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24 tháng 8 2019

C, \(x^2+2xy+y^2-x-y-12\)

\(=\left(x+y\right)^2-\left(x+y\right)-12\)

\(=\left(x+y\right)^2-4\left(x+y\right)+3\left(x+y\right)-12\)

\(=\left(x+y\right)\left(x+y-4\right)+3\left(x+y-4\right)\)

\(=\left(x+y-4\right)\left(x+y+3\right)\)

28 tháng 8 2019

a) (x^2-3)^2+16

=(x4-2.x2.3+32)+16

=(x4-6x2+9)+16

=x4-6x2+9+16

=x4-6x2+25

19 tháng 9 2018

7(x - 3) - x(3 - x)

= (x - 3)(7 + x)

chỉ bt có v mà k bt có đúng k 

19 tháng 9 2018

1 ) 7 ( x - 3 ) - x ( 3 - x ) 

= 7 ( x - 3 ) + x ( x - 3 )

= ( x - 3 ) ( 7 + x )

2 ) 4x2 - 6x + 3 - 2x

= 4x2 - 2x - 6x + 3

= 2x ( 2x - 1 ) - 3 ( 2x - 1 )

= ( 2x - 1 ) ( 2x - 3 )

3 ) ( 4 - x ) - 4x + x2

= ( 4 - x ) - x ( 4 - x )

= ( 4 - x ) (  1 - x )

4 ) x2 - 2xy + y2

= ( x - y )2

1 tháng 1 2022

`a. =4(x^2+4x+3)=4(x^2+3x+x+3)=4(x+3)(x+1)`

`b. =x^2+8x-7x-56=x(x+8)-7(x+8)=(x+8)(x-7)`

`c. =x^2-9x+8x-72=x(x-9)+8(x-9)=(x-9)(x+8)`

`d. =(x-y)^2-9=(x-y-3)(x-y+3)`

1 tháng 9 2018

( 3y - 2 ) 2 = ( 3y - 2 )6

TH1  : ( 3y - 2 ) = 0

=> 02 = 02  ( luôn đúng )

=> 3y = 2

=> y = \(\frac{2}{3}\)

TH2 : 3y - 2 khác 0

=> ( 3y - 2 )2 : ( 3y - 2 )2 = ( 3y - 2 )6 : ( 3xy - 2 )2

=> ( 3y - 2 )4 = 1

=> 3y - 2 = + 1

=> y = 1 hoặc y = \(\frac{1}{3}\)

21 tháng 10 2021

a) \(x^2-xy+x-y\)

\(=x\left(x-y\right)+\left(x-y\right)\)

\(=\left(x+1\right)\left(x-y\right)\)

21 tháng 10 2021

b) \(x^2+5x+6\)

\(=x^2+2x+3x+6\)

\(=x\left(x+2\right)+3\left(x+2\right)\)

\(=\left(x+3\right)\left(x+2\right)\)

19 tháng 9 2018

1 ) 3yx - 6xy2 

= 3xy ( 1 - 2y )

2 ) 5ab2 - 20a3b2

= 5ab2 ( 1 - 4a2 )

= 5ab2 ( 1 - 2a ) ( 1 + 2a )

3 ) 3x - 3b - y ( b - x )

= 3 ( x - b ) + y ( x - b )

= ( x - b ) ( 3 + y ) 

19 tháng 9 2018

1)3xy-6xy2=3xy(1-2y)

2)5ab2-20a3b2=5ab2(1-4a2)=5ab2[12-(2a)2]=5ab2(1+2a)(1-2a)

3)3x-3b-y(b-x)=3x-3b-by+xy=(3x+xy)-(3b+by)=3x(1+y)-3b(1+y)=3(1+y)(x-b)

18 tháng 2 2021

 a) 3x2 – 7x + 2

\(=3x^2-6x-x+2\)

\(=\left(3x^2-6x\right)-\left(x-2\right)\)

\(=3x\left(x-2\right)-\left(x-2\right)\)

\(=\left(x-2\right)\left(3x-1\right)\)

 b) a(x2 + 1) – x(a2 + 1)

\(=ax^2+a-\left(a^2x+x\right)\)

\(=a\left(x^2+1\right)-x\left(a^2+1\right)\)

.......?

 

 

 

 

a) Ta có: \(3x^2-7x+2\)

\(=3x^2-6x-x+2\)

\(=3x\left(x-2\right)-\left(x-2\right)\)

\(=\left(x-2\right)\left(3x-1\right)\)

b) Ta có: \(a\left(x^2+1\right)-x\left(a^2+1\right)\)

\(=x^2a+a-a^2x-x\)

\(=\left(x^2a-a^2x\right)+\left(a-x\right)\)

\(=xa\left(x-a\right)-\left(x-a\right)\)

\(=\left(x-a\right)\left(xa-1\right)\)

c) Ta có: \(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)

\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)

\(=\left(x^2+7x\right)^2+22\left(x^2+7x\right)+120-24\)

\(=\left(x^2+7x\right)^2+22\left(x^2+7x\right)+96\)

\(=\left(x^2+7x\right)^2+16\left(x^2+7x\right)+6\left(x^2+7x\right)+96\)

\(=\left(x^2+7x\right)\left(x^2+7x+16\right)+6\left(x^2+7x+16\right)\)

\(=\left(x^2+7x+16\right)\left(x^2+7x+6\right)\)

\(=\left(x^2+7x+16\right)\left(x+1\right)\left(x+6\right)\)

d) Ta có: \(\left(a+1\right)\left(a+3\right)\left(a+5\right)\left(a+7\right)+15\)

\(=\left(a^2+8a+7\right)\left(a^2+8a+15\right)+15\)

\(=\left(a^2+8a\right)^2+22\left(a^2+8a\right)+105+15\)

\(=\left(a^2+8a\right)^2+22\left(a^2+8a\right)+120\)

\(=\left(a^2+8a\right)^2+12\left(a^2+8a\right)+10\left(a^2+8a\right)+120\)

\(=\left(a^2+8a\right)\left(a^2+8a+12\right)+10\left(a^2+8a+12\right)\)

\(=\left(a^2+8a+12\right)\left(a^2+8a+10\right)\)

\(=\left(a+2\right)\left(a+6\right)\left(a^2+8a+10\right)\)

23 tháng 12 2022

bài 11

a) \(x^2-xy+x\\ =x\left(x-y+1\right)\)

b)

\(x^2-2xy-4+y^2\\ =\left(x^2-2xy+y^2\right)-4\\ =\left(x-y\right)^2-4\\ =\left(x-y-2\right)\left(x-y+2\right)\)

c)

\(x^3-x^2-16x+16\\ =x^2\left(x-1\right)-16\left(x-1\right)\\ =\left(x-1\right)\left(x-4\right)\left(x+4\right)\)

bài 12

\(2x\left(x-5\right)-x\left(3+2x\right)=26\)

\(2x^2-10x-3x-2x^2=26\)

\(-13x=26\\ x=-2\)

b)

\(2\left(x+5\right)-x^2-5x=0\\ 2\left(x+5\right)-x\left(x+5\right)=0\\ \left(x+5\right)\left(2-x\right)=0\\ \left[{}\begin{matrix}x+5=0\\2-x=0\end{matrix}\right.\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)