CHƯƠNG 1: CÁC BÀI TOÁN VỀ HÀM SỐ
BÀI 1. PHƯƠNG PHÁP HÀM SỐ
I. TÍNH ĐƠN ĐIỆU, CỰC TRỊ HÀM SỐ, GIÁ TRỊ LỚN NHẤT & NHỎ NHẤT
CỦA HÀM SỐ
1.y=fxab⇔
( )
x x a b∀ < ∈
( ) ( )
f x f x<
2.y=fxab⇔
( )
x x a b∀ < ∈
( ) ( )
f x f x>
3.y=fxab⇔ƒ′x≥∀x∈abƒ′x=
∈ab
4.y=fxab⇔ƒ′x≤∀x∈abƒ′x=
∈ab
5.Cực trị hàm số: !"#
( )
k
x x f x
′
= ⇔
$%&
k
x
6. '(#)*&+!,&-!
• './y=ƒx)01#02ab3"#
( )
n
x x a b∈
45
[ ]
( )
( )
( )
( ) ( )
{ }
67 67 8
n
x a b
f x f x f x f a f b
∈
=
[ ]
( )
( )
( )
( ) ( )
{ }
6 6
n
x a b
f x f x f x f a f b
∈
=
• 9y=fx2ab3:
[ ]
( ) ( )
[ ]
( ) ( )
6 8 67
x a b
x a b
f x f a f x f b
∈
∈
= =
• 9y=fx2ab3:
[ ]
( ) ( )
[ ]
( ) ( )
6 8 67
x a b
x a b
f x f b f x f a
∈
∈
= =
b
j j j
x x x
− ε + ε
i i i
x x x
− ε + ε
a
x
• !;&
( )
f x x= α + β
#0<
[ ]
8a b
(#)*&(#,&(=>
a; b
??@ AB9'@ C@ D6EFG?H9IJK9@ AB9'LMN9 GOL@ AB9'LMN9
1.9P-QRS#:ux=vx)!<!<-
( )
y u x=
+*
( )
y v x=
2.9P-&QRS#:ux≥vx)!
Q=<!RST+*Q=
( )
y u x=
UVQW#0
<+*Q=
( )
y v x=
3.9P-&QRS#:ux≤vx)!
Q=<!RST+*Q=
( )
y u x=
UVQW%R*<+*Q=
( )
y v x=
4.9P-QRS#:ux=m)!<!
<-RXy=m+*
( )
y u x=
5.G@Lux≥m>∀x∈?⇔
( )
?
6
x
u x m
∈
≥
6.G@Lux≤m>∀x∈?⇔
( )
?
67
x
u x m
∈
≤
7.G@Lux≥mPx∈?⇔
( )
?
67
x
u x m
∈
≥
8.G@Lux≤mPx∈?⇔
( )
?
6
x
u x m
∈
≤
III. Các bài toán minh họa phương pháp hàm số
Bài 1. Y<!
( )
Zf x mx mx= + −
a.L:mQRS#:ƒx=Px∈283
b.L:m&QRS#:ƒx≤P>∀x∈28[3
c.L:m&QRS#:ƒx≥Px∈
[ ]
8Z−
Giải: a.G$QRS#:ƒx=5
( )
( )
( )
( )
Z Z
Z Z
f x mx mx m x x g x m
x x
x
= + − = ⇔ + = ⇔ = = =
+
+ −
\ƒx=Px∈283:
[ ]
( )
[ ]
( )
8
8
6 67
x
x
g x m g x
∈
∈
≤ ≤
Z
]
m⇔ ≤ ≤
b.L∀x∈28[3:
( )
Z f x mx mx= + − ≤
⇔
( )
Zm x x+ ≤ ⇔
( )
[ ]
Z
8[
g x m x
x x
= ≥ ∀ ∈
+
[ ]
( )
8[
6
x
g x m
∈
⇔ ≥
^<
( )
( )
Z
g x
x
=
+ −
.#028[30_⇔
[ ]
( )
( )
8[
6 [
]
x
g x g m
∈
= = ≥
c.L+*x∈
[ ]
8Z−
:
( )
Z f x mx mx= + − ≥
⇔
( )
Zm x x+ ≥
\`
( )
[ ]
Z
8Z
g x x
x x
= ∈ −
+
ab(c.de_5
f9
x =
:&QRS#:#V!
Zm = ≥
0+gP
f9
(
]
8Zx∈
:G@L
⇔
( )
g x m≤
P
(
]
8Zx∈
(
]
( )
8Zx
Min g x m
∈
⇔ ≤
^<
( )
( )
Z
g x
x
=
+ −
.
(
]
8Z
0_
(
]
( ) ( )
8Z
Z
h
x
Min g x g m
∈
⇔ = = ≤
α
β
b
x
a
v(x)
u(x)
a
b
x
y = m
f9
[
)
8x∈ −
:
x x+ <
0G@L
( )
g x m⇔ ≥
P
[
)
8x∈ −
[
)
( )
8
Max g x m
−
⇔ ≥
L
( )
( )
( )
[ ]
Z
8
x
g x x
x x
− +
′
= ≤ ∀ ∈ −
+
^<
( )
g x
0
[
)
( )
( )
8
ZMax g x g m
−
= − = − ≥
Kết luận:ƒx≥Px∈
[ ]
8Z−
(
]
)
8 Z 8
h
m
⇔ ∈ −∞ − +∞
U
Bài 2. L:m&QRS#:5
Z
Z
Z x mx
x
−
− + − <
P>∀x≥
Giải:G@L
( )
Z
Z [
Z Z mx x x m x f x x
x
x x
⇔ < − + ∀ ≥ ⇔ < − + = ∀ ≥
L
( )
h h
[
[ [
f x x x
x x x x x
−
′
= + − ≥ − = >
÷
_#
( )
f x
d
iYGL
( ) ( )
( )
Z Z
Z
x
f x m x f x f m m
≥
⇔ > ∀ ≥ ⇔ = = > ⇔ >
Bài 3. L:m&QRS#:
( )
[
x x
m m m
+
+ − + − >
>
x∀ ∈ ¡
Giải: \`
x
t = >
:
( )
[
x x
m m m
+
+ − + − >
>
x∀ ∈ ¡
( ) ( )
( )
[ [ [ m t m t m t m t t t t⇔ + − + − > ∀ > ⇔ + + > + ∀ >
( )
[
[
t
g t m t
t t
+
⇔ = < ∀ >
+ +
L
( )
( )
[
[
t t
g t
t t
− −
′
= <
+ +
0
( )
g t
#0
[
)
8+∞
_
#_⇔
( ) ( )
t
Max g t g m
≥
= = ≤
Bài 4. L:mQRS#:5
( )
h [x x x m x x+ + = − + −
P
Giải: \jcP
[x≤ ≤
G$@L
( )
h [
x x x
f x m
x x
+ +
⇔ = =
− + −
Chú ý:9W
( )
f x
′
#7b%&:<(#&QTQ%k=)l
Thủ thuật:\`
( ) ( )
Z
g x x x x g x x
x
′
= + + > ⇒ = + >
+
( ) ( )
h [
h [
h x x x h x
x x
−
′
= − + − > ⇒ = − <
− −
E_#5
( )
g x >
+!d8
( )
h x
m+!._
( )
h x
>
+!d
⇒
( )
( )
( )
g x
f x
h x
=
dE_#
( )
f x m=
P
[ ]
( )
[ ]
( ) ( )
( )
[ ]
( )
8[
8[
87 8 [ h 8m f x f x f f
⇔ ∈ = = −
Bài 5.L:m &QRS#:5
( )
Z
Z
Z x x m x x+ − ≤ − −
P
Giải:\jcP
x ≥
9e.+G@L+*
( )
Z
x x+ − >
;Rn
&QRS#:
( )
( )
( )
Z
Z
Z f x x x x x m= + − + − ≤
\`
( ) ( )
( )
Z
Z
Z 8 g x x x h x x x= + − = + −
L
( ) ( )
( )
Z o 8 Z
g x x x x h x x x
x x
′ ′
= + > ∀ ≥ = + − + >
÷
−
^<
( )
g x >
+!d
x∀ ≥
8
( )
h x >
+!d0
( ) ( ) ( )
f x g x h x=
d
x∀ ≥
4&QRS#:
( )
f x m≤
P
( )
( )
Z
x
f x f m
≥
⇔ = = ≤
Bài 6. L:m
( ) ( )
[ o x x x x m+ − ≤ − +
P>
[ ]
[ox∀ ∈ −
Cách 1. G@L
( ) ( ) ( )
[ of x x x x x m⇔ = − + + + − ≤
>
[ ]
[ox∀ ∈ −
( )
( ) ( )
( )
( ) ( )
[ o [ o
x
f x x x x
x x x x
− +
′
= − + + = − + = ⇔ =
÷
+ − + −
I;Q.0_#67
[ ]
( )
( )
[o
oMax f x f m
−
= = ≤
Cách 2.\`
( ) ( )
( ) ( )
[ o
[ o h
x x
t x x
+ + −
= + − ≤ =
L
[t x x= − + +
4&QRS#:#V!
[ ]
( )
[ ]
[ 8h [ 8 8ht t m t f t t t m t≤ − + + ∀ ∈ ⇔ = + − ≤ ∀ ∈
L5
( )
f t t
′
= + >
⇒
( )
f t
d0
( )
[ ]
8 8hf t m t≤ ∀ ∈ ⇔
[ ]
( ) ( )
8h
7 h of t f m= = ≤
Bài 7. L:m
Z o ] Z x x x x m m
+ + − − + − ≤ − +
>
[ ]
Zox∀ ∈ −
Giải:
\`
Z o t x x= + + − >
⇒
( )
( ) ( )
Z o p Z ot x x x x= + + − = + + −
⇒
( ) ( ) ( ) ( )
p p Z o p Z o ]t x x x x≤ = + + − ≤ + + + − =
( ) ( )
( )
] Z Z o p 8 Z8Z
x x x x t t
⇒ + − = + − = − ∈
ab
( ) ( ) ( ) ( )
Z8Z
p
8 8 Z8Z 7 Z Z
f t t t f t t t f t f
′
= − + + = − < ∀ ∈ ⇒ = =
_
( )
Z8Z
7 Z q f t m m m m m
⇔ = ≤ − + ⇔ − − ≥ ⇔ ≤ − ≥
Bài 8. (Đề TSĐH khối A, 2007)
L:mQRS#:
[
Z x m x x− + + = −
P"
Giải: \45
x ≥
$QRS#:
[
Z
x x
m
x x
− −
⇔ − + =
+ +
\`
[
)
[
[
x
u
x x
−
= = − ∈
+ +
4
( )
Z g t t t m= − + =
L
( )
o
Z
g t t t
′
= − + = ⇔ =
^<_0=
Z
m⇔ − < ≤
Bài 9. (Đề TSĐH khối B, 2007):YT#U5q*r
m >
QRS#:
( )
] x x m x+ − = −
)g>PQeP
Giải:\jcP5
x ≥
G$QRS#:5
( ) ( ) ( )
o x x m x⇔ − + = −
( ) ( ) ( )
o x x m x⇔ − + = −
( )
( )
( )
Z Z
o Z q 7 o Zx x x m x x x m⇔ − + − − = ⇔ = = + − =
_
( )
g x m⇔ =
>Pc<.
( )
8+∞
L;+;_5
( ) ( )
Z [ g x x x x
′
= + > ∀ >
^<
( )
g x
!
( )
g x
)01+!
( )
( )
8 )
x
g g x
→+∞
= = +∞
0
( )
g x m=
>P∈
( )
8+∞
q;_
m∀ >
QRS#:
( )
] x x m x+ − = −
PQeP
fss
7f
Bài 10. (Đề TSĐH khối A, 2008)L:mQRS#:>P"QeP5
[ [
o ox x x x m+ + − + − =
Giải:\`
( )
[ ]
[ [
o o 8 8o f x x x x x x= + + − + − ∈
L5
( )
( ) ( )
( )
Z Z
[ [
8o
o
o
f x x
x x
x x
′
= − + − ∈
÷
÷
−
−
\`
( )
( ) ( )
( )
( )
Z Z
[ [
8 o
o
o
, xu x v x
x x
x x
= − = − ∈
−
−
( ) ( )
( )
( ) ( )
( ) ( )
( )
o
u x v x x
u v
u x v x x
> ∀ ∈
⇒ = =
< ∀ ∈
( )
( )
o
f x x
f x x
f
′
> ∀ ∈
′
⇒ < ∀ ∈
′
=
9:GGL@LPQeP
⇔
[
o o Z om+ ≤ < +
Bài 11. (Đề TSĐH khối D, 2007):
L:mPQRS#:P
Z Z
Z Z
h
h
x y
x y
x y m
x y
+ + + =
+ + + = −
Giải:\`
8u x v y
x y
= + = +
(
)
(
)
Z
Z
Z
Z Zx x x x u u
x x x
x
+ = + − × + = −
+!
8 u x x x v y y
x x x y y
= + = + ≥ = = + ≥ =
4P#V!
( )
Z Z
h
h
]
Z h
u v
u v
uv m
u v u v m
+ =
+ =
⇔
= −
+ − + = −
⇔
u v
)!P-QRS#:;
( )
h ]f t t t m= − + =
PP
( )
f t m⇔ =
P
t t
,t
8 t t≥ ≥
xofsf(x)
[
o o
+
I;QG.0-!
( )
f t
+*
t ≥
−∞
s h f
∞
( )
f t
′
– –
+
( )
f t
f
∞
u[
f
∞
9:.0PP
u
[
m⇔ ≤ ≤ ∨ ≥
Bài 12.(Đề 1I.2 Bộ đề TSĐH 1987-2001):
L:x&QRS#:
( )
< x x y y+ + + ≥
>+*
y∀ ∈ ¡
Giải: \`
< u y y
= + ∈ −
G@L
( ) ( )
( )
( )
6
u
g u x u x u g u
∈ −
⇔ = + + ≥ ∀ ∈ − ⇔ ≥
^<
( )
y g u=
)!<X+*
u
∈ −
0
( )
6
u
g u
∈ −
≥
( )
( )
g x x x
x x x
g
− ≥ − + ≥ ≥ +
⇔ ⇔ ⇔
+ + ≥ ≤ −
≥
Bài 13.Y<
Z
a b c
a b c
≥
+ + =
YT#U5
[a b c abc+ + + ≥
Giải:G\L
( ) ( ) ( )
[ Z [a b c bc abc a a a bc⇔ + + − + ≥ ⇔ + − + − ≥
( ) ( )
o h f u a u a a⇔ = − + − + ≥
#<
(
)
( )
Z
[
b c
u bc a
+
≤ = ≤ = −
9R
( )
y f u=
)!<X+*
( )
8 Z
[
u a
∈ −
L
( )
(
)
( )
(
)
( ) ( )
Z
o h 8 Z
[ [
f a a a f a a a= − + = − + ≥ − = − + ≥
0_#
( )
8f u ≥
( )
8 Z
[
u a
∀ ∈ −
q;_
[a b c abc+ + + ≥
\XT7._#
a b c⇔ = = =
Bài 14. (IMO 25 – Tiệp Khắc 1984):
Y<
a b c
a b c
≥
+ + =
YT#U5
u
u
ab bc ca abc+ + − ≤
Giải:
( ) ( ) ( ) ( ) ( ) ( ) ( )
a b c a bc a a a bc a a a u f u+ + − = − + − = − + − =
\
( ) ( ) ( )
y f u a u a a= = − + −
+*
(
)
( )
[
a
b c
u bc
−
+
≤ = ≤ =
)!<X+*(
#=>
( ) ( )
( )
u
[ u
a a
f a a
+ −
= − ≤ = <
+!
( )
(
)
( )
(
)
(
)
Z
u u
[ [ u [ Z Z u
f a a a a a− = − + + = − + − ≤
^<
( )
y f u=
)!<X+*
( )
8
[
u a
∈ −
+!
( )
u
u
f <
8
( )
(
)
u
[ u
f a− ≤
0
( )
u
u
f u ≤
\XT7._#
Z
a b c⇔ = = =
Bài 15. YT#U5
( ) ( )
[a b c ab bc ca+ + − + + ≤ ∀
[ ]
a b c∈
Giải: G$&XT+j!;&ab, c
( ) ( ) ( )
[ ]
[ f a b c a b c bc a b c= − − + + − ≤ ∀ ∈
\
( )
y f a=
)!<X+*
[ ]
a∈
0
( ) ( )
( )
{ }
67 8 f a f f≤
L
( ) ( ) ( )
( )
( )
[ ]
[ [8 [ [ [ f b c f bc f a a b c= − − − ≤ = − ≤ ⇒ ≤ ∀ ∈
Bài 16. Y6M5
( ) ( ) ( ) ( )
[ ]
a b c d a b c d a b c d− − − − + + + + ≥ ∀ ∈
Giải: G%k&XT+j!;&a, b, c, d, 5
( ) ( ) ( ) ( )
[ ]
( ) ( ) ( )
[ ]
f a b c d a b c d b c d a b c d
= − − − − + − − − + + + ≥ ∀ ∈
\
( )
[ ]
y f a a= ∀ ∈
)!<X0
[ ]
( ) ( )
( )
{ }
6 6
a
f a f f
∈
=
L
( )
[ ]
f b c d b c d= + + + ≥ ∀ ∈
( ) ( ) ( ) ( ) ( ) ( ) ( )
[ ]
( ) ( )
f b c d b c d g b c d b c d c d
= − − − + + + ⇔ = − − − + − − + +
\
( )
[ ]
y g b b= ∀ ∈
)!<X0
[ ]
( ) ( )
( )
{ }
6
b
g b Min g g
∈
=
L
( )
( ) ( ) ( )
8 g c d g c d c d cd
= + + ≥ = − − + + = + ≥
⇒
( ) ( )
[ ]
f g b b= ≥ ∀ ∈
q;_
( )
f a ≥
_Q
BÀI 2. TÍNH ĐƠN ĐIỆU CỦA HÀM SỐ
A. TÓM TẮT LÝ THUYẾT.
1.y=fxab⇔ƒ′x≥∀x∈abƒ′x=
∈ab
2.y=fxab⇔ƒ′x≤∀x∈abƒ′x=
∈ab
Chú ý: L#<RS#:Q$gc/%11. 2.<(!v_w,
jcPƒ′x=∈ab
CÁC BÀI TẬP MẪU MINH HỌA
Bài 1. L:m
( ) ( )
o h Z
mx m x m
y
x
+ + − −
=
+
#02+∞
Giải: !#02+∞⇔
( )
u
mx mx
y x
x
+ +
′
= ≤ ∀ ≥
+
⇔
( )
u u mx mx m x x x
+ + ≤ ⇔ + ≤ − ∀ ≥
⇔
( )
u
u x m x
x x
−
= ≥ ∀ ≥
+
( )
6
x
u x m
≥
⇔ ≥
L5
( )
( )
u
x
u x x
x x
+
′
= > ∀ ≥
+
⇒ux#02+∞⇒
( )
( )
u
6
Z
x
m u x u
≥
−
≤ = =
Bài 2. L:m
( ) ( )
Z
Z [
Z
y x m x m x
−
= + − + + −
#0Z
Giải. !d#0Z⇔
( ) ( )
( )
Z Zy x m x m x
′
= − + − + + ≥ ∀ ∈
^<
( )
y x
′
)01x=+!x=Z0⇔y′≥∀x∈2Z3
⇔
( )
[ ]
Z Zm x x x x+ ≥ + − ∀ ∈
⇔
( )
[ ]
Z
Z
x x
g x m x
x
+ −
= ≤ ∀ ∈
+
[ ]
( )
Z
67
x
g x m
∈
⇔ ≤
L5
( )
( )
[ ]
]
Z
x x
g x x
x
+ +
′
= > ∀ ∈
+
⇒gx#02Z3⇒
[ ]
( ) ( )
Z
67 Z
u
x
m g x g
∈
≥ = =
Bài 3. L:m
( ) ( )
Z
Z
Z Z
m
y x m x m x= − − + − +
#0
[
)
+∞
Giải: !d
[
)
+∞
⇔
( ) ( )
Z y mx m x m x
′
= − − + − ≥ ∀ ≥
⇔
( )
o m x x x
− + ≥ − + ∀ ≥
⇔
( )
( )
o
x
g x m x
x
− +
= ≤ ∀ ≥
− +
L5
( )
( )
o Z
Z
x x
g x
x x
− +
′
= =
− +
Z o
Z o
x x
x x
= = −
⇔
= = +
8
( )
)
x
g x
→∞
=
LxGGL⇒
( )
( )
67
Z
x
g x g m
≥
= = ≤
Bài 4.
( )
( ) ( )
Z
u u Zy x mx m m x m m= − − − + + − −
[
)
+∞
Giải: !d#0
[
)
+∞
( )
Z u u y x mx m m x
′
⇔ = − − − + ≥ ∀ ≥
L
( )
u Z Zm m
′
= − +V
(
)
Z Z
u
[
m
= − + >
0
y
′
=
P
x x<
G@Lgx≥SjP')!5
L
( )
y x
′
≥
>
x∀ ≥
⇔
[
)
G+∞ ⊂
( )
( )
h
h
Z Z Z h
o
Z
m
x x y m m m
S m
m
′
∆ >
− ≤ ≤
′
⇔ < ≤ ⇔ = − + + ≥ ⇔ ⇔ − ≤ ≤
<
= <
Bài 5. L:m
( )
x m x m
y
x m
+ − + +
=
−
#0
( )
+∞
Giải: !#0
( )
+∞
⇔
( )
[
x mx m m
y x
x m
− + − −
′
= ≥ ∀ >
−
⇔
( )
( )
[
g x x
g x x mx m m x
m
x m
≥ ∀ >
= − + − − ≥ ∀ >
⇔
≤
− ≠
Cách 1:Phương pháp tam thức bậc 2
L5
( )
m
′
∆ = + ≥
_#gx=P
x x≤
G@Lgx≥SjP')!5
Lgx≥>∀x∈+∞⇔
( )
G+∞ ⊂
( )
( )
o Z
Z
Z
m
m
x x g m m m
m
S
m
′
≤
≤ ∆ ≥
⇔ ≤ ≤ ⇔ = − + ≥ ⇔ ⇔ ≤ −
≤ −
≥ +
= − ≤
Cách 2:Phương pháp hàm số
L5g′x=[x−m≥[x−m∀xm⇒gx#02+∞
x
x
x
x
xyfYL
^<
( )
( )
( )
o
Z
6
Z
Z
x
g m m
m
g x
m
m
m
m
m
≥
= − + ≥
≤ −
≥
⇔ ⇔ ⇔ ≤ −
⇔
≥ +
≤
≤
≤
Bài 6. L:m
( ) ( )
[ h < Z Z y m x m x m m= − + − + − +
.
x∀ ∈ ¡
Giải:i0=!<(
( )
h [ Z y m x m x
′
⇔ = − + − ≤ ∀ ∈ ¡
( ) ( )
[ ]
h [ Z 8g u m u m u⇔ = − + − ≤ ∀ ∈ −
^<
( )
[ ]
8y g u u= ∈ −
)!<X0_
( )
( )
o ]
[
Z
g m
m
g m
− = − ≤
⇔ ⇔ ≤ ≤
= − + ≤
Bài 7. L:m!
Z
[ p
y mx x x x= + + +
d+*r
x∈ ¡
Giải: i0=!<(
< < <Z
Z
y m x x x x
′
⇔ = + + + ≥ ∀ ∈ ¡
⇔
( ) ( )
Z
< < [< Z<
Z
m x x x x x+ + − + − ≥ ∀ ∈ ¡
( )
[ ]
Z
[
Z
m u u g u u⇔ ≥ − − + = ∀ ∈ −
+*
[ ]
< u x= ∈ −
L
( ) ( )
[ 8
g u u u u u u u
′
= − − = − + = ⇔ = − =
I;QGGL_#_0=!<(⇔
[ ]
( )
( )
h
67
o
x
g u g m
∈ −
= − = ≤
Bài 8. Y<!
( ) ( ) ( )
Z
Z
Z
y m x m x m x m= + + − − + +
L:mc<.-!%!U[
Giải. ab
( ) ( ) ( )
Z y m x m x m
′
= + + − − + =
^<
u Z m m
′
∆ = + + >
0
y
′
=
P
x x<
4<.-!%!U[
[ ]
8 8 8 [y x x x x x
′
⇔ ≤ ∀ ∈ − =
m⇔ + >
+!
[x x− =
L
[x x− = ⇔
( ) ( )
( )
( )
( )
[ [ Z
o [
m m
x x x x x x
m
m
− +
= − = + − = +
+
+
( ) ( ) ( ) ( )
[ Z m m m m⇔ + = − + + +
u o
Z u
o
m m m
±
⇔ − − = ⇔ =
cnQ+*
m + >
_#
u o
o
m
+
=
B. ỨNG DỤNG TÍNH ĐƠN ĐIỆU CỦA HÀM SỐ
I. DẠNG 1: ỨNG DỤNG TRONG PT, BPT, HỆ PT, HỆ BPT
Bài 1. '.QRS#:5
h Z
Z [ x x x+ − − + =
Giải. \jcP5
Z
x ≤
\`
( )
h Z
Z [ f x x x x= + − − + =
L5
( )
[
Z
h Z
Z
f x x x
x
′
= + + >
−
⇒fx#0
(
Z
−∞
6`c(f −=0QRS#:fx=P%_&x=−
Bài 2. '.QRS#:5
h Z ]x x x+ = − + +
Giải. G&QRS#:⇔
( )
Z ] hf x x x x= − + + − +
=
f9
Z
x ≤
:fxz⇒+gP
f9
Z
x >
:
( )
Z
Z
] h
f x x x
x x
′
= + − > ∀ >
÷
+ +
⇒fx#0
(
)
Z
+∞
!f =0>Px=
Bài 3. '.&QRS#:5
Z h
[
h u u h Z u ]x x x x+ + − + − + − <
{
Giải. \jcP
h
u
x ≥
\`
( )
Z h
[
h u u h Z uf x x x x x= + + − + − + −
L5
( )
( ) ( )
Z [
h
Z
[
h u Z
h Z u
Z h u [ u h
f x
x
x
x x
′
= + + + >
+
× −
× − × −
⇒fx#0
)
h
u
+∞
6!fZ=]0{⇔fxzfZ⇔xzZ
q;_P-&QRS#:t<)!
h
Z
u
x≤ <
Bài 4. '.@L5
Z
h [ Z h u u
Z o
x x x x
x x x
x x x+ + + = + + − + − +
{
Giải. {
( )
(
)
( ) ( )
( )
Z
h [ Z h u u
Z o
x x x
x x x x
f x x x x g x
⇔ = + + + − − − = − + − + =
Lfx+!g′x=−ox
+x−uz∀x⇒gx
9P-fx=gx)!<!<-
( ) ( )
+!y f x y g x= =
^<fxd8gx.+!
( ) ( )
Zf g= =
0{P%_&x=
Bài 5. L:m67
( )
< < m x x x x x x
+ + ≤ + + + ∀
{
Giải. \`
( )
< < t x x t x x x= + ≥ ⇒ = + = +
⇒
t≤ ≤
⇒
t≤ ≤
c{
⇔
( )
m t t t t
+ ≤ + + ∀ ∈
⇔
( )
t t
f t m t
t
+ +
= ≥ ∀ ∈
+
⇔
( )
6
t
f t m
∈
≥
^<
( )
( )
t t
f t
t
+
′
= >
+
0ft
⇒
( )
( )
Z
6
t
f t f
∈
= =
⇒
Z
m ≤
⇒
Z
67
m =
Bài 6. '.QRS#:
<
] ] <
x x
x− =
< <
] ] < ] ] <
x x x x
x x x x− = − ⇔ + = +
{
ab
( )
]
u
f u u= +
L
( )
] )
u
f u u
′
= + >
E_#
( )
f u
{
( ) ( )
< < < f x f x x x x⇔ = ⇔ = ⇔ =
[
k
x k
π π
⇔ = + ∈ ¢
Bài 7. L:
( )
x y∈ π
,tP
< <
Z h
x y x y
x y
− = −
+ = π
Giải.
< < < < x y x y x x y y
− = − ⇔ − = −
ab!`#R
( )
( )
< f u u u u= − ∈ π
L
( )
f u
u
′
= + >
E_#
( )
f u
#0
( )
π
4
( )
( )
[
Z h
f x f y
x y
x y
=
π
⇔ = =
+ = π
Bài 8. '.PQRS#:
Z
Z
Z
x y y y
y z z z
z x x x
+ = + +
+ = + +
+ = + +
{
Giải. ab
( )
Z
f t t t t= + +
+*
t ∈ ¡
⇒
( ) ( )
f t t t
′
= + + >
⇒ftd
4g&W$v(./x≤y≤z
⇒
( )
( )
( )
f x f y f z≤ ≤
⇒
z x y z x y+ ≤ + ≤ + ⇔ ≤ ≤
⇒x=y=z=±
Bài 9. '.P&QRS#:
Z
Z
Z
x x
x x
+ − <
− + >
Giải.
Z
Z
x x x+ − < ⇔ − < <
\`
( )
Z
Z f x x x= − +
L5
( ) ( ) ( )
Z f x x x
′
= − + <
⇒
( )
f x
.+!
( )
(
)
(
)
Z u Z
f x f x> = > ∀ ∈ −
II. DẠNG 2: ỨNG DỤNG TRONG CHỨNG MINH BẤT ĐẲNG THỨC
Bài 1.YT#U5
Z Z h
Z| Z| h|
x x x
x x x− < < − +
∀xm
Giải
Z
Z|
x
x x− <
∀xm⇔
( )
Z
Z|
x
f x x x= − + >
∀xm
L
( )
<
|
x
f x x
′
= − +
⇒
( )
f x x x
′′
= −
⇒
( )
< f x x
′′′
= − ≥
∀xm
⇒
( )
f x
′′
2f∞⇒
( ) ( )
f x f
′′ ′′
> =
∀xm
⇒
( )
f x
′
2f∞⇒
( ) ( )
f x f
′ ′
>
}∀xm
⇒
( )
f x
2f∞⇒fxmf}∀xm⇒Q
Z h
Z| h|
x x
x x< − +
∀xm⇔gx}
h Z
h| Z|
x x
x x− + − >
∀xm
Lg′x}
[
<
[| |
x x
x− + −
⇒g′′x}
Z
Z|
x
x x− +
}fxm∀xm
⇒g′x2f∞⇒g′xmg′}∀xm
⇒gx2f∞⇒gxmg}∀xm⇒Q
Bài 2.YT#U5
x
x x
π
> ∀ ∈
÷
π
Giải.
x x
x f x
x
> ⇔ = >
π π
∀x∈
π
÷
abT<!
<
g x
x x x
f x
x x
−
′
= =
Ve_cWPgx}x<x−x
Lg′x}<x−xx−<x}−xxz∀x∈
π
÷
⇒gx.#0
π
÷
⇒gxzg}
⇒
( )
g x
f x
x
′
= <
∀x∈
π
÷
⇒f x.#0
π
÷
⇒
( )
(
)
f x f
π
> =
π
⇔
x
x x
π
> ∀ ∈
÷
π
Bài 3.YT#U5
) )
x y x y
x y
+ −
>
−
∀xmym
Giải. ^<xmym)xm)y⇔)x−)ym0$&XT
⇔
) ) )
x
x y yx
x y
x
x y y
y
−
−
− > × ⇔ > ×
+
+
⇔
)
t
t
t
−
> ×
+
+*
x
t
y
=
m
⇔
)
t
f t t
t
−
= − × >
+
∀tmL
( )
( )
( )
( )
[
t
f t
t
t t t
−
′
= − = >
+ +
∀tm
⇒ft2f∞⇒ftmf}∀tm⇒Q
Bài 4.YT#U5
) ) [
y x
y x y x
− >
÷
− − −
( )
x y
x y
∀ ∈
≠
Giải. abc.de_5
f9ymx:⇔
( )
) ) [
y
x
y x
y x
− > −
− −
⇔
) [ ) [
y x
y x
y x
− > −
− −
f9yzx:⇔
( )
) ) [
y
x
y x
y x
− < −
− −
⇔
) [ ) [
y x
y x
y x
− < −
− −
ab!`#Rft}
) [
t
t
t
−
−
+*t∈
L
( )
( )
[
t
f t
t t t t
−
′
= − = >
− −
∀t∈⇒ft
⇒fymfxymx+!fyzfxyzx ⇒Q
Bài 5.YT#U5
b a
a b<
∀amb≥~
Giải. a
b
zb
a
⇔)a
b
z)b
a
⇔b)aza)b⇔
) )a b
a b
<
ab!`#Rfx}
) x
x
∀x≥~
L
) )
x e
f x
x x
− −
′
= ≤ =
⇒fx2~f∞
⇒fazfb⇔
) )a b
a b
<
⇔a
b
zb
a
Bài 6. (Đề TSĐH khối D, 2007)
YT#U
( ) ( )
b a
a b
a b
a b+ ≤ + ∀ ≥ >
Giải. G$&XT
( ) ( )
[ [
b a
b a
a b
a b
a b a b
+ +
+ ≤ + ⇔ ≤
÷ ÷
( ) ( ) ( ) ( )
( ) ( )
) [ ) [
[ [ ) [ ) [
a b
b a b a
a b a b
a b
+ +
⇔ + ≤ + ⇔ + ≤ + ⇔ ≤
ab!`#R<+
( )
( )
) [
x
f x
x
+
=
+*
x >
L
( )
( ) ( )
( )
[ ) [ [ ) [
[
x x x x
x
f x
x
− + +
′
= <
+
( )
f x⇒
.#0
( )
( ) ( )
f a f b+∞ ⇒ ≤
Bài 7. (Bất đẳng thức Nesbitt)
YT#U5
Z
a b c
b c c a a b
+ + ≥
+ + +
∀abm
Giải. 4g&W$v(./a≥b≥\`x}a⇒x≥b≥m
L⇔f x}
x b c
b c c x x b
+ +
+ + +
+*x≥b≥m
⇒
( ) ( ) ( ) ( )
b c b c
f x
b c b c
x c x b b c b c
′
= − − > − − =
+ +
+ + + +
⇒fx2bf∞⇒
b c
f x f b
b c
+
≥ =
+
\`x}b⇒x≥m7b!gx}
x c
x c
+
+
+*x≥m
⇒
( )
c
g x
x c
′
= >
+
∀m⇒gx2f∞⇒
Z
g x g c
≥ =
Z
LxZ_#
Z
a b c
b c c a a b
+ + ≥
+ + +
∀abm
Bình luận:G&XTNesbitt#d1905+!)!&XT#&$#<
c•L#0e_)!(T&XT!_#<45 (T
Gr7~c.<=_-((T#<(5“Những viên kim
cương trong bất đẳng thức Toán học”-(.%<NXB Tri thứcQ(!(3/2009
BÀI 3. GIÁ TRỊ LỚN NHẤT, NHỎ NHẤT CỦA HÀM SỐ
A. GIÁ TRỊ LỚN NHẤT, NHỎ NHẤT CỦA HÀM SỐ
I. TÓM TẮT LÝ THUYẾT
1. Bài toán chung:L:(#,&<`)*&-!
( )
f x
GR*5^"<(+!T
( ) ( )
8f x c f x c≥ ≤
GR*5Y€#jcP-
( )
f x c=
2. Các phương pháp thường sử dụng
@RSQ(Q5G$!$(:QRS
@RSQ(Q5LT;
@RSQ(QZ5E/%1&XT$5Côsi; Bunhiacôpski
@RSQ(Q[5E/%1<!
@RSQ(Qh5E/%1$)Rn(
@RSQ(Qo5E/%1QRSQ(Q+bS+!Pr
@RSQ(Qu5E/%1QRSQ(Q:r+!Pr
II. CÁC BÀI TẬP MẪU MINH HỌA:
Bài 1.
L:(#,&-Pxy}x
fy
−oxyf]x−]yf
Giải. G$T%R*%
Pxy}x−Zyf[
fy−
fZ≥Z
Lx_#6@xy}Z⇔
Z [
y y
x y x
− = =
⇔
− + = = −
Bài 2.
Y<xymL:(#,&-5S}
[
[
[ [
y y y
x x x
y x
y x y x
+ − − + +
Giải.
y y yx
x x
S
y x
y x y x
= − + − − + + + +
÷
÷
S
y y yx x
x
y x y x
y x
= − + − + − + + − +
÷
÷ ÷
÷
S
y y x yx
x
y x xy
y x
−
= − + − + − + + ≥
÷
÷
÷
q*x}ym:6E}
Bài 3.
L:(#)*&-!
S x y x y
= + + +
Giải .
S x y x y
= + + +
}
< <
<
yx
x y
−−
+ + − +
S
p
< < < < < <
[ [
x y x y x y x y x y x y
= − + − − + = − + + − + +
S
p p
< <
[ [ [
x y x y x y
= − − + + − − ≤
q*
Z
x y k
π
= = + π k∈:
p
67
[
S =
Bài 4. L:(#,&-T
Z ] Z o u u ] ]
S x x x x x x x x x x x x x= + + + + − + + + + +
Giải.
Z Z [ [ h
Z [ Z h [
[ Z o [ ] h
S x x x x x x x x
= − + − + − + − +
÷ ÷
÷ ÷
h o o u u ] ]
o h u o ] u p ] [ [
o u [ ] o p p p
x x x x x x x
+ − + − + − + − − ≥ −
÷ ÷ ÷ ÷
q*
Z o u u ] ]
o u ]
8 88 8 8
Z u ] p
x x x x x x x x x= = = = =
:
[
6
p
S = −
Bài 5.Y<
x y z ∈ ¡
L:(#,&-T5
E}px
fh[y
foz
−oxz−[yfZoxy
Giải. G$E⇔fx}px
−]z−]yxfh[y
foz
−[y
L∆′
x
}gy}]z−]y
−h[y
foz
−[y}−uy
fo]zy−[z
⇒∆′
y
}][z
−u[z
}−o[[z
≤∀z∈M⇒gy≤∀yz∈M
E_#∆′
x
≤∀yz∈M⇒fx≥q*
x y z= = =
:
MinS =
Bài 6.Y<x
fxyfy
}ZL:(#)*&+!,&-T5
E}x
−xyfy
Giải aby}⇒x
}Z⇒E}Z)!(#-!
aby≠c$T%R*%e_
( )
Z
x y x y
x xy yS t t
u u
x xy y x y x y t t
− +
− + − +
= = = = =
+ + + + + +
+*
x
t
y
=
⇔ut
ftf}t
−tf⇔u−t
fuftfu−}{
f9u}:t}⇒x}y}
Z±
⇒u})!(#-!
f9u≠:u;Q(#!⇔QRS#:{Pt
⇔∆}Zu−Z−u≥⇔
Z
Z
u≤ ≠ ≤
q;_;Q(#-u)!
Z
Z
⇒
6
Z
u =
867u}Z
6E}⇔
6
Z
u =
⇔t}⇒
Z
x y
x y
x xy y
=
⇔ = = ±
+ + =
67E}p⇔67u}Z⇔t}−⇒
Z Z
Z
Z Z
x y
x y
x xy y
x y
= −
= = −
⇔
+ + =
= − =
Bài 7.Y<xy∈M,tjcP
( ) ( )
[ x y x y x y− + + − + =
L:(#)*&,&-TE}
x y+
Giải. G$
( ) ( ) ( )
[ x y x y x y x y− + − + + − + =
⇔
( ) ( )
Z [ x y x y x+ − + + + =
⇔
( ) ( )
Z [x y x y x+ − + + =−
^<−[x
≤0
( ) ( )
Z x y x y
+ − + + ≤
⇔
Z h Z h
x y
− +
≤ + ≤
q*x}y}
Z h
−
±
:
Z h
6
x y
−
+ =
q*x}y}
Z h
+
±
:
Z h
67
x y
+
+ =
Bài 8
.
L:(#,&-!
( )
[ f x x x x= + + +
Giải.
'ry
)!(#-!fx
⇒x
<<y
}
[ x x x+ + +
⇔
[ [ y x x x y y x x x x− = + + ⇒ − + = + +
⇔gx
}
Z x y x y+ + + − =
Lgx}Px
⇔∆′}
Z y y y y+ − − = + −
}
y y+ − ≥
^<y
}
Z Z Z x x x x x x x+ + + ≥ + = + ≥
0
∆′≥⇔y
−≥⇔
y ≥
q*x}
−
:6fx}
Bài 9.Y<
( )
h [ y f x x x mx
= = − + +
L:((#-m<<
6 y
>
Giải. L
( )
( )
( )
( )
( )
h [ 8 7 [ 5
h [ 8 [ 5
x m x x P
f x
x m x x P
+ − + ≤ ∨ ≥
=
− + + − ≤ ≤
'rP)!-y}fx⇒P}P
∪P
cP#<(:%
e_
Hoành độ của các điểm đặc biệt trong đồ thị (P):
<!<P
P
x
A
}8x
B
}[8 <!€P
5
h
C
m
x
−
=
9:+!<7b(c.d5
9x
C
∈2x
A
x
B
3⇔m∈2−ZZ3:6fx}6{ff[}
46fxm⇔
Z Z
[ [
m
f m
f m
− ≤ ≤
= >
= >
⇔zm≤Z
9x
C
∉2x
A
x
B
3⇔m∉2−ZZ3:6fx}
( )
h
C
m
f x f
−
=
÷
}
p
[
m m
− + −
46fxm⇔
2 ZZ3
Z h Z
Z
m
m
m m
∉ −
⇔ < < +
− + <
Kết luận
:
Lx+!_#6fxm⇔
325m1
+<<
Bài 10. (Đề thi TSĐH 2005 khối A)
Y<
x y z >
8
[
x y z
+ + =
L:6-E
x y z x y z x y z
= + +
+ + + + + +
Giải:E/%1&XTYg<(a, b, c, d > 5
( )
(
)
[
[
o
[ [ oa b c d abcd
a b c d abcd a b c d a b c d
+ + + + + + ≥ = ⇒ + + + ≥
+ + +
o o
o o
o o
o [ o 6
x x y z x x y z x y z
x y y z x y y z x y z
x y z z x y z z x y z
S
x y z x y z x y z x y z
+ + + ≥ =
+ + + + +
+ + + + ≥ =
+ + + + +
+ + + ≥ =
+ + + + +
= + + ≥ + + ⇒ =
÷ ÷
+ + + + + +
•
G
Y
P
P
•
G
Y
P
P
•
G
Y
P
P
Bài 11. (Đề thi TSĐH 2007 khối B)
Y<
x y z >
L:6-E
y
x
z
x y z
yz zx xy
= + + + + +
÷
÷ ÷
Giải: E/%1&XTYg<p
E
[ [ [
p
[ [ [
p p p
6
y y x y z
x x
z z
x y z S
yz yz zx zx xy xy
x y z
= + + + + + + + + ≥ = ⇒ =
÷
Bài 12.
Y<
x y
x y
>
+ =
L:(#,&-S}
yx
x y
+
− −
Giải:
( ) ( ) ( )
yx
S y x x y x y x y x y
y x
= + + + − + ≥ + − + = +
÷ ÷
6`c(S}
yx
x y
+
− −
}
y x
y x
− −
+
}
( )
x y
x y
+ − +
÷
÷
E_#S≥
x y
+
≥
[
xy x y
≥ =
+
⇒
S ≥
⇒6S}
Bài 13.
Y<xy‚mL:67-5S}
( )
( )
xyz x y z x y z
x y z xy yz zx
+ + + + +
+ + + +
Giải: E/%1&XTCôsi +!BunhiaCôpski Z((5
Z
Zx y z x y z+ + ≥ ×
8
ƒ6Gƒ^ƒv<Z
Z
Z
Z Zxy yz zx xy yz zx x y z+ + ≥ =
( )
( )
Zx y z x y z x y z+ + ≤ + + + + = + +
Lx_#
( )
( )
Z Z
Z
Z
Z
Z Z Z Z
Z Z p
Z
Z
xyz x y z xyz xyz
S
xyz
x y z x y z x y z
+ + +
+ + +
≤ = × ≤ × =
+ + + +
Bài 14. (Đề thi TSĐH 2003 khối B)
L:(#)*&,&-!
[y x x= + −
Cách 1: L;Q7(
[ ]
8D = −
8
8 [
[
x
y y x x
x
′ ′
= − = ⇔ = −
−
[
x
x
x x
≥
⇔ ⇔ =
= −
⇒
7
y
y
=
= −
Cách 2: \`
8
x u u
π π
= ∈ −
⇒
( )
(
)
< 8
[
y u u u
π
= + = + ∈ −
8
7 8 y y= = −
Bài 15. (Đề dự bị TSĐH 2003 khối B)
L:(#)*&,&-
( )
Z
o
[ y x x= + −
#0<
[ ]
8−
Cách 1.\`
[ ]
8u x= ∈
L
( )
Z
Z Z
[ Z [y u u u u u= + − = − + − +
[ ]
p [ 8 8
Z
y u u u u
′
= − + − = ⇔ = ∈ = >
9:.0
[
7 [8
p
y y= =
Cách 2. \`
o o
[ <x u y u u= ⇒ = +
xy′−+
y[
x−y′f −
y
−
( ) ( )
o o o
< Z< < Z [u u u u u= + + ≤ + + =
q*
x =
:
7 [y =
E/%1&XTYg5
o o
Z
o o
Z
] ] ] ]
[
Z
u u u u Z
[ [ [ [ [
[< Z [< <
u u u u Z
u u u
u u u
+ + ≥ × × × =
+ + ≥ × × × =
( )
o o
]
[ [ [
[ < <
p Z Z p
y u u u u y= + + ≥ + = ⇒ ≥
q*
[
Z p
x y= ⇒ =
Bài 16.I;Q.0+!:(#)*&-!
Z
x
y
x
+
=
+
Y<
a b c+ + =
YT#U5
a b c+ + + + + ≥
Giải. La\5
D = ¡
8
( )
(
)
Z
Z Z
x
y x y
x x
−
′
= = ⇔ = ⇒ =
+ +
( ) ( )
Z Z
) ) ) )
x x x x
x x x x
x
y
x
x
x
x
→∞ →∞ →∞ →∞
+ +
= = =
+
+
E_#
) 8 )
x x
y y
→+∞ →−∞
= = −
9:GGL
Z
7
x
y y
x
+
= ≤ ⇒ =
+
L~<Q=:
y x≤ ∀
⇔
Z x x x+ ≤ + ∀
\`P&XT!_((#
x a x b x c= = =
5
5 Z
5 Z
5 Z
x a a a
x b b b
x c c c
= + ≤ +
= + ≤ +
= + ≤ +
( )
p a b c a b c
+ + + ≤ + + + + +
⇔
a b c
≤ + + + + +
Cách 2. L#0`QXr„7_`
( ) ( ) ( )
8 8 8 8 8OA a AB b BC c= = =
uur uuur uuur
4
( )
8 ZOC OA AB BC a b c= + + = + +
uuur uur uuur uuur
^<
OA AB BC OA AB BC OC+ + ≥ + + =
uur uuur uuur uur uuur uuur uuur
Lx_#
a b c+ + + + + ≥
Bài 17. (Đề 33 III.2, Bộ đề thi TSĐH 1987 – 1995)
Y<
x y+ =
L:676-A=
x y y x+ + +
Giải. 1. L:MaxA: E/%1&XTBunhiaCôpski
A≤
( )
( )
( )
( )
x y y x x y x y
+ + + + = + + ≤ + + = +
q*
x y= =
:67A=
+
2. L:MinA:ab#RnQe_
…Trường hợp 159
xy ≥
7bc.d5
xZ y′f −
y
−
a
a+ba+b+c
C
A
B
Z
O x
y
f9
x y≥ ≥
:•m⇒
6 A >
f9x≤y≤:
|A|≤
[ ]
x y x y x y+ + + + = + +
}
( )
x y x y− − ≤ − + =
Lxc.dt7b_#+*
xy ≥
:6A}−
…Trường hợp 25ab
xy <
5\`
x y t+ =
⇒
t
xy
−
= <
⇒
( )
t ∈ −
( )
( )
( )
( )
( )
A x y xy x y y x xy x y xy x y xy
= + + + + + + = + + + + + +
=
t t t
t t
− − −
+ × + × + +
( )
t
t
−
= + + +
⇔
( )
( ) ( )
Z
A f t t t t
= = + + − + + −
L5
( )
( )
Z
8
Z
f t t t t t t t
+
+ +
′
= + − = ⇔ = = − = = −
L
t t
+!<Q=%R-
( )
f t
<
( )
f t
′
⇒
( )
( )
( )
p Z
8
u
f t f t
−
= =
9:.0_#5
( ) ( )
A f t A f t≤ ⇒ ≥ −
_#
( )
( )
p Z
6
u
A f t
−
= − = − < −
7._#⇔
x y t+ =
8
t
xy
−
=
⇒xy)!P-
Z
Z p
u u
+ −
+ + =
⇒
( )
h
o
x y
− + ± −
=
Kết luận: 67A=
+
8
( )
p Z
6
u
A
−
= −
Bài 18.Y<
[ ]
x y z ∈
<.tjcP5
Z
x y z+ + =
L:676-T5
( )
<S x y z= + +
Giải. ^<
[ ]
x y z ∈
0
Z
x y z x y z
π
< + + < + + = <
q:!
<y = α
#0
(
)
π
0!<(#V!
1. L:MaxS_:Min
( )
x y z+ +
( )
( )
( )
Z
Z [
x y z x y z x y z+ + = + + + + ≥ + + =
q*
x y z= = =
:67E}
Z
<
[
2. L:MinS _:Max
( )
x y z+ +
Cách 1: Phương pháp tam thức bậc hai:
4g&W$v(./
{ }
8
z Max x y z z
= ⇒ ∈
G$+!((R+j
T;‚
( )
(
)
( )
Z p
Z
[
x y z z x y xy z z z z f z+ + = + + − ≥ + − = − + =
t−t
t
ƒ′+−+ƒ
^<!_}f‚)!Q#<)v_j)†)0#005
( )
(
)
( )
{ }
(
)
( )
h
67 67 8
[
f z f f f f= = = =
q*
8 8
z x y= = =
:6E}
h
<
[
Cách 2: Phương pháp hình học
abPr\j(+g„7_‚L;QnQ(
( )
M x y z
<.tjcP
[ ]
x y z ∈
U#<:);QQRS•GY^•′G′Y′„+*•8G8Y
8^8•′8G′8Y′
6`c(%<
Z
x y z+ + =
0
( )
M x y z
U#0`QX@5
Z
x y z+ + =
q;_;QnQ(
( )
M x y z
<.tjcP.U#0%Pƒ?‡4I9+*
(ƒ?‡4I9)!#(:);QQRS'r„′)!:-„)0
ƒ?‡4I9:„′)!e-:);QQRS+!ˆ)!e-)1(jƒ?‡4I9L„′6
)!:-„6)0ƒ?‡4I9^<„6
}
x y z+ +
0„6)*&⇔„′6)*&
⇔6#‰+*#<o€ƒ?‡4I9
Lx_#5
(
)
h
[ [
x y z OK+ + ≤ = + =
( )
(
)
h
< <
[
x y z⇒ + + ≥
q*
8 8
z x y= = =
:6E}
h
<
[
Bài 19. Y<
a,b,c 0
>
,tjcP
3
a b c
2
+ + ≤
L:(#,&-
S a b c
b c a
= + + + + +
Giải. Sai lầm thường gặp:
Z
o
Z ZS a b c a b c
b c a b c a
≥ + × + × + = + + +
÷ ÷ ÷
o
o
Z Z ] Z 6 Z a b c S
b c a
≥ × × × × × × = = ⇒ =
÷ ÷ ÷
÷ ÷ ÷
• Nguyên nhân:
Z
6 Z Z
S a b c a b c
a b c
= ⇔ = = = = = = ⇒ + + = >
el+*.
• Phân tích và tìm tòi lời giải :
^<E)!T7T+*abc0%"<(6E
a b c
= = =
Sơ đồ điểm rơi:
y
Z
„
ƒ
4
Z
‡
6
z
x
?
I
9
Z
„′
a b c
= = =
⇒
[
[
a b c
a b c
= = =
= = =
α
α α α
⇒
[
[
=
α
⇒
16
α =
Cách 1:G$+!/%1&XTCôsi ta có
o o o
o o o o o o
S a b c
b b c c a a
= + + + + + + + + + + +
1 4 4 2 4 4 3 1 4 4 2 4 4 3 1 4 44 2 4 4 43
u u u
u u u
o Z o Z o Z ] o ] o ] o
u u u u
o o o o o o
a b c a b c
b c a b c a
≥ × + × + × = + +
Z
u u u u
] o ] o ] o ] h h h
u Z Z u
o o o o
a b c
b c a a b c
≥ × × × =
(
)
u
h h
u
Z u Z u Z u
Z
a b c
a b c
= ≥ ≥
×
+ +
×
q*
a b c
= = =
:
Z u
6
S
=
Cách 2:G$+!/%1&XTBunhiaCôpski
( )
( )
( )
[
[
u u
[
[
u u
[
[
u u
a a a
b
b b
b b b
c
c c
c c c
a
a a
+ = × + + ≥ × +
÷ ÷
+ + = × + + ≥ × +
÷ ÷
+ = × + + ≥ × +
÷ ÷
⇒
[ [ [
u
S a b c
a b c
≥ × + + + + +
÷
h
[ [ [ [
u
a b c
a b c a b c
= × + + + + + + + +
÷
o Z
Z
h [h
o Z Z
[ [ [ [ [
u u
abc
a b c a b c
abc
≥ × × × × + × × × = + ×
÷
÷
[h [h Z u
Z Z
[ [
u u
Z
a b c
≥ + × ≥ + × =
÷
÷
+ +
÷
q*
a b c
= = =
:
Z u
6
S
=
Cách 3:\`
(
)
(
)
(
)
8 8 u a v b w c
b c a
= = =
uur uur uuur
^<
u v w u v w+ + ≥ + +
uur uur uuur uur uur uuur
0_#5
( )
S a b c a b c
a b c
b c a
= + + + + + ≥ + + + + +
÷
=
( )
h
o o
a b c
a b c a b c
+ + + + + + + +
÷ ÷
≥
( )
(
)
Z
h
Z
[
o
a b c
a b c a b c
+ + × × + + + × × ×
÷
≥
( )
Z
Z
Z
Zh
Z Z
o
abc
a b c
abc
× × × × × × + ×
≥
(
)
p Zh
o
Z
a b c
+ ×
+ +
≥
p Zh ] Zh hZ Z u
[
o [ [ [
+ × = + = =
q*
a b c= = =
:
Z u
6
S =
B. CÁC ỨNG DỤNG GTLN, GTNN CỦA HÀM SỐ
I. ỨNG DỤNG TRONG PHƯƠNG TRÌNH, BẤT PHƯƠNG TRÌNH
Bài 1. '.QRS#:5
[ [
[ x x− + − =
Giải. \`
( )
[ [
[f x x x= − + −
+*
[x≤ ≤
( )
( ) ( )
Z Z
[ [
Z
[
[
f x x
x x
′
= − = ⇔ =
− −
9:GGL_#5
( ) ( )
[ ]
Z [f x f x≥ = ∀ ∈
⇒@RS#:
( )
[ [
[ f x x x= − + − =
P%_&x=Z
Bài 2. '.QRS#:5
Z h o
x x
x+ = +
Giải. @L ⇔
( )
Z h o
x x
f x x= + − − =
L5
( )
Z ) Z h ) h o
x x
f x
′
= + −
⇒
( ) ( ) ( )
Z ) Z h ) h
x x
f x
′′
= + >
x∀ ∈ ¡
⇒ƒ′x
6`c(ƒ′x)01+!
( )
) Z ) h o f
′
= + − <
( )
Z) Z h) h o f
′
= + − >
⇒@RS#:ƒ′x=>Px
9:.0_#5
@RS#:
( )
Z h o
x x
f x x= + − − =
cgv(P
6!
( )
( )
f f= =
0QRS#:>P
x =
+!
x =
Bài 3. L:mG@L5
pm x x m+ < +
P>
x∀ ∈ ¡
Giải.
pm x x m+ < +
⇔
( )
p m x x+ − <
⇔
( )
p
x
m f x
x
< =
+ −
L5
( )
( )
p p
p p
x
f x
x x
− +
′
=
+ + −
=⇔
p p ox x+ = ⇔ = ±
( )
) )
p
x x
f x
x
x
→+∞ →+∞
= =
+ −
8
( )
) )
p
x x
f x
x
x
→−∞ →−∞
− −
= =
+ +
9:GGL
( )
f x m>
x∀ ∈ ¡
⇔
( ) ( )
Z Z
6 o
[ [
x
f x f m m
∈
−
= − = − > ⇔ <
¡
x−∞x
+∞f′−+
f
ƒx
x−∞−oo+∞f′−f−
ƒ
xZ[ƒ′−+
ƒ
Bài 4. L:m@L5
( )
<x m x+ = +
P
x
π π
∈ −
Giải. ^<
x
π π
∈ −
⇒
[ [
x −π π
∈
0`
[ ]
x
t = ∈ −
⇒
<
t
x
t
−
=
+
8
t
x
t
=
+
4⇔
( ) ( )
< <x x m x+ = +
⇔
( )
( )
t t t
m f t t t m
t t
+ − −
= + ⇔ = + − =
÷ ÷
+ +
L5
( )
( )
( )
8 f t t t t t t
′
= + − − = ⇔ = = −
⇒G.0
9:.0_#5
\P
[ ]
t ∈ −
:
[ ]
( )
[ ]
( )
6 67
t
t
f t m f t
∈ −
∈ −
≤ ≤
⇔
[ m m≤ ≤ ⇔ ≤ ≤
q;_P
x
π π
∈ −
:
[ ]
8 m∈
Bài 5. L:mPG@L5
Z
Z
[
x x
x x x m m
− ≤
− − − + ≥
P
Giải. ⇔
( )
Z
Z
[
x
f x x x x m m
≤ ≤
= − − ≥ −
L5
( )
[
)
(
]
Z [ [ 8
Z [ [ 8Z
x x x
f x
x x x
+ − ∀ ∈
′
=
− + ∀ ∈
8
ƒ′x=⇔
Z
x =
9:GGL_#5
[ ]
( ) ( )
8Z
67 Z
x
f x f
∈
= =
\P:
[ ]
( )
8Z
67 [
x
f x m m
∈
≥ −
⇔
[ m m− ≤
⇔−Z≤m≤u
Bài 6. L:m≥P5
Z
Zh
< o
[
ZZ
< o
[
x y m m m
x y m m
= − − +
= − +
P
Giải
⇔
Z
Z
< < u
< <
x y x y m m
x y x y m m
+ = − +
− = − +
⇔
( )
( )
Z
Z
u
x y m m
x y m m
+ = − +
− = − +
ab
( )
Z
uf m m m= − +
L5
( )
Z f m m m
′
= − = ⇔ = >
9:GGL_#5ƒm≥ƒ=∀m≥
cnQ+*
( )
x y+ ≤
_#P
P:m=cP#V!5
( )
( )
x y
x y
+ =
− =
P
8
Z o
x y
π π
= =
q;_P
⇔
m=
xZf′−++fYL]
t−ƒ′t−+
ƒt[
[
m+∞ƒ′−+
ƒu+∞
II. ỨNG DỤNG GTLN, GTNN CHỨNG MINH BẤT ĐẲNG THỨC
Bài 1. YT#U5
( )
) x x x x+ + + ≥ +
x∀ ∈ ¡
G\L⇔
( )
( )
) f x x x x x= + + + − + ≥
x∀ ∈ ¡
L5
( )
( )
) f x x x x
′
= + + = ⇔ =
⇒G.0
9:.0_#5
( ) ( )
f x f≥ =
⇒Q
Bài 2. Y<
a b c
a b c
>
+ + =
Y6M5T=
Z Z
a b c
b c c a a b
+ + ≥
+ + +
L5T=
( ) ( ) ( )
a b c a b c
a b c
a a b b c c
+ + = + +
− − −
− − −
ab!
( )
( )
f x x x= −
+*xm
L
( )
Z
Z
f x x x
′
= − = ⇔ = >
9:.0⇒
( )
Z Z
f x x≤ ∀ >
45
( ) ( ) ( )
( )
Z Z Z Z
a b c
T a b c
f a f f c
= + + ≥ + + =
\XT7._#
Z
a b c⇔ = = =
Bài 3. Y<Z≤n)ŠYT#U5∀x≠5
( ) ( )
Z
| | | Z| |
n n
x x x x x
x x
n n
+ + + + − + − + − <
\`
( ) ( )
Z
8
| | | Z| |
n n
x x x x x
u x x v x x
n n
= + + + + = − + − + −
L=T
( ) ( ) ( )
f x u x v x=
z
L5
( )
( )
( )
( )
( )
( )
| |
|
| |
|
n n
n n
x x x
u x x u x
n
n
x x x
v x x v x
n
n
−
−
′
= + + + + = −
−
′
= − + − + − = − −
−
⇒
( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )
| |
n n
x x
f x u x v x u x v x u x v x u x v x
n n
′ ′ ′
= + = − − +
⇒
( ) ( ) ( )
[ ]
|
n
x
f x u x v x
n
−
′
= +
( )
[
| | [|
|
n n
x x x x
n
n
−
−
= + + + +
−
^<Z≤n)Š0ƒ′x‰%&+*−x
9:.0_#5
x−∞+∞f′−+
f
x−∞+∞f′+−
f
x−∞+∞f′+−
f
( ) ( )
f x f x< = ∀ ≠
⇒Q
Bài 4.YT#U5
Z Z [ [
Z [
a b a b+ +
≤
∀abm
(
)
(
)
[
[
[ [
[ [ [
[ [
Z Z
Z Z
Z Z Z Z
Z
a
a b t
b
a b t
a
b
+
+ +
≥ ⇔ = ≥
+ +
+
abft}
( )
( )
[
[ [
[
Z
Z
Z
Z
t t
t
t
+ +
=
+
+
+*
a
t
b
= >
f′t}
( ) ( ) ( ) ( )
( )
Z
[ Z Z [ Z
[ [ Z
Z
Z
Z
t t t t t t
t
−
−
+ + − + +
+
( )
( )
( )
( )
Z
Z
[
[
Z
Z
Z
t t t t
t
−
−
+ + −
=
+
f′t}⇔t}⇒G.0-ft
LxGGL⇒
[
Z
≤ftz∀tm⇒
[
[ [
[
Z
Z
Z Z
a b
a b
+
≤
+
⇒
Z Z [ [
Z [
a b a b+ +
≤
^&U7._#⇔a}bm
III. BÀI TẬP VỀ NHÀ
Bài 4. Y<∆•GY
A B C> >
L:(#,&-!5
( )
x A x B
f x
x C x C
− −
= + −
− −
Bài 5. L:676-5 y=
o o
< <x x a x x+ +
Bài 6. Y<ab≠L:6-
[ [
[ [
a b a b a b
y
b a
b a b a
= + − + + +
÷
Bài 7. Y<
x y+ >
L:676-
[
x y
S
x xy y
+
=
+ +
Bài 8. './QRS#:
x px
p
+ + =
Px
x
L:p≠<<
[ [
S x x= +
,&
Bài 9. L:6-
( ) ( ) ( ) ( )
Z Z ] Z Z
x x x x
y
= + + − − + + −
Bài 10. Y<xy≥+!
x y+ =
L:676- Z p
x y
S = +
Bài 11. Y<
x y z+ + =
L:676-
P x y z xy yz zx= + + + + +
Bài 12. L:m@L5
( ) ( )
x x x x m− + + − − + =
P
Bài 10 L:m@L5
p px x x x m+ − = − + +
P
Bài 11 L:m@L5
( )
Z
[ [x x x x x x m− + − − + = − +
[PQeP
Bài 12 L:m@L5
Z
x
x mx
x
−
= − +
−
P%_&
Bài 13 L:m@L5
< [ < m x x x m− + − =
P
(
)
[
x
π
∈
tf∞f
′
−ff