can thuc va cac phep toan ve can thuc

7/23/2019 Can Thuc Va Cac Phep Toan Ve Can Thuc

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Gi?i:

a)23

2300 =23

100.23 =23

100.23 = 100 = 10

b)x

x

7

63 3 =x

xx

7

.7.9 2 =x

xx

7

73 = 3x v2i x > 0

III. BI TLP #$NGHN

A C

B D= n@u AD = BC

A A.M=

B B.M;

A A:N=

B B:N (M 0; N 0; B 0)

A - A A A - A;

B - B B - B B= = =
http://www.xuctu.com/mailto:[email protected]:[email protected]://www.xuctu.com/


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b)3 2

3 2 2 3

3 2 1

2 2

x xy y

x x y xy y x y

+ +=

+

TI&T 3: PHN TCH #A TH*C THNH NHN TU

I. KI&N TH*C C5B6N1. #Vnh nghXa:

Phn tch 1a th,c thnh nhn tD (hay th$a s') l bi@n 1:i 1a th,c 1 thnh m0ttch c7a nhRng 1a th,c.

V d):

a) 2x2+ 5x - 3 = (2x - 1).(x + 3)

b) x - 2 x y +5 x - 10y = [( x )2 2 y x ] + (5 x - 10y)

= x ( x - 2y) + 5( x - 2y)= ( x - 2y)( x + 5)

2. Cc ph>Ing php phn tch Ha thJc thnh nhn tZ

a) Ph!ng php "#t nhn t'chung :

N@u tt c?cc h6ng tDc7a 1a th,c c m0t nhn tDchung th 1a th,c 1 1%5c biBudiSn thnh m0t tch c7a nhn tDchung v2i m0t 1a th,c khc.

Cng th,c:

V d):

1. 5x(y + 1) 2(y + 1) = (y + 1)(5x - 2)

2. 3x + 12 x y = 3 x ( x + 4y)

b) Ph!ng php dng h)ng "*ng th,c:

AB + AC = A(B + C)


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"#$% '( ) "#* +, -.# / 012 3 4 5 6678768 9 7:;:;6Ing php phn tch Ha thJc thnh nhn tZ:

d. Ph!ng php tch m/t h.ng t':(tr$%ng h(p *+c bi,t c/a tam th2c b4c 2 c nghi,m)

Tam th,c bUc hai c d6ng: ax2+ bx + c = ax2+ b1x + b2x + c ( 0a ) n@u1 2

1 2

b b ac

b b b

= + =

V d):

a) 2x2-3x + 1 = 2x2-2x-x +1

= 2x(x -1) -(x -1) = (x -1)(2x -1)

( ) ( )

( )( )

3 2 2 2

1 2 1

2 1

y y y y y

y y y

y y

+ = +

=

=

b)

e. Ph!ng php thm, b2t cng m/t h.ng t':

V d):

a) y4+ 64 = y4+ 16y2+ 64 -16y2

= (y2

+ 8)2

- (4y)2

= (y2+ 8 -4y)(y2+ 8 + 4y)

b) x2+ 4 = x2+ 4x + 4 -4x = (x + 2)2-4x

= (x + 2)2- ( )2

2 = ( ) ( )2 2 2 2x x x x + + +

g. Ph!ng php ph3i h5p nhi6u ph!ng php:


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Gi?i: a) KiEu ki=n A xc 1Lnh: 0; 1a a>

Ta c: 1 1 2:1 ( 1) 1 ( 1)( 1)

aA

a a a a a a

= + + +

. 1 1 2 1 1 1: .

( 1) ( 1)( 1) ( 1) 1

a a a a a a

a a a a a a a a

+ = = =

+ +


16/20

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> > >

III. BI TLP #$NGHN

Bi 1: Rt gAn: 3 23 1 3 1

B= ++

Bi 2: Cho biBu th,c:2 2 2 2 2 2a a bQ 1 :

a b a b a a b = +

a) Rt gAn Q.

b) Tm gi trLc7a Q khi a = 3b

Bi 3: Cho biBu th,c P 2 x 2 x 4x x 3:x 42 x 2 x 2 x x

+ = +

a) Rt gAn P.

b) Tm cc gi trL c7a x 1BP > 0, P < 0.

c) Tm gi trLc7a x sao cho P 1= .

TI&T 10:BI&N #(I #5N GI6N BIjU TH*C

CH*A CkN TH*C BLC HAI (Ti0p )

I. KI&N TH*C C5B6N

a) ( ) ( )2 2A B A B A 0,B 0 ; A B A B A 0, B 0= < = ;

b) ( )A 1 AB AB 0,B 0B B=

c) ( )A A B

B 0BB

= > ;

d)( )

( )C A BC

A 0,B 0,A BA BA B

=

+.

( )( )

C A BC A 0,B 0,A B

A BA B

+=

V d): Rt gAn r9i so snh gi trLc7a M v2i 1 bi@t:

1a2a

1a:

1a

1

aa

1M

++

+

= v2i 1a,0a >

Gi?i:


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( ) ( )( )( )

( ) ( ) ( )( )

( ) ( )

2 22 2

3

x 1 x 1x x x x 1 x 1 x x. x xP . .

4 4 x x 1 x 1 4 x x 1 x 1

x 1 x 1 . x 1 x 1 x x 1 x 1x x 2 x 2. .

x 1 x 14 x 4 x

x x 1 x 1

2 x

+ + = = + +

+ + + + + = =

+ =

b) KB P = 0 ( )x x 1 0 + = x 0

x 1

=

=

Cc gi trLny khng thYa mn 1iEu ki=n, do 1 khng c gi trLno c7a x 1B P = 0.

III. BI TLP #$NGHN

Bi 1: Rt gAn biBu th,c: 5 5 5 55 5 5 5

+ + +


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Gi?i:

a) KiEu ki=n: 0 : 1x x

( ) ( )( ) ( )

( )

2 22 2

22

1 11 1 1 2 2

21 1 2 41 1

14 1 4 1

1 1 42

x xx x x x

P x x x xx x

xx x x x

x x xx x

+ + = = + +

= = =


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"#$% '( ) "#* +, -.# / 012 3 4 5 6678768 9 7:;:;6 > < . K@t h5p v2i 1iEu ki=n ta 1%5c: 10

3x< <

Cu 3: Gi?i ph%3ng trnh:2

14 11

x 3x 9= +

Gi?i: Ta c ph%3ng trnh2

14 11

x 3x 9

= + ( )( )

14 11

x 3 x 3 x 3

= ++

KKXK: x X 3 .

( )( ) ( )( ) ( )

14 11 14 x 3 x 3 x 3

x 3 x 3 x 3= + = + + +

+

2

2

14 x 9 x 3

x x 20 0

= + +

+ =

= 1 + 4.20 = 81 > 0, 81 9 = =

1 21 9 1 9

x 4;x 52 2

+ = = = = ,

x1= 4; x2= -5 1Eu tho?mn KKXKVUy ph%3ng trnh c hai nghi=m x1= 4; x2= -5.

TI&T 12: KIjM TRA

#$S/1

Cu 1: Rt gAn cc phn th,c sau:

a)2 3

2

x 4x 3

x 5x 6

+ +

b)2 24 4x 9y 12xy

2x 2 3y

+ +

c)2 3 2 3

xy 4y 2xy 4y

x y x y

++

Cu 2: Tnh: 2 1 3 1:2 14 2 3

+ +

Cu 3: Cho biBu th,c

1 x xA x

x x 1 x 1

= + +

a) Tm 1iEu ki=n xc 1Lnh c7a A? Rt gAn A.

b) Tnh gi trLc7a biBu th,c A khi x = 34

c) Tm x 1BA < 8.


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can thuc va cac phep toan ve can thuc

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