cong thuc toan THPT can nho - Pdf 23

KIN THC TON CN NH
GV: Lấ VN LAI

A.I S V GII TCH

1.Hng ng thc cn nh:

( )
( )( )
( )
( )
( )
( )
2
2 2
2 2
3
3 2 2 3
3 3 2 2
2
2 2 2
1) a b a 2ab b
2) a b a b a b
3) a b a 3a b 3ab b
4)a b a b a ab b
5) a b c a b c 2ab
2bc 2ca
= +
- = + -
= +
= +

= + +

( )
2 2
b 4ac hay ' b ' ac
D = - D = -

0
D <
:
( )
a.f x 0 x R
> " ẻ

0
D =
:
( )
a.f x 0 x R
" ẻ


0
D >
: f(x) cú 2 nghim
1 2
x x
<
4.PT v BPT vụ t:
PT cha n di du cn bc 2:

2
c) A B A B
= =
BPT cha n di du cn bc 2:


< >
<

ù
ù
ù
ù

ù
ù
ù
ù

2
A 0
a) A B B 0
A B


<
<

ù
ù


2
B 0
A B

5.PT v BPT cha n trong du
GTT:
PT cha du GTT:

B 0
a) A B
A B
A B

=
=
= -

ù
ù
ù
ù


ù

ù

ù
ù


> <
>

ù
ù

ù
ù

2 2
B 0
b) A B B 0 hay
A B

6.Cụng thc lng giỏc:
H thc c bn:

+ =
=
=
+ =
+ =
=
2 2
2
2
2
2
1) sin a cos a 1


( )
1) sin a sina
p - =

( )
( )
( )
2) cos a cos a
3) tan a tan a
4) cot a cot a
p - = -
p - = -
p - = -
Cung ph:
p
- =
ổ ử






ố ứ
1) sin a cos a
2

p
- =

3) tan a co t a
2
4) cot a tan a
2
Khỏc
p
:

( )
1) sin a sin a
p + = -( )
( )
( )
2) cos a cos a
3) tan a tan a
4) cot a cot a
p + = -
p + =
p + =
Chỳ ý:
( )
2
sin a cos a sin a / 4 = p( )
2 a


=
m
m
Cụng thc nhõn:
2 2
2
1) sin 2a 2 sin a cos a
1
2) sin a cos a sin 2a
2
3) cos 2a cos a sin a
4) cos 2a 2 cos a 1
=
=
= -
= -
2
3
3
5) cos 2a 1 2 sin a
6) sin 3a 3 sin a 4 sin a
7) cos 3a 4 cos a 3 cos a
= -
= -
= -
Cụng thc h bc :

2
2

sin
sin
+ -
+ =
+ -
- = -
+ -
+ =
+ -
- =
Cụng thc bin i tớch -> tng:
( ) ( )
[ ]
( ) ( )
[ ]
( ) ( )
[ ]
1
1) cos a cos b cos cos
2
1
2) sin a sin b cos cos
2
1
3) sin a cos b sin sin
2
a b a b
a b a b
a b a b
-


p
p
p
( )
( )
( )
( )
( )
( )
( )
( )
( )
( )
( )
( )
'
2
2
2
n n 1
x x
x x
a
1. C ' 0
2. x ' 1
1 1
3.
x x
4.

=
=
=
ổ ử






ố ứ
=
x
f(x)


cựng du vi a
x


f(x)
cựng du vi a
b / 2a
-
0
cựng du vi a
f(x)
0
2
x


2
2
3) a sin u b sin ucos u c cos d
u
+ + =
Cỏch gii:

B1: thay
cos u 0
=
vo pt
B2: chia 2 v ca pt cho
2
cos u

( )
*
4)a sinu cos u b sinu cos u c
+ =
Cỏch gii:

t
t sinu cos u 2 sin u
4
= + = +
p
ổ ử



m
n m
n
n m n m
m
n n.m
a.b a .b
1
7
a
a a
1)a .a a 5)
b b
a
2) a 6)
a
3) a a )a
4) a
a
-
-
+
=
ổ ử


= =




a
2)
4) a
6)
7)
8)
9)
10)
11)
12)
13)
1)log b b a
log 1 0
3) log a 1
b
5) log a
log b b log b log b
b
log log b log b
b
1
log log b
b
log b log b
1
log b log b
n
log b
log b
log a

'
2
1) u v w ' u ' v ' w '
2) uv ' u ' v uv '
3) ku ' k.u '
u u ' v uv '
4)
v v
1 u '
5)
u u
+ - = + -
= +
=
-
=
= -
ổ ử






ố ứ
ổ ử





cx d
cx d
ax bx c
2)y
dx e
adx 2aex be dc
y '
dx e
a x b x c
3)y
a x b x c
y '
a x b x c
+ +
+
= ị =
+
+
+ +
=
+
+ + -
ị =
+
+ +
=
+ +
ị =
+ +
o hm ca hm hp:

u '
7. cot u '
sin u
8. e ' e u
9. a ' a u ln a
u '
10. ln u '
u
u '
11. log u '
uln a
u ' n.u .u'
. '
. '.
-
= -
=
=
= -
=
= -
=
=
=
=
ổ ử





1
8) du tan u C
cos u
1
9) du cot u C
sin u
1
10) e du e C
k
a
11) a du C
ln a
1 1
12) du ln au b C
au b a
a+
a
=
= +
= + a ạ -
a +
= +
= - +
= - +
= +
= +
= - +
= +
= +
= + +

2
n n 1
x x
x x
a
1. C ' 0
2. x ' 1
1 1
3.
x x
4.
1
5. x '
2 x
6. sin x ' cos x
7. cos x ' sin x
1
8. tan x '
cos x
1
9. cot x '
sin x
10. e ' e
11. a ' a lna
1
12. ln x '
x
1
13. log x '
x ln a


f(x)
cựng du vi a
b / 2a
-
0
cựng du vi a
f(x)
0
2
x
0
cựng dutrỏi du
x


1
x
cựng du


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