I HC QUI
I HC KHOA HC T
,
LUC
i
I HC QUI
I HC KHOA HC T
,
:
: 604640
LUKHOA HC
NG DN KHOA HC:
TS.
i
MC LC
5
7
8
8
9
9
1.1
: 9
1.2
: 9
1.3
: 9
1.4 : 10
1.5 : 10
1.6
: 11
1.7 : 11
1.8
: 11
1.9
11
1.10 Mt s bng thn 17
1.10.1 Bng thc Cauchy 17
1.10.2 Bng thc Bunhiacopxki (B.C.S) 17
1.10.3 Bng th 18
20
20
20
2.1.1
20
2.1.2 Nhng biu dic (2.1.4) thng b
(2.1.1) 22
2.1.3. Nhng min con cng vi nh
24
u thc ca nh p,x,y
26
gia nhng trong m 28
2.2
35
2.2.1
35
60
NG MINH BNG TH 60
ng minh bng thc d c 60
dng bng th chng thc trong tam
66
dng bng th chng thc trong
74
3.4
81
KT LUN 94
liutham
k
h
o
95
g
.
, .
,
.
:
1:
-
,
,
, ,,
,
-
1.9
-
1.10
.
2:
.
.
p cho
=
+ +
2
=
4(+ + )
+ +
=
8
2
3
2
+
2
+
2
+ 2(+ + )
(+ + )
2
.
,
, , , .
.
2.2
,
R, r, p.
(, , )
.
3
,
, ,
,
.
, .
, Ts.
.
, ,
.
!
.
,
,
,
.
,
.
.
, 10\05\2013
A, B, C :
a, b, c : , B, C
,
,
: , B, C
:
:
,
,
: A, B,
:
:
1:
1.1
sin:
=
=
= 2.
1.2
cos:
2
=
2
+
2
2.
2
=
2
+
2
2 .
2
=
2
+ 2.
1.3
tan:
+
=
2
+
2
+
=
2
+
2
+
=
2
+
2
1.4 :
=
1
2
=
1
2
=
1
2
=
1
2
. =
1
2
. =
1
2
.
=
4
= = ()
= ()
= ()
=
( )()( ) ( -ron).
1.5 :
=
2
=
2
=
2
=
4
= ()
2
= ()
2
= ()
2
=
:
=
2
=
=
2
=
=
2
=
.
1.6
:
2
=
2
+
2
2
2
4
2
=
2
+
2
2
2
4
2
=
2
+
2
2
2
4
.
1.7 :
=
2
+
2
=
2
+
2
=
2
+
2
1.8
:
= . + . = (
2
+
2
)
= . + . = (
2
+
2
)
= . + . = (
2
+
2
).
1.9
Bi tp 1.9.1C
:
1.9.1.1+ + = 4
2
2
2
.
1.9.1.2 2+ 2+ 2= 4.
1.9.1.3
2
+
2
+
2
= 2(1 + ).
1.9.1.4 + + = 1 + 4
2
2
2
.
1.9.1.52+ 2+ 2= 1 4 .
1.9.1.6
2
+
2
+
2
= 1 2.
Chứng minh
,
. A + B + C = . Ta
1.9.1.3
2
+
2
+
2
=
1 cos2A
2
+
1 cos2B
2
+ 1 cos
2
C
= 2 (+ )()
2
= 2 +
()
= 2 + (2)
+
2
2
= 2(1 + ).
1.9.2
ABC :
1.9.2.1
2
+
2
+
2
=
2
2
2
.
1.9.2.2
2
2
+
2
2
+
2
2
= 1.
1.9.2.3+ + = 1.
1.9.2.4 + + =
. ( )
Chứng minh
,
1.9.2.3 .
(+ ) =
1
+
=
+ + = 1.
1.9.3
ABC k Z :
1.9.3.1(2+ 1)+ (2+ 1)+ (2+ 1)
= (1)
4(2+ 1)
2
(2+ 1)
2
(2+ 1)
2
.
1.9.3.2 2+ 2+ 2= (1)
+1
4
1.9.3.3(2+ 1)+ (2+ 1)+ (2+ 1)
= 1 + (1)
4(2+ 1)
2
(2+ 1)
2
(2+ 1)
2
.
1.9.3.42+ 2+ 2= 1 + (1)
4 .
1.9.3.5 + + = .
1.9.3.6(2+ 1)
2
+ (2+ 1)
2
+ (2+ 1)
2
=
(2+ 1)
2
(2+ 1)
2
(2+ 1)
2
.
1.9.3.7(2+ 1)
2
(2+ 1)
2
+ (2+ 1)
2
(2+ 1)
2
+
(2+ 1)
2
(2+ 1)
2
= 1.
1.9.3.8+ + = 1.
1.9.3.9
2
+
2
+
2
= 1 + (1)
2.
1.9. 3.10
2
+
2
+
2
= 2 + (1)
+1
2.
Chứng minh
1.9.3.1 1.91.1
: (2+ 1)+ (2+ 1)+ (2+ 1)=
= 2(2+ 1)
+
2
(2+ 1)
2
+ 2(2+ 1)
2
(2+ 1)
2
= 2(1)
(2
+ 1)
2
(2+ 1)
2
+ (2+ 1)
+
2
= (1)
4(2+ 1)
2
(2+ 1)
2
(2+ 1)
2
.
1.9.3.2 ; 1.9.3.3 ; 1.9.3.4:
1.9.1.2;
1.9.1.4; 1.9.1.5 1.9.3.1.
1.9.3.5 1.9.2.4
: =
(+ )
= (+ )
=
+
1
.
+ + = .
1.9.3.6 1.9.2.1
1.9.3.5.
1.9.3.7 1.9.2.2
(2+ 1)
2
= (2+ 1)(
2
+
2
)
=
(2+ 1)
2
+ (2+ 1)
2
=
1
(2+ 1)
2
+ (2+ 1)
2
=
1 (2+ 1)
2
(2+ 1)
2
(2+ 1)
2
+ (2+ 1)
2
(2+ 1)
2
(2+ 1)
2
+ (2+ 1)
2
(2+ 1)
2
+ (2
+ 1)
2
(2+ 1)
2
= 1.
1.9.3.8 1.9.2.3
1.9.3.7.
1.9.3.9 1.9.1.6.
:
2
+
2
+
2
=
1
2
(1 + 2) +
1
2
(1 + 2) + (1)
(+ )
= 1 + (1)
() + (+ )
= 1 + (1)
2.
1.9.3.10 1.9.1.3
1.9.3.9.
1.9.4
, y,
1.9.4.1 + + (+ + )
= 4
+
2
+
2
+
2
.
1.9.4.2+ + + (+ + )
= 4
+
2
+
2
+
2
.
1.9.4.3+ + (+ + )
=
(+ )(+ )(+ )
(+ + )
.
1.9.4.4 + + (+ + )
=
(+ )(+ )(+ )
(+ + )
.
Chứ ng minh
1.9.4.1+ + (+ + )
= 2
+
2
2
2
+ + 2
2
+
2
= 2
+
2
2
+ + 2
2
= 4
+
2
+
2
+
2
.
1.9.4.2 1.9.4.1.
1.9.4.3+ + (+ + )
=
(+ )
(+ )
(+ + )
=
(+ )
(+ + )
.
(+ + )
=
(+ )
2 (+ + )
.
(+ + 2 ) + (+ )
(+ ) ()
=
(+ )(+ )(+ )
(+ + )
.
1.9.4.4 1.9.4.3.
: Thay{ , , } 1.9.4
{, , }; {(2+ 1), (2+ 1), (2+ 1)}; {2, 2, 2};
{(2+ 1)
2
, (2+ 1)
2
, (2+ 1)
2
}
, ,
1.9.1; 1.9.2; 1.9.3.
1.10 Mt s bng thn
1.10.1 Bng thc Cauchy
Cho n s
1
,
2
, .
. ng thc :
1
+
2
+ +
=
1
2
.
Dng thc x khi
1
=
2
= =
.
1.10.2 Bng thc Bunhiacopxki (B.C.S)
Cho n cp s bt k
1
,
2
, ,
;
1
,
2
, ,
ng thc:
(
1
1
+
2
2
+ +
)
2
(
1
2
+
2
2
+ +
2
)(
1
2
+
2
2
+ +
2
).
Hay g :
(
=1
)
2
(
2
=1
)(
2
=1
).
Dng thc x khi :
:
=
()
Vi = 1,2, , (Nu
0 () c vit :
1
1
=
2
2
= =
)
1.10.3 Bng th
Cho hai sp th t ging nhau:
1
2
1
2
ng thc sau:
1
+
2
+ +
1
+
2
+ +
(
1
1
+
2
2
+ +
)
Du bng thc xy ra khi:
1
=
2
= =
hoc
1
=
2
= =
Chứng minh
. .
1
=1
(
1
=1
)(
1
=1
)
=1
(
=1
)(
=1
)
=1
=1
(
=1
)(
=1
)
(
)
,
0
(
+
)
<
0
(
) +
(
) 0
=1
(
) +
(
) 0
=1
vi >
1.10.4
.
.
1.10.4.1
= () [a, b].
:
1,
2
(, ), , 0: + = 1,
(
1
+
2
) (
1
) + (
2
).
1.10.4.2= () [a, b] ,
() .
()
Cho f(x)
m
(a, b).
*
> 0( , ) (x) (a, b).
*
(x)< 0(, ) (x) (a, b).
1.10.4.3
Cho f (x) [a, b].
1,
2,,
[, ]
> 0, =
1, 2, , ;
1
+
2
+ +
= 1.
=1
(
).
=1
:
1
=
2
= =
=
1
,
f(x) [a, b],
:
1
+
2
+ +
1
[(
1
) + (
2
) + + (
)
].
:
2
u ng cnh ln nh nht ca mt tam
nh th ba c
0 < < + (2.1.1)
2.1.1
=
+ +
2
,
=
4(+ + )
+ +
,
=
8
2
3
2
+
2
+
2
+ 2
+ +
+ +
2
. .
Chng minh rng nhng bng th
> 0 ; 0 < < 1; < 2
2
(. . )
Li gii.
1. Bng thc > 0hi
2. T (2.1.1) ta nhc
4 (+ + ) = 3 (+ ) + 2 (+ )
= (+ ) > 0
> 0.
3. 4 (+ + ) = + 2(+ ) < + 2(+ ) (+ ) =
(+ + ) ,
< 1.
4ng thc sau:
+ > 0, + > 0, 2 > 0
ta nhc 2b
2
2a
2
2c
2
+ 4ac > 0 vii dng
5
2
3
2
3
2
+ 2
+ +
>
+ +
(3)
Hoc
8b
2
3
a
2
+ b
2
+ c
2
+ 2(ab + bc + ca)
(a + b + c)
2
>
3b a c
a + b + c
> .
5ng thc sau 0, 0, 8 > 0, ta nhn
c 8
2
+
888
0 vit lc
5
2
3
2
3
2
+ 2
+ +
2 (+ + ) (3) (3)
2
Ho
8
2
3
2
+
2
+
2
+ 2(+ + )
(+ + )
2
4
+ +
+ +
3
2
+ +
2
2
2
.
Ta gii h (2.1.2) i vi a, c
=
1
4
3
; =
1
2
+ 1
; =
1
4
(3 + ), (2.1.4)
=
2
+ 10+ 1 8(2.1.5)
Biu thc (2.1.5) t vy, t (2.1.3)
8168
2
=
2
+ 10+ 1 (31)
2
2
+ 10+ 1(2.1.6)
2.1.2 Nhng biu dic (2.1.4) thng
b2.1.1)
Li gii.
18> 8 = 168 + 8(1 ) > 168 vit li biu
th(3 )
2
>
2
+ 10+ 1 8 ho(2.1.3) (2.1.6)
dn 3 > > 0
2 8168
2
ho
(31)
2
2
(2.1.7)
Vi <
1
3
b(2.1.7) vii dng
(31) (2.1.8)
Bng th
1
3
. Ta vit (2.1.8)
1
4
2+ 2
1
4
3
, ta nhc .
3. Vi >
1
3
b(2.1.7) vii dng
(31) (2.1.9)
Bng th
1
3
. Ta vit (2.1.9)
1
4
2+ 2
1
4
3 +
, ta nhn c b .
4. Ta vit b8> 8 ng (+ 1)
2
>
2
+ 10+ 1
8 ho(2.1.3) (2.1.5) ng + 1 > vit li
1
4
(3 ) +
1
4
(2+ 2) >
1
4
(3 + ) , + > .
2.1.1 2.1.2t lp quan h gia nhng cp s
(x,y, p) th(2.1.1) (2.1.3). Mi quan h ng mt - mt.
ng nn t tuy
i s c nh mt t th
0 < < 1; < 2
2
(2.1.10)
t c nh(, , ) ng dng. Bng thc (2.1.10)
nh trong h t mt min gii hn bng thng =
parabol = 2
2
m n
nhm n u thuc tng
vn nhng dng v
Nhm ca minh tt c nhng tam gii nhng lng
dng.
2.1.3. Nhng min con cng vi nh
u n nht c
=
2
+
2
2
2
.
thuu
thc
2
+
2
2
> 0, = 0 hoc< 0.
T ph thuc
7
2
101
3
2
hoc <
7
2
101
(3)
2
Ph th c=
7x
2
10x1
(x3)
2
, 0 < < 1, nm trong min G, th
hing parabol = 2
2
ti
- 2
2 , 8
2 - 11).
1. Nhc vi tt c nhm trong min
3 2
2 < 1, =
7
2
101
(3)
2
Vi nhcung PM ca T tr m M.
2. Nhc t nhm trong min
0 < < 3 2
2, < 2
2
,
= 3 2
2, 3 2
2 < < 8
2 1. (2.1.12)
3 2
2 < < 1, < <
7
2
101
(3)
2
t nhm gii hn bng thng =
parabol = 2
2
.
3. Nhn nhc t nhm trong min
3 2
2 < 1,
7
2
101
(3)
2
< 2
2
. (2.1.13)
t c nhm gii hn ba parabol
= 2
2
.
4ng vi nh
T = c 1 3=
2
+ 10+ 1 8
1
3
ng thc tc dng
= 2
2
y ta nhc cung parabol:
OQ : = 2
2
, 0 <
1
3
c nh
QM: = 2
2
,
1
3
< 1.
= = ch tm Q(
1
3
,
5
9
).
X
1
0
O
1
Y
M
Q
P