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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
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
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
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
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
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
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
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
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
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)
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
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
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
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
*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
   
    
   
       
       
       
       
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0,+ 促 suy ra pt c坦 nghi畛m d動董ng duy nh畉t .

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  • 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 .