề THPT Quốc Gia Môn Toán – Thầy Đặng Thành Nam – Mathlinks.vn!
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„!


AB
! "!!
,AC
! "!!
⎡
⎣
⎢
⎤
⎦
⎥
= (−6;3;2)
R&'!5uB!.I!G$-G!.Dm23!343!BP!
G$HI3J!.K?3$!R&!

−6(x −1)+ 3y + 2z = 0 ⇔ 6x − 3y − 2z −6 = 0
*!
y;!F`1!]:ononE;!5<1!E‡o!.E!BP!

S
ABC
=
1
2
AB
! "!!
,AC
! "!!
⎡
⎣

a = 3(t / m)
a = −9(l )
⎡
⎣
⎢
⎢
*!
rZm!71@'!Bd3!.?'!R&!]:ononq;*!!!!!
Bi=),)j\c*)>%d6kF)>K%3J!'s.!G$f3J!5<1!.KkB!.%A!7X!lWm!B$%!$?3$!B$Y!3$Z.![\#]!BP!

AC = 2BC
G$HI3J!.K?3$!7Hj3J!B$u%![#!R&!

3x − y − 3 = 0
*!F`1!F!R&!.K`3J!.M'!BiE!.E'!J1-B!
[#]!5&!71@'!

H 3;
2
3
⎛
⎝
⎜
⎜
⎜
⎜
⎞
⎠
⎟
⎟

+ b
2
) = ( 3a − b)
2
⇔ 2b
2
+ 2 3ab = 0 ⇔ b(b + 3a) = 0 ⇔
b = 0
b = − 3a
⎡
⎣
⎢
⎢
⎢
*!
y;!b2D!

b = 0 ⇒ CD : x − 3 = 0
*!!
>%A!7X!71@'!#!R&!3J$1Q'!BiE!G$HI3J!.K?3$!

x −3 = 0
3x − y − 3 = 0
⎧
⎨
⎪
⎪
⎩
⎪
⎪

⎜
⎜
⎜
⎞
⎠
⎟
⎟
⎟
⎟
= 0;−2 3
( )
⇒ D(3;0)
*!
xHj3J!.$f3J![]!71!tDE!]!5&!5DU3J!JPB!5<1!#]!BP!G$HI3J!.K?3$w!

y = 0
*!!!!
y;!b2D!

b = − 3a ⇒ CD : x − 3y −1 = 0
*!
>%A!7X!71@'!#!R&!3J$1Q'!BiE!$Q!G$HI3J!.K?3$!

x − 3y −1= 0
3x − y − 3 = 0
⎧
⎨
⎪
⎪
⎪

( )
⇒ D(4; 3)
.!!
Đường!thẳng!AD!đi!qua!điểm!D!và!vuông!góc!với!CD!có!phương!trình:

3x + y −5 3 = 0
.!
HC#)$=;&(!Vậy!có!hai!đường!thẳng!thoả!mãn!yêu!cầu!bài!toán!là

3x + y −5 3 = 0; y = 0
.!!
BI=)J)KLM*)>%N6OF)Giải!hệ!phương!trình!

x
y +1
x +1
+ y
x +1
y +1
=
x + y + 2xy
2
x
2
+ y
2
−3xy + 5y = 4 x ( 5y −1 − x )
⎧
⎨
⎪

+1
⎛
⎝
⎜
⎜
⎜
⎜
⎞
⎠
⎟
⎟
⎟
⎟
+ y
x +1
y +1
+1
⎛
⎝
⎜
⎜
⎜
⎜
⎞
⎠
⎟
⎟
⎟
⎟
⎡

x + y + 2
2
.
x + y + 2xy
(x +1)(y +1)
=
x + y + 2
2(xy + x + y +1)
.(x + y + 2xy)
.!
Dấu!bằng!xảy!ra!kh!và!chỉ!khi!

x = y
.!!
Kết!hợp!với!phương!trình!thứ!nhất!của!hệ!ta!có:!

x + y + 2xy
2
≤
x + y + 2
2(xy + x + y +1)
.(x + y + 2xy) ⇔
x + y + 2
xy + x + y +1
≥1 ⇔ xy ≤1 (1)
.!
Dấu!bằng!xảy!ra!khi!và!chỉ!khi!

xy = 1
.!!

xy = 1
.!Vì!vậy!

x = y
xy =1
(x − y)
2
+ (2 x − 5y −1)
2
= 0
⎧
⎨
⎪
⎪
⎪
⎪
⎩
⎪
⎪
⎪
⎪
⇔
x = 1
y = 1
⎧
⎨
⎪
⎪
⎩
⎪

2
+ ( y + z)
2
100
.!
Lời$giải:$
Khoá giải đề THPT Quốc Gia Môn Toán – Thầy Đặng Thành Nam – Mathlinks.vn!
!"#$%&'()*+,-) ).*.)) /0&1)23)&456)7)489):%&4)&4;&)<=)>?%)489)@4A)))
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7!
Sử!dụng!bất!đẳng!thức!Cauchy!–!Schwarz!ta!có:!
!

x
2
+ ( y + z)
2
≥
1
2
x + y + z
( )
2
.!!
Sử!dụng!bất!đẳng!thức!AM!–GM!ta!có:!
!

2x
2
4x

xyz
2
≤
x
x + 4 −
xyz
2
.!!
Sử!dụng!bất!đẳng!thức!Cauchy!–!Schwarz!ta!có:!
!

x.yz + 2.y + 2.z + 2.1
( )
2
≤ (x
2
+12)( y
2
z
2
+ y
2
+ z
2
+1)
= x
2
+12
( )
y

2
+ 28
3
⎛
⎝
⎜
⎜
⎜
⎜
⎞
⎠
⎟
⎟
⎟
⎟
⎟
3
= 64
⇒ xyz + 2y + 2z ≤ 6 ⇒
x
x + 4 −
xyz
2
≤
x
x + y + z +1
.!
Vì!vậy!

P ≤

f '(t) =
1
(t +1)
2
−
t
100
; f '(t) = 0 ⇔ t(t +1)
2
= 100 ⇔ t = 4
.!
Vì!f’(t)!đổi!dấu!từ!dương!sang!âm!khi!đi!qua!t=4!nên!

P ≤ f (t ) ≤ f (4) =
18
25
.!
Dấu!bằng!đạt!tại!

x = 2; y = z = 1
.!Vậy!giá!trị!lớn!nhất!của!P!bằng!18/25.!!
BP94).()
Sử dụng bất đẳng thức AM-GM cho hai số dương ta được:

x
2
yz ≤ x
2
.
y

2
8
=
3
4
x
2
+ 5;
2x ≤
x
2
2
+ 2 ≤ (
x
4
16
+1) + 2 =
x
4
16
+ 3

Cộng theo vế ba bất đẳng thức cùng chiều ta thu được:

x
2
yz + 2xy + 2xz + 2x ≤ 2x
2
+ 8
.

200
.!Ta!có!kết!quả!tương!tự!trên.
Khoá giải đề THPT Quốc Gia Môn Toán – Thầy Đặng Thành Nam – Mathlinks.vn!
!"#$%&'()*+,-) ).*.)) /0&1)23)&456)7)489):%&4)&4;&)<=)>?%)489)@4A)))
B4%)#%C#()DE#4$%&2:FG&)!
8!
Q4;&)RS#F))
+)!Mấu!chốt!bài!toán!tìm!được!điểm!rơi!

x = 2; y = z = 1
.!
+)!Đánh!giá!

xyz + 2 y + 2z ≤ 6
.!
TU%)#;@)#<V&1)#W)X!Cho!x,y,z!là!các!số!thực!dương!thoả!mãn!

x
2
+ y
2
+ z
2
= 26
.!Tìm!giá!trị!lớn!
nhất!của!biểu!thức!

P = xyz + 32 y + 45z
.!
Đ/S:!