1. I-B畉t 畉ng th畛c c担 si
1.Ch畛ng minh r畉ng
2 2 2
2
a b c a b c
b c c a a b
v畛i a,b,c>0
2.Ch畛ng minh r畉ng
3 3 3
1 1 1 3
2
a b c b c a c a b
v畛i a,b,c>0 v abc =1
3.Cho a,b,c>0 v abc=1.Cm:
3 3 3
a 3
1 1 1 1 1 1 4
b c
b c c a a b
4.Cho k s畛 kh担ng 但m 1 2
, ,..., k
a a a tho畉 1 2... 1
k
a a a
Cm: 1 2 1 2
... ...
m m m n n n
k k
a a a a a a
v畛i ; ,
m n m n N
5.Cho 3 s畛 th畛c x,y,z tho畉 m達n: 2004 2004 2004
3
x y z
.T狸m GTLN c畛a bi畛u th畛c
3 3 3
A x y z
6.Cho a+b+c =0 .Ch畛ng minh r畉ng 8 8 8 2 2 2
a b c a b c
7.Cho s畛 t畛 nhi棚n 2
k . 1 2
, ,..., k
a a a l c叩c s畛 th畛c d動董ng
Cmr: 1 2
1 2
2 3 1
... ...
m
m m
m n m n m n
k
n
n n n
a
a a
a a a
a a a
8.Cho x,y,z l ba s畛 th畛c tho畉 m達n
1 1 1
1
x y z
.T狸m GTNN c畛a bi畛u th畛c
2006 2006 2006
2007 2007 2007
x y z
A
y z x
9.T狸m GTNN c畛a
20 20 20
11 11 11
x y z
A
y z x
v畛i 1
x y z
10.Cho n s畛 th畛c 1 2
, ,..., n
x x x thu畛c o畉n
, , 0
a b a
Cmr:
2
1 2
1 2
1 1 1
... ...
4
n
n
n a b
x x x
x x x ab
11.Cho n l s畛 nguy棚n d動董ng;l畉y
2000;2001
i
x v畛i m畛i i=1,2,n
T狸m GTLN c畛a
1 2 1 2
2 2 ... 2 2 2 ... 2
n n
x x
x x x x
F
12.X辿t c叩c s畛 th畛c 1 2 2006
, ,...,
x x x tho畉 1 2 2006
, ,...,
6 2
x x x
T狸m GTLN c畛a bi畛u th畛c
1 2 2006
1 2 2006
1 1 1
sin sin ... sin ...
sin sin sin
A x x x
x x x
13.Cho n s畛 d動董ng 1 2
, ,..., n
a a a 畉t :
1 2 1 2
min , ,..., , ax , ,...,
n n
m a a a M M a a a
1 1
1
,
n n
i
i i i
A a B
a
.Cmr:
1
B n m M A
mM
2. 14.Cho 0, 0, 1,
i i
a b i n
.Ch畛ng minh r畉ng:
1 1 2 2 1 2 1 2
... ... ...
n n
n
n n n n
a b a b a b a a a bb b
15.Cho 0, 1,
i
a i n
.Ch畛ng minh r畉ng:
1 2 1 2
1 1 ... 1 1 ...
n
n
n n
a a a a a a
16.Ch畛ng minh
1.2... 1 1 1.2...
n
n n n
v畛i 2,
n n N
17.Ch畛ng minh trong tam gi叩c ABC ta c坦 :
1/
3
1 1 1 2
1 1 1 1
sin sin sin 3
A B C
駈 駈
件 件
醐 醐
2/
3
1 1 1 2
1 1 1 1
B C 3
os os os
2 2 2
A
c c c
駈 駈
件 件
件 件
件 件
醐 醐
3/
3
1 1 1 2
1 1 1 1
3
a b c
m m m R
駈 駈
件 件
醐 醐
18.Cho a,b,x,y,z > 0 v x+y+z = 1.Ch畛ng minh:
4
4 4
4
3 3
b b c
a a a a b
x y z
19.Cho
1
, 0, 0 1,.. ; 1
n
i i
i
a b x i n x
. Cmr:
1 2
...
m
m m
m
n
b b b
a a a n a nb
x x x
v畛i m > 0
20.Cho , , 0, 1
a b c a b c
.Ch畛ng minh r畉ng: 3
1 1 1
1 1 1 8
ab bc ca
駈 駈
件 件
醐 醐
21.Cho
;
x a b .T狸m GTLN c畛a bi畛u th畛c ( ) ( ) ( )
m n
F x x a b x
= - - v畛i *
,
m n
22.Cho 0
2
;
x
辿 湛
棚 炭
棚 炭
谷 短
.T狸m GTLN c畛a bi畛u th畛c ( ) p
sin . os
q
F x x c x
= v畛i *
,
p q
23.Cho a,b,c kh担ng 但m v c坦 a + b + c =1.T狸m GTLN c畛a bi畛u th畛c ( ) 30 4 2004
, ,
F a b c a b c
=
24.Cho , 0, 6
x y x y
続 + 贈 .T狸m GTLN c畛a c叩c bi畛u th畛c sau :
1/ ( ) ( )
2002
, . . 6
F x y x y x y
= - -
2/ ( ) ( )
2002
, . . 4
F x y x y x y
= - -
25.X辿t c叩c s畛 th畛c d動董ng th畛a m達n a + b +c =1.T狸m GTNN c畛a bi畛u th畛c
2 2 2
1 1 1 1
P
ab bc ca
a b c
= + + +
+ +
26.X辿t c叩c s畛 th畛c d動董ng th畛a m達n a +b +c + d =1.T狸m GTNN c畛a bi畛u th畛c
2 2 2 2
1 1 1 1 1
P
acd abd abc bcd
a b c d
= + + + +
+ + +
27.Gi畉 s畛 1 2
, ,..., n
x x x >0 th畛a m達n i畛u ki畛n
1
1
1
n
i
i i
x
x
=
=
奪
+
. Cmr:
( )
1
1
1
n
i n
i
x
n
=
贈
-
28.Gi畉 s畛 a,b,c >0 th畛a m達n
2 3
1
1 1 1
a b c
a b c
+ + =
+ + +
. Cmr:
2 3
6
1
5
ab c 贈
3. 29. Gi畉 s畛 1 2
, ,..., n
x x x >0 th畛a m達n i畛u ki畛n
1
1
n
i
i
x
=
=
奪 .Cmr:
( )
1
1
1 1
n
i
n
i i
x
x n
=
贈
- -
30. (QG-98) Gi畉 s畛 1 2
, ,..., n
x x x >0 th畛a m達n i畛u ki畛n
1
1 1
1998 1998
n
i i
x
=
=
奪
+
Cmr: 1 2
. ...
1998
1
n
n
x x x
n
続
-
31.Cho n s畛 d動董ng th畛a m達n i畛u ki畛n
1
1
n
i
i
a
=
<
奪
Cmr:
( )
( )( )( ) ( )
1
1 2 1 2
1 2 1 2
... 1 ... 1
... 1 1 ... 1
n
n n
n n
a a a a a a
a a a a a a n
+
辿 湛
- + + + 脱 旦
谷 短 歎
巽
贈 歎
巽 歎
巽
竪 淡
+ + + - - -
33.Cmr: , 2
n N n
" 続 ta c坦 1 1 2
n n
n n
n n
n n
- + + <
34.Cho [ ]
, , 0;1
x y z .Cmr: ( ) ( )
3 3 3 2 2 2
2 3
x y z x y y z z x
+ + - + + 贈
35. Cho [ ]
, , 0;2
x y z .Cmr: ( ) ( )
6 6 6 4 2 4 2 4 2
2 192
x y z x y y z z x
+ + - + + 贈
36.Cho [ ]
1;2
i
x v畛i i=1,,2000.Th畛a m達n
2000
1
2005
i
i
x
=
=
奪 T狸m GTLN c畛a
2000
3
1
i
i
A x
=
= 奪
37.Ch畛ng minh : 2 2 2
1 1 1
3.2
a b c
ab bc ca
Trong 坦 , , , 0
a b c
38.Cho s畛 d動董ng a .X辿t b畛 s畛 d動董ng x,y,z th畛a m達n i畛u ki畛n:xy + yz + zx = 1
T狸m GTNN c畛a bi畛u th畛c
2 2 2
P a x y z
39.X辿t c叩c s畛 th畛c x,y,z th畛a m達n : 2 2 2 2
16
25
x y z xy a
.Trong 坦 a l m畛t s畛 d動董ng
cho tr動畛c .T狸m GTLN c畛a bi畛u th畛c :P = xy + yz + zx
40.X辿t c叩c s畛 th畛c a,b,c,d th畛a m達n : 2 2 2 2
1
1
2
a b c d
T狸m GTLN v GTNN c畛a :
2 2 2 2
2 2 2 2
P a b c b c d b a c d
41.Cho hm s畛 ( )
f x th畛a m達n pt ( ) 4 4
2 cot
f tg x tg x g x
= +
Cmr: ( ) ( )
sinx cosx 196
f f
+ 続 ( OLP-30-4-99)
II. PH働NG PHP HNH H畛C
1.Cho a,b,c,d l c叩c s畛 th畛c tho畉 m達n 2 2
4
a b
v c+d=4.
T狸m gi叩 tr畛 l畛n nh畉t c畛a bi畛u th畛c P=ac+bd+cd
2.Cho a,b,c,d l c叩c s畛 th畛c tho畉 m達n 2 2
1
a b
v c+d=3Cmr:
9 6 2
ac+bd+cd
4
3(HSG-NA-2005) a,b,c,d l c叩c s畛 th畛c tho畉 m達n 2 2
1
a b
v c-d=3
Cmr:
9 6 2
ac+bd-cd
4
4.Cho c叩c s畛 a,b,c,d,x,y th畛a m達n : 2 2 2 2
40 8 10 ; 12 4 6 ;3 2 13
a b a b c d c d x y
T狸m GTNN c畛a
2 2 2 2
P x a y b x c y d
4. 5.Cho hai s畛 a,b th畛a m達n i畛u ki畛n a - 2b + 2 = 0
Ch畛ng minh r畉ng : 2 2 2 2
6 10 34 10 14 74 6
a b a b a b a b
6.Cho b畛n s畛 a,b,c,d th畛a m達n i畛u ki畛n:a + 2b = 9;c + 2d = 4
Cmr: 2 2 2 2 2 2 2 2
12 8 52 2 2 4 8 20 4 5
a b a b a b c d ac bd c d c d
7.Cho b畛n s畛 th畛c a,b,c,d th畛a m達n : 2 2
6; 1
c d a b
Cmr: 2 2
2 2 18 6 2
c d ac bd
8.Cho a,b,c,d l b畛n s畛 th畛a m達n i畛u ki畛n :
2 2 2 2
2 ; 4 1
a b a b c d c d
Cmr:
4 2 2 2 4 2 2
a b c d
9. .Cho a,b,c,d l b畛n s畛 th畛a m達n i畛u ki畛n : 2 2 2 2
5
a b c d
Cmr:
3 30
5 2 5 2 5
2
a b c d ac bd
.X辿t d畉u b畉ng x畉y ra khi no?
10.Cmr v畛i m畛i x,y ta 畛u c坦: 2 2 2 2
4 6 9 4 2 12 10 5
x y x x y x y
11.Cho a,b,c,d l b畛n s畛 th畛c th畛a m達n
2 2 2 2
1 2 ; 36 12
a b a b c d c d
Cm:
6 6
2 2
2 1 2 1
a c b d
12.Cho x,y l hai s畛 th畛c th畛a m達n :
2 3 2
3 9
0, 0
x y
x y
x y
Cmr: 2 2
35
4 8 45
2
x y x y
13.Cho c叩c s畛 x,y th畛a m達n :
2 8 0
2 0
2 4 0
x y
x y
y x
Cm: 2 2
16
20
5
x y
III. 畛NG D畛NG 畉O HM
1Ch畛ng minh r畉ng v畛i m畛i ta c坦
2 2
17 os 4 os +6 os 2 os +3 2 11
c c c c
2.T狸m GTNN c畛a hm s畛 2 2
4 12 2 3
y x x x x
3.a)Ch畛ng minh b畉t 畉ng th畛c: sin 2 ; 0;
2
tgt t t t
b)Cho tam gi叩c ABC c坦 c叩c g坦c l A,B,C .
Ch畛ng minh :
A B C
1 os 1 os 1 os
2 2 2 3 3
A B C
c c c
( A,B,C o b畉ng raian)
4.Cho
, 0;1
a b Ch畛ng minh r畉ng
1 1 1 1
1 1 1
x b a
x a b
a b x a x b
v畛i
0;1
x
5.Cho hm s畛
2
2
os -2x+cos
x 2 os +1
x c
y
xc
v畛i
0;
Ch畛ng minh : 1 1;
y x
5. 6.Ch畛ng minh sin sin sin 2
A B C tgA tgB tgC
.v畛i A,B,C l ba g坦c
c畛a m畛t tam gi叩c.
7.Ch畛ng minh sinx 1
2 2 2 ;0
2
tgx x
x
8.Gi畉 s畛 f(x) l m畛t a th畛c b畉c n th畛a m達n i畛u ki畛n 0,
f x x
Cmr:
, ,,
... 0,
n
f x f x f x f x x
9.Ch畛ng minh r畉ng trong tam gi叩c ABC ta c坦
1 1 1
cot cot cot 3 3 2
sin sin sin
gA gB gC
A B C
10.Cho tam gi叩c ABC kh担ng t湛 ,th畛a m達n h畛 th畛c:
1 1 5
os3A+cos3B os2A+cos2B osA+cosB=
3 2 6
c c c
.Ch畛ng minh tam gi叩c ABC 畛u
11.Cho 0
2
a b
.Ch畛ng minh r畉ng :
a.sina-bsinb>2 cosb-cosa
12.Cho
a 1
0 q p q+1
.Ch畛ng minh r畉ng
1
p q p q
a p q a a
13.Cho
0
2
x .Ch畛ng minh r畉ng :
3
sinx
osx
x
c
14.Cho tam gi叩c ABC nh畛n .Cmr:
6 sin sin sin 12 3
tgA tgB tgC A B C
15.Cho a,b,c l c叩c s畛 kh担ng 但m th畛a 2 2 2
1
a b c
.
Ch畛ng minh r畉ng:
2 2 2 2 2 2
3 3
2
a b c
b c c a a b
16.Ch畛ng minh trong tam gi叩c nh畛n ABC ta c坦
2 1
sin sin sin
3 3
A B C tgA tgB tgC
17.Cho
0
2
x .Cmr:
3
1
2sinx 2
2 2 2
x
tgx
18Cho s畛 nguy棚n l畉 3
n .Cmr: 0
x
ta lu担n c坦 :
2 3 2 3
1 ... 1 ... 1
2! 3! ! 2! 3! !
n n
x x x x x x
x x
n n
駈
件
件
醐
19.v畛i gi叩 tr畛 no c畛a m th狸 3 3
sin os ,
x c x m x
20.Cho x,y >0 .Ch畛ng minh r畉ng :
2
3
2 2
4 1
8
4
xy
x x y
21.Cho 0, 0
x y
l hai s畛 th畛c thay 畛i th畛a m達n 2 2
x y xy x y xy
T狸m GTLN c畛a bi畛u th畛c
3 3
1 1
A
x y
22.Cho a,b,c l c叩c s畛 th畛a m達n i畛u ki畛n
3
, ,
4
a b c
Ch畛ng minh ta c坦 b畉t 畉ng th畛c
2 2 2
9
10
1 1 1
a b c
a b c
6. 23.(HSG B R畛a12-04-05)
1/T狸m c畛c tr畛 c畛a hm s畛
2
1
1
x
y
x x
2/ Cho c叩c s畛 x,y,z th畛a m達n x + y + z = 3
T狸m GTNN c畛a 2 2 2
1 1 1
P x x y y z z
24.T狸m GTNN c畛a
2 2 2
3 1 1 1 2
P x y z x y z
25. Cho , , 0
a b c v 6
a b c
. Cmr: 4 4 4 3 3 3
2( )
a b c a b c
26. Cho , , 0
a b c v
2 2 2
1
a b c
. Cmr:
1 1 1
( ) ( ) 2 3
a b c
a b c
27Cho a,b,c>0 .Cmr :
2 2 2
9
4( )
( ) ( ) ( )
a b c
a b c
b c c a a b
28. (Olp -2006)Cho , , 0
a b c .Cmr:
2 2 2 2 2 2
( ) ( ) ( ) 6
5
( ) ( ) ( )
a b c b c a c a b
a b c b c a c a b
39.(Olp nh畉t 1997)Cho , , 0
a b c .Cmr:
2 2 2
2 2 2 2 2 2
( ) ( ) ( ) 3
5
( ) ( ) ( )
b c a c a b a b c
b c a c a b a b c
40.x辿t c叩c s畛 th畛c d動董ng x,y,z th畛a m達n i畛u ki畛n :
4
2
x y z
xyz
.
T狸m GTLN v NN c畛a bi畛u th畛c 4 4 4
P x y z
(QG -B-2004)
41. x辿t c叩c s畛 th畛c d動董ng x,y,z th畛a m達n i畛u ki畛n 3
32
x y z xyz
T狸m GTLN v GTNN c畛a
4 4 4
4
x y z
P
x y z
(QG-A-2004)
42.C叩c s畛 th畛c d動董ng a,b,c,d th畛a m達n a b c d
v bc ad
.Ch畛ng minh r畉ng
b c d a d a b c
a b c d a b c d
43.X辿t c叩c s畛 th畛c x,y th畛a m達n i畛u ki畛n: 3 1 3 2
x x y y
T狸m GTLN v GTNN c畛a P = x + y ( QG B-2005)
44.Cho hm s畛 f x叩c 畛nh tr棚n R l畉y gi叩 tr畛 tr棚n R v th畛a m達n ( )
cotgx sin2 os2x
f x c
= + ,
( )
0;
x
T狸m GTNN v GTLN c畛a hm s畛 ( ) ( ) ( )
2 2
sin os
g x f x f c x
= QG B-2003 )
45.Cho hm s畛 f x叩c 畛nh tr棚n R l畉y gi叩 tr畛 tr棚n R v th畛a m達n ( )
cotgx sin2 os2x
f x c
= + ,
( )
0;
x
T狸m GTNN v GTLN c畛a hm s畛 ( ) ( ) ( ) [ ]
1 , 1;1
g x f x f x x
= - - ( QG A-2003)
46.Cho x>0 v , 0; ;
2
a b a b
脱 旦
歎
巽
孫
歎
巽 歎
巽
竪 淡
Cmr:
sin sin
sina sin
sin sin
x b b
x a
x b b
+
脱 旦 脱 旦
+ 歎 歎
巽 巽
>
歎 歎
巽 巽
歎 歎
巽 巽
竪 淡 竪 淡
+
IV-畛NG D畛NG 畛NH L LA GRNG
1.Ch畛ng minh r畉ng n畉u 0 < b < a th狸 ln
a b a a b
a b b
2.Ch畛ng minh r畉ng n畉u 0
2
a b
th狸 2 2
os os
b a b a
tgb tga
c a c b
7. 3.Ch畛ng minh
1
1 ; 0;1
2
n
x x x
ne
4.Cho m > 0 c嘆n a,b,c l 3 s畛 b畉t k畛 th畛a m達n i畛u ki畛n
0
2 1
a b c
m m m
.Ch動ng minh pt 2
0
ax bx c
c坦 鱈t nh畉t m畛t nghi畛m
thu畛c kho畉ng
0;1
5.Cho pt b畉c n: 1
1 1 0
... 0
n n
n n
a x a x a x a
trong 坦 1 1 0
0, ,..., ,
n n
a a a a
l s畛 th畛c th畛a m達n : 1 1
0
... 0
1 2
n n
a a a
a
n n
.Ch畛ng minh pt 達 cho c坦
鱈t nh畉t m畛t nghi畛m thu畛c kh畛ang
0;1
6.Cho c叩c s畛 th畛c a,b,c v s畛 nguy棚n n > 0 th畛a m達n
5 2 6 0
c n a b
Ch畛ng minh pt : sin cos sin 0
n n
a x b x c x c
c坦 nghi畛m thu畛c kho畉ng 0;
2
7.Cho hm s畛 li棚n t畛c :
: 0;1 0;1
f c坦 畉o hm tr棚n kho畉ng
0;1 Th畛a m達n
0 0, 1 1
f f
.Ch畛ng minh t畛n t畉i
, 0;1
a b sao cho a b
v
, ,
1
f a f b
8.Gi畉i c叩c pt sau :
a) 3 5 2.4
x x x
b) osx osx
3 2 osx
c c
c
c)
osx osx
1 osx 2 4 3.4
c c
c
d) 2003 2005 4006 2
x x
x
9.X辿t ph動董ng tr狸nh :
2 2
1 1 1 1 1
... ...
1 4 1 2
1 1
x x k x n x
Trong 坦 n l tham s畛 nguy棚n d動董ng
a)Cmr v畛i m畛i s畛 nguy棚n d動董ng n ,pt n棚u tr棚n c坦 duy nh畉t nghi畛m l畛n h董n 1
K鱈 hi畛u nghi畛m 坦 l n
x
b)Cmr d達y s畛
n
x c坦 gi畛i h畉n b畉ng 4 khi n
(QG-A-2002)
10.Cho hm s畛 ( )
f x v ( )
,
f x 畛ng bi畉n tr棚n o畉n [ ]
;
a b ,v畛i
( ) ( ) ( ) ( )
,
1 1
,
2 2
f a a b f b b a
= - = -
Ch畛ng minh r畉ng t畛n t畉i , ,
留 硫 隆 ph但n bi畛t trong ( )
;
a b sao cho ( ) ( ) ( ) 1
, , ,
f f f
留 硫 隆 =
11.Cho [ ] [ ]
0 1 0 1
: ; ;
f 速 tho畉 m達n c叩c i畛u ki畛n ( ) [ ]
0 0 1
,
; ;
f x x
> " v ( ) ( )
0 0 1 1
,
f f
= =
Cm:t畛n t畉i d達y s畛 1 2
0 ... 1
n
a a a
贈 < < < 贈 sao cho ( )
1
1
,
n
i
i
f a
=
続
(n l s畛 nguy棚n d動董ng 2
n 続 )
12.Cho a,b,c,d l 畛 di c叩c c畉nh c畛a m畛t t畛 gi叩c
CMR: 3
4 6
abc abd bcd acd ab ac ad bd cd
+ + + + + + +
贈
8. V .DNG 畛NH NGH懲A 畛 TNH 畉O HM T畉I M畛T I畛M
1.T鱈nh 畉o hm c叩c hs sau t畉i c叩c i畛m 達 ch畛 ra:
a) 2
1 osxcos2x...cosnx
khi x 0
0 khi x=0
c
f x x
t畉i x=0
b)
ln osx
khi 0
x
0 khi 0
c
x
f x
x
t畉i x=0
2.X叩c 畛nh a,b 畛 hm s畛 :
2
khi 0
1 khi 0
bx
x a e x
f x
ax bx x
c坦 畉o hm t畉i x=0
3.Cho hm s畛
p cosx +qsinx khi 0
px+q+1 khi 0
x
f x
x
Ch畛ng t畛 r畉ng m畛i c叩ch ch畛n p,q hm s畛 f(x) kh担ng th畛 c坦 畉o hm t畉i x=0
VI. 畛NG D畛NG TNH N I畛U C畛A HM S畛
1.Gi畉i bpt : 3 2
2 3 6 16 2 3 4
x x x x
2.X叩c 畛nh a 畛 b畉t pt sau c坦 nghi畛m duy nh畉t
2 2
1
2
log 11 log ax 2 3.log ax 2 1 1 0
a a
x x
3. X叩c 畛nh a 畛 b畉t pt sau c坦 nghi畛m duy nh畉t
2 2
1 5
log 3 log x ax+5 1 .log ax+6 0
a
a
x
4.T狸m m畛i gi叩 tr畛 c畛a tham s畛 a soa cho v畛i m畛i gi叩 tr畛 坦 pt sau c坦 炭ng 3 nghi畛m
ph但n bi畛t.
2
2 2
1
3
3
4 log 2 3 2 log 2 2 0
x a x x
x x x a
5.T狸m nh畛ng gi叩 tr畛 c畛a a 畛 v畛i m畛i gi叩 tr畛 坦 pt:
2 2 2
3 1 9 2
x a a x
c坦 s畛 nghi畛m kh担ng nhi畛u h董n s畛 nghi畛m c畛a pt
2 3
3
1
3 2 .3 8 4 log 3 3
2
x a a
x a x
6. T狸m nh畛ng gi叩 tr畛 c畛a a 畛 pt:
2 2 4 2
15 2 6 1 3 2 0
x m x m m
c坦 s畛 nghi畛m kh担ng nhi畛u
h董n s畛 nghi畛m c畛a pt :
2 3 6 8
3 1 .12 2 6 3 9 2 0,25
x m m
a x x
7.Gi畉i pt :
3
3 2
3log 1 2log
x x x
8.Gi畉i h畛 5
2 3
4
tgx tgy y x
x y
9.Gi畉i b畉t pt
7 3
log log 2
x x
9. 10.Gi畉i pt :
2 2
1 1
1
2 2
x x
a a
a a
v畛i tham s畛
0;1
a
11. Gi畉i h畛:
(1)
1 1 8 (2)
tgx tgy y x
y x y
12 Gi畉i pt:
2
osx=2
tg x
e c v畛i ;
2 2
x
13 Gi畉i pt:
2 2
3 (2 9 3) (4 2)( 1 1) 0
x x x x x
14.Gi畉i pt:
3
3 1 log (1 2 )
x
x x
VII.TM I畛U KI畛N 畛 PT C NGHI畛M
1.T狸m m 畛 pt sau c坦 nghi畛m :
2 2
1 1
x x x x m
2. T狸m t畉t c畉 c叩c gi叩 tr畛 c畛a a 畛 pt:
2
1 cos
ax x
c坦 炭ng m畛t nghi棚m 0;
2
x
3.Cho hm s畛
( )( )
y x x a x b v畛i a,b l hai s畛 th畛c d動董ng kh叩c nhau cho tr動畛c .Cmr
v畛i m畛i
0;1
s 畛u t畛n t畉i duy nh畉t s畛 th畛c
1
0: ( )
2
s s s
a b
f
(QG-A-2006)
4.Cho pt : 2
cos2x= m+1 cos 1
x tgx
a)Gi畉i khi m = 0
b)T狸m m 畛 pt c坦 nghi畛m trong o畉n 0;
3
5.T狸m m 畛 pt sau c坦 nghi畛m:
4 3 3 3 4 1 1 0
m x m x m
6.T狸m m 畛 t畛n t畉i c畉p s畛 (x;y) kh担ng 畛ng th畛i b畉ng 0 th畛a m達n pt:
2 2
4 3 3 4 1 0
m x m y m x y
7.T狸m m 畛 pt :
1 cos8
6 2cos4
x
m
x
c坦 nghi畛m.
8.T狸m a pt : 2
2cos 2
ax x
+ = 炭ng 2 nghi畛m thu畛c 0
2
;
辿 湛
棚 炭
棚 炭
谷 短
9.Cho hm s畛: ( )
2
2
x
sinx+
x
f x e
= -
a) T狸m GTNN c畛a hm s畛
b) Cm pt ( ) 3
f x = c坦 炭ng hai nghi畛m.
10.Ch畛ng minh pt ( )
1
1
x
x
x x
+
= + c坦 m畛t nghi畛m d動董ng duy nh畉t
11. Cho ( ) ( )
3 2
x 0; 0
f x ax bx c a
= + + + = 孫 c坦 3 nghi畛m ph但n bi棚t
a)H畛i pt: ( ) ( ) ( )
2
,, ,
2 0
f x f x f x
辿 湛
- =
棚 炭
谷 短
c坦 bao nhi棚u nghi畛m
10. b)Ch畛ng minh r畉ng: ( )
3
3 2
27 2 9 2 3
c a ab a b
+ - < -
12.Cho pt :
2
... 0
2 2 2n
tg x tg x tg x
脱 旦 脱 旦
脱 旦 歎 歎
巽 巽
歎
巽 + + + + + + =
歎 歎
歎 巽 巽
巽 歎 歎
歎
巽 巽 巽
歎 歎
竪 淡 竪 淡 竪 淡
( n l tham s畛)
a) Cmr v 畛i m畛i s畛 nguy 棚n 2
n 続 ,pt c 坦 m畛t nghi畛m duy nh畉t trong kho畉ng
0
4
;
脱 旦
歎
巽 歎
巽 歎
巽
竪 淡
.k 鱈 hi棚畛 ng 坦 l n
x
b)Cm d達y s畛 ( n
x ) c坦 gi畛i h畉n
13.Ch畛ng minh pt ( ) 4 3 2
4 2 12 1 0
f x x x x x
= + - - + = c坦 4 nghi畛m ph但n bi畛t 1 4
; ,
i
x i =
v h達y t鱈nh t畛ng
( )
2
4
2
1
2 1
1
i
i i
x
S
x
=
+
= 奪
-
VIII M畛T S畛 BI TON V畛 H畛 PH働NG TRNH
1.T狸m a 単e奪 he辰 sau co湛 nghie辰m duy nha叩t:
2 3 2
2 3 2
4 ax
x 4
y x x
y y ay
2. T狸m m 畛 h畛 pt sau c坦 nghi畛m
2x+ y-1
2 1
m
y x m
3.Gi畉i h畛
2
2
2
1
2
1
y
x
y
x
y
x
4.Ch畛ng t畛 r畉ng v畛i m畛i 0
a th狸 h畛 sau c坦 nghi畛m duy nh畉t
2
2
2
2
2
2
a
x y
y
a
y x
x
5.T狸m a 畛 h畛
sinx=a
sin
x
y
y
y a
x
c坦 nghi畛m duy nh畉t 0 2 ,0 2
x y
6.Gi畉i h畛:
3 2
3 2
3 2
3 3 ln( 1)
3 3 ln( 1)
3 3 ln( 1)
x x x x y
y y y y z
z z z z x
7.Gi畉i h畛:
2
3
2
3
2
3
2 6 log (6 )
2 6 log (6 )
2 6 log (6 )
x x y x
y y z y
z z x z
( QG A- 2006)
11. 8.T狸m a 畛 h畛 c坦 nghi畛m duy nh畉t (HSG12-2006)
2 3 2
1 2 2 2
2 3 2
2 3 3 3
2 3 2
1 1 1
4 ax
4 ax
............................
4 ax
n
x x x
x x x
x x x
6.Gi畉i h畛:
2 1 2 2 1
2 2
1 4 .5 1 2
4 1 ln 2 0
x y x y x y
y x y x
( HSGQG 1999)
7.Gi畉i h畛:
2 3
2 3
log 1 3 osx log sin 2
log 1 3sin log osx 2
c y
y c
(THTT)
8.G畛i ( )
;
x y l nghi畛m c畛a h畛 pt:
2 4
3 1
x my m
mx y m
狸 - = -
誰
誰
鱈
誰 + = +
誰
樽
( m l tham s畛)
T狸m GTLN c畛a bi畛u th畛c 2 2
2
A x y x
= + - ,khi m thay 畛i
H働畛NG D畉N GI畉I
I.B畉t 畉ng th畛c
12. 4. , 1,..,
m n
i i
na m n ma i k
7.
1
2 1
2
1
1 2
2
* :
...
* :
* :
...
m
m n m n
n
m
m n m n
n
a
m n m n na ma
a
m n csi
a
m n n m ma na
a
20.
2
1 1 1
1 1 1
1 1 1
ab bc ca
A
ab bc ca abc
駈 駈
件 件
醐 醐
Ta c坦:
2
1 1 1 1 1 1
2 2
1 1
4 4 4 2
a b c c a b
a b a b a b
ab
T動董ng t畛 suy ra:
2
1 1 1 1
1 1 1
8
A
a b c
駈 駈
件 件
醐 醐
M:
3
3
3
1 1 1 1
1 1 1 1 4
a b c abc
駈 駈
件 件
醐 醐
V畉y:
3
8
A dpcm
26. 2 2 2 2
1 1 1 1 1 1 1
2
a b c d
P
ab ac ad bc bd cd bcd cda abd bca
a b c d
2 2
1 1 1 1 1 1 1
*
...
1 1 1 1 1 1
*
*
A B C
A
ab ac ad bc bd cd
a d
B
ab ac ad bc bd cd
a b c d
C
bcd acd dab abc
Ta cm: 100, 96, 64 260
A B C P
29.畉t: , 1,...,
1
i
i
i
x
X i n
x
ta c坦 1
1
1
... ... 1
1 1
n
n
n
X
X
x x
X X
T畛 坦 suy ra:
1 2
1
1 1 1
... 1 . ...
1 1 1
n n
n
n X X X
X X n
(pcm)
30. 畉t: , 1,
1998
i
i
x
X i n
.Ta c坦:
1
1 1
... 1
1 1 n
X X
T畛 坦 suy ra:
1... 1
n
n
X X n
.v畉y c坦 (pcm)
31.t:
1
1
1
1
1 ...
; 1,..., ;
1 ...
n
i n
i n
a a
a
X i n X
a a a
13. Ta c坦:
1 1
1 1 1
...
1 1 1
n n
n
X X X
.v畉y
1
1 1
1
...
n
n n
X X X
n
38.
2 2
2 2 2 2 2 2 2
2 2
2 2 1
2
z z
P a x y z x y a x y
xz yz xy
Ch畛n
2
a
39.
2
2 2
2 2 2 2 2 2 2
16 16
1
25 2 2 25
16
2 2 1
2 25
z z
P x y z xy qx qy q x y xy
q
xz yz q xy
Ch畛n
16 18
2 2 1
2 25 25
q
q q
2
ax
5
6
M
a
P khi 3
3
5 3
a
x y
a
z
39Do vai tr嘆 c畛a a v d,bv c trong bi畛u th畛c tr棚n ta d畛 o叩n i畛m c畛c tr畛
s畉 畉t 動畛c t畉i c叩c b畛 s畛 th畛a k: 2 2 2 2
,
a d c d
.v畛i p>0x叩c 畛nh sau ta c坦
c畛ng theo v畉 :
2 2 2 2
5 10
5 5
p
P p a d b c
p
Ch畛n p th畛a :
1 2 1 5
1
2
p
p p
p
V畉y
ax
5 3 5
2
m
P
43.畛ng dung k c坦 nghi畛m c畛a hpt x
II PH働NG PHP HNH H畛C
1.G畛i
; , ;
M a b N c d T畛 gt suy ra M,N n畉m tr棚n 動畛ng tr嘆n 2 2
4
x y
v 動畛ng
th畉ng
4
x y
.D畛 th畉y
2 2 2
2 20 20
ac bd cd a c b d MN
M 2
12 8 2
MN n棚n
2 8 8 2 4 4 2
ac bd cd ac bd cd
V畉y axP=4+4 2
m khi 2; 2
a b c d
14. 2.v 3 t動董ng t畛
4.G畛i
; , , , ;
N a b Q c d M x y T畛 gt suy ra N,Q,M l畉n l動畛t thu畛c c叩c 動畛ng tr嘆n
2 2 2 2
1 2
: 4 5 1, : 2 3 1
C x y C x y
v 動畛ng th畉ng
:3 2 13 0
x y
Khi 坦 P MQ MN
G畛i 1
,
I R v 2
,
J R l畉n l動畛t l t但m v b叩n k鱈nh c畛a
1 2
,
C C
L畉y
;
K u v 畛ix畛ng v畛i I qua
th狸 118 21
;
13 13
K
1 2
2 13 1
P MQ MN MJ JQ MI IN MJ MK R R
畉ng th畛c x畉y ra khi v ch畛 khi 1 1 1
, ,
M M Q Q N N
.Trong 坦 1 1
,
M Q l giao
C畛a JK v畛i
v
2
C c嘆n
1 1 1
N M I C
V畉y
min 2 3 1
P
III 畛NG D畛NG 畉O HM 畛 CM BT
3.T畛 c但u a) ta c坦
1 ost ost
cot
2t sin
c c
gt
t
.v v狸 cot cot 3 3
2 2 2
A B C
g cogt g
n棚n c坦 pcm
4.Hm s畛
1 1 1
1 1 1
x b a
f x x a b
a b x a x b
v畛i
0;1
x
c坦 畉o hm c畉p hai kh担ng 但m n棚n 畉o hm c畉p m畛t c坦 nhi畛u nh畉t 1 nghi畛m
1
TH :
,
0
f x VN Th狸
ax f 0 ; 1 1
f x M f
2
TH :
,
0
f x c坦 nghi畛m duy nh畉t x
th狸 v狸
,
f x 畛ng bi畉n n棚n l i畛m
c畛c ti畛u v狸 v畉y
0;1
ax 0 ; 1 1
axf x m f f
m (pcm)
8.畉t
,
...
n
F x f x f x f x
th狸
, , ,
...
n
F x f x f x f x F x f x
(1)
v狸 f l a th畛c b畉c n n棚n
1
0
n
f x
.T畛 gt bi to叩n suy ra f l a th畛c b畉c ch畉n
c坦 h畛 s畛 cao nh畉t d動董ng do 坦 F 畉t GTNN.Gi畉 s畛 F 畉t GTNN t畉i 0
x Th狸
,
0 0
F x
v畉y t畛 (1) suy ra
,
0 0 0 0 0
F x F x f x f x
(pcm)
12.
1 p+q 1 0
p q p q p q p q
a a a a p q a a
Hm s畛: 1
p q p q
f x x p q x x
畛ng bi畉n tr棚n
1;
V c坦
1 0
f n棚n t畛 1
a ta c坦 (pcm)
13.C担 l畉p x v x辿t d畉u 畉o hm c畛a 2 3
sin .
f x x tgx x
Ch炭 箪:
2 2
2 2 1 1
2sin 2sinx+tgx 3
3 3
x tg x x
15. *C滴ng c坦 th畛 x辿t 畉n 畉o hm c畉p 3 畛 kh動 x
15.T畛 d畛 o叩n i畛m r董i d畉n 畉n x辿t hm s畛 c坦 i畛m c畛c tr畛 1
3
x l
3 2
1
y x x x x
23.
2
1
1
x
y
x x
畉t c畛c 畉i duy nh畉t b畉ng 2 t畉i x=1
n棚n 2 2 2
1 1 1
P x x y y z z
nh畛 nh畉t b畉ng 3
*c坦 th畛 d湛ng bunhia ho畉c hm l畛i
40.
2
4 4 4 2 2 2 2 2 2 2 2 2
2
2 2
2 2
2
2 2 2
16 2 2 16
P x y z x y z x y y z z x
x y z xy yz zx xy yz zx xyz x y z
t t
v畛i t=xy + yz +zx
2
4
t x y z yz x x
x
V狸
2
4 2 4
3 5;2
2 2 2
y z x x
yz x
x
do (0<x<4)
T畛 坦 t狸m 動畛c min v max c畛a P
41.T動董ng t畛40
42. L畉y ln hai v畉 ta c坦
ln ln ln ln
d b c a c a d b
(1)
N畉u a c
ho畉c d b
th狸 hi畛n nhi棚n 炭ng
X辿t a c
v d b
.Khi 坦 (1)
ln ln
ln ln ln ln
1
1 1
c d
c a d b a b
c d
c a d b
a b
a b
X辿t hm s畛 :
ln
, 1,
1
x
f x x
x
ngh畛ch bi畉n tr棚n
1, Suy ra:
ln ln ln ln
1 1 1 1
c d c d
c d a b a b
f f
c d c d
a b
a b
a b a b
44,45. Bi畛u di畛n sin 2 , os2x
x c theo cotgx ta 動畛c ( )
2
2
2 1
1
t t
f t
t
+ -
=
+
IV NG D畛NG 畛NH L LAGRANG
6. x辿t hm s畛 ( )
2 2
3 2
2 sin 2 sin 2
sin os
2 2 3
n n
a x b x c
f x x cc x
n n
+ +
= - + -
+ +
8.a) 3 5 2 4 5 4 4 3
.
x x x x x x x
+ = 束 - = - (1) .Gi畉 s畛 pt c坦 nghi畛m x 留
=
X辿t hm s畛 ( ) ( )
1 0
,
f t t t t
留 留
= + - > c坦 ( ) ( )
4 3
f f
= .Do 坦 t畛n t畉i ( )
3 4
;
c
16. Sao cho ( ) ( ) 1 1 0
0 1 0
1
,
f c c c
留 留 留
留
留
- - 辿 =
辿 湛 棚
= 束 + - = 束
棚 炭 棚
谷 短 =
谷
Th畛 l畉i th畉y 0
x = v 1
x = 畛u th畛a m達n (1)
V畉y pt c坦 hai nghi畛m 0
x = , 1
x =
b) 2 3 3 2 2
t
t=cosx 3 t t t
t t t
速 - = 束 - = - . Gi畉 s畛 pt c坦 nghi畛m x 留
=
X辿t ( )
f t t t
留
留
= - th狸 ( ) ( )
3 2
f f
= suy ra pt ( ) 0
,
f t = c坦 nghi畛m c坦
nghi畛m ( )
2 3
;
c .
( ) ( ) ( )
, 1 , 1 0
1 0
1
留 留 留
f t 留t 留 f c 留 c
留
- - 辿 =
棚
= - 速 = - =
棚 =
谷
c)畉t 1 1
cos ,
t x t
= - 贈 贈
Ta c坦 pt: ( )( ) ( )
3 4
1 2 4 3 4 1 0
2 4
.
.
t
t t
t
t f t t
+ + = 束 = - - =
+
( )
( )
( ) ( )
2
2
6 4 4
1 0 6 4 4 2 4
2 4
, ,
ln .
, ln .
t
t t
t
f t f t
= - = 束 = +
+
.但y l pt b畉c hai theo 4t
n棚n c坦 kh担ng qu叩 hai nghi畛m do 坦 pt ( ) 0
f t = c坦 kh担ng qu叩 3 nghi畛m
Ta th畉y
1
0 1
2
, ,
t t t
= = = l 3 nghi畛m c畛a pt
C) X辿t ( ) 2003 2005 4006 2
x x
f x x
= + - - c坦 畉o hm c畉p hai d動董ng
V ( ) ( )
0 1 0
f f
= = .v畉y pt c坦 hai nghi畛m l 0 v 1
9)Vi畉t l畉i pt d動畛i d畉ng ( ) 2
1 1 1 1
0
2 1 4 1 1
...
n
f x
x x n x
= - + + + + =
- - -
(1)
D畛 th畉y ,v畛i m畛i *
n hm ( )
n
f x li棚n t畛c v ngh畛ch bi畉n tr棚n ( )
1;+ 促
H董n n畛a ( )
n
f x 速 + 促 khi 1
x +
速 v ( )
1
2
n
f x 速 - khi x 速 + 促 .T畛 坦 suy ra
V畛i m畛i *
n ,pt(1) c坦 duy nh畉t nghi畛m 1
n
x >
V畛i m畛i *
n ,ta c坦
( )
( )
( )
( )
2 2 2
1 1 1 1
4
2 2 1 4 1 2 1
1 1 1 1 1 1 1 1
1 1
2 3 3 5 2 1 2 1 2 1 2 1
1
0
2 2 1
...
... ...
n
f
n
k k n n
f x
n
= - + + + +
- - -
脱 旦
歎
巽
= - + - + - + + - + + - 歎
巽 歎
巽
竪 淡
- - - +
= - < =
+
T畛 坦, dohm ( )
n
f x tr棚n ( )
1;+ 促 n棚n 4
n
x < v畛i m畛i *
n (2)
M畉t kh叩c hm ( )
n
f x c坦 畉o hm tr棚n [ ]
4
,
n
x n棚n theo 畛nh l鱈 Lagrange
V畛i m畛i *
n t畛n t畉i ( )
4
;
n
t x
sao cho
( ) ( )
( )
( ) ( ) ( )
2
2 2 2
4 1 4 1
4 9
1
1 4 1
, *
...
n n n
n
f f x n
f t n
x n t
t t
- - - -
= = + + + < - "
- -
- -
17. Hay
( )( ) ( )
1 1 9
4
2 2 1 4 9 2 2 1
* *
n
n
n x n
n x n
-
< - " > - "
+ - +
(3)
t畛 (2) v (3) :
( )
9
4 4
2 2 1
*
,
n
x n
n
- < < "
+
suy ra 4
lim n
x = (pcm)
III .畛NG D畛NG 畉O HM TM K 畛 PT C NGHI畛M
2. ( )
1 0
2
2
2
osx-1
ax osx , ;
x
c
c a f x x
脱 旦
歎
巽
+ = = = " 歎
巽 歎
巽
竪 淡
T狸m mi畛n gi叩 tr畛 c畛a f(x) ta 動畛c a c畉n t狸m
3.Hm s畛 ( )( )
y x x a x b
= - + + + c坦 mi畛n gi叩 tr畛 tr棚n ( )
0;+ 促 l
2
;
a b
ab
脱 旦
+ 歎
巽 歎
巽 歎
巽
竪 淡
Do 坦 ch畛 c畉n cm:
1
2 2
s s s
a b a b
ab
脱 旦
+ +
歎
巽 歎
巽
< <
歎
巽 歎
巽 歎
巽
竪 淡
,v畛i m畛i ( )
0 1
;
s
4
.
( ) ( )
4 3 3 3 4 1 1 0
3 3 4 1 1
4 3 3 1 1
m x m x m
x x
m
x x
- + + - - + - =
+ + - +
=
+ + - +
Ch炭 箪:
2 2
3 1
1
2 2
x x
脱 旦 脱 旦
+ -
歎 歎
巽 巽
歎 歎
+ =
巽 巽
歎 歎
巽 巽
歎 歎
歎 歎
巽 巽
竪 淡 竪 淡
.Do 坦 l動畛ng gi叩c h坦a v 動a v畛 畉n ph畛
2
t tg
留
=
R畛i kh畉o s叩t hm s畛 thu 動畛c theo t
5.T動董ng t畛 4
10. ( ) ( ) ( ) ( )
1
1 1 1 0
ln ln
x
x
x x f x x x x x
+
= + = + - + =
Ta c坦 ( )
1 1 1 1 1 1
1 0
1 1
,
ln
f x
x x x x x x
脱 旦
歎
巽
= + - - < - - <
歎
巽 歎
巽
竪 淡 + +
v畛i x>0 v畉y f Nb
M ( )
1 2 0
ln
f = > v
( ) ( ) ( )
( )
1
1
1 1 1
1
1 1
ln ln
lim lim
ln ln
lim
x x
x
x
f x x x
x
x
x
速 + 促 速 + 促
+
速 + 促
辿 湛
脱 旦
歎
巽
棚 炭
= + + - +
歎
巽 歎
巽
棚 炭
竪 淡
谷 短
辿 湛
脱 旦
棚 炭
歎
巽
= + - + = - 促
歎
巽
棚 炭
歎
巽
竪 淡
棚 炭
谷 短
K畉t h畛p f li棚n t畛c trong ( )
0,+ 促 suy ra pt c坦 nghi畛m d動董ng duy nh畉t .