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30 tháng 5 2019

#)Giải :

a, 16( x - y )2 - 49( x + y )2

= [4( x - y )]2 - [7( x + y )]2

= [ 4x - 4y ]2 - [ 7x + 7y ]2

= ( 4x - 4y + 7x + 7y )[( 4x - 4y ) - ( 7x + 7y )]

= ( 11x - y )( x - 11y )

           #~Will~be~Pens~#

26 tháng 11 2017

a) \(x^2-xy+x-y\)

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

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

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

b) \(2xy-x^2-y^2+16\)

\(=16-\left(x^2-2xy+y^2\right)\)

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

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

c) \(x^2-6x-16\)

\(=x^2-6x+9-25\)

\(=\left(x^2-6x+9\right)-25\)

\(=\left(x-3\right)^2-5^2\)

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

\(=\left(x-8\right)\left(x+2\right)\)

26 tháng 11 2017

a)  x2 - xy + x - y = x(x - y) + (x - y) = (x - y)(x + 1)

b) 2xy - x2 - y2 + 16 = 16 - (x - y) = (4 - x + y)(4 + x - y)

c) x2 - 6x - 16 = (x - 3)2 - 25 = (x - 3 - 5)(x - 3 + 5) = (x - 8)(x + 2)

2 tháng 11 2018

a) A = (x + 1)(y - 2) - (2 - y)2

= -[(x + 1)(2 - y) + (2 - y)2]

= -[(x + 1 - 2 + y)(2 - y)]

= -[(x - 1 + y)(2 - y)]

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

2 tháng 11 2018

Bài 2:

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

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

\(A=\left(y-2\right)\left(x+1-y+2\right)\)

\(A=\left(y-2\right)\left(x-y+3\right)\)

b) \(B=x^2-6xy+9y^2+4x-12y\)

\(B=\left[x^2-2\cdot x\cdot3y+\left(3y\right)^2\right]+4\left(x-3y\right)\)

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

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

Bài 3:

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

\(\left(3x^2+3x-18\right)-\left(3x^2+x\right)-2=0\)

\(3x^2+3x-18-3x^2-x-2=0\)

\(2x-20=0\)

\(x=10\)

b) \(6x^2+13x+5=0\)

\(6x^2+10x+3x+5=0\)

\(2x\left(3x+5\right)+\left(3x+5\right)=0\)

\(\left(3x+5\right)\left(2x+1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}3x+5=0\\2x+1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{-5}{3}\\x=\frac{-1}{2}\end{cases}}}\)

a: \(2x^3+x^2-13x+6\)

\(=2x^3-4x^2+5x^2-10x-3x+6\)

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

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

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

b: \(2x^2+y^2-6x+2xy-2y+5=0\)

\(\Leftrightarrow x^2+2xy+y^2+x^2-4x+4-2x-2y+1=0\)

\(\Leftrightarrow\left(x+y\right)^2+\left(x-2\right)^2-2\left(x+y\right)+1=0\)

\(\Leftrightarrow\left(x-2\right)^2+\left(x+y-1\right)^2=0\)

=>x-2=0 và x+y-1=0

=>x=2 và y=-1

21 tháng 7 2015

a,  x^2 - x  - y^2 - y

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

= ( x- y)(x+y) - ( x+y )

= ( x - y  - 1 )(x+ y)

 

b, x^2+ 6x + 9 - y^2 

= ( x+ 3)^2 - y^2 

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

8 tháng 9 2019

\(x^2-y^2+4x+4\)

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

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

\(4x^2-y^2+8\left(y-2\right)\)

\(=4x^2-\left(y^2-8y+16\right)\)

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

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

\(x^3+6x^2-13x-42\)

\(x^3+6x^2-13x-42\)

\(=\left(x+7\right)\left(x-3\right)\left(x+2\right)\)

2 tháng 8 2016

b, \(2x^3-x^2+3x+6\)

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

\(=2x^2\left(x+1\right)-3x\left(x+1\right)+6\left(x+1\right)\)

\(=\left(x+1\right)\left(2x^2-3x+6\right)\)

20 tháng 11 2016

a) \(\left(x+8\right)^2-2\left(x+8\right)\left(x-2\right)+\left(x-2\right)^2\)

\(=\left[\left(x+8\right)-\left(x-2\right)\right]^2\)

\(=\left(x+8-x+2\right)^2\)

\(=10^2\)

\(=2^2.5^2\)

b)\(x^3-4x^2-12x+27=\left(x^3+27\right)-\left(4x^2+12x\right)\)

\(=\left(x+3\right)\left(x^2-3x+9\right)-4x\left(x+3\right)\)

\(=\left(x+3\right)\left(x^2-3x+9-4x\right)\)

\(=\left(x+3\right)\left(x^2-7x+9\right)\)

c)\(x^3+6x^2+11x+6=x^3+x^2+5x^2+5x+6x+6\)

\(=x^2\left(x+1\right)+5x\left(x+1\right)+6\left(x+1\right)\)

\(=\left(x+1\right)\left(x^2+5x+6\right)\)

\(=\left(x+1\right)\left(x^2+2x+3x+6\right)\)

\(=\left(x+1\right)\left[x\left(x+2\right)+3\left(x+2\right)\right]\)

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

d)\(x^3+6x^2-13x-42=x^3-3x^2+9x^2-27x+14x-42\)

\(=x^2\left(x-3\right)+9x\left(x-3\right)+14\left(x-3\right)\)

\(=\left(x-3\right)\left(x^2+9x+14\right)\)

\(=\left(x-3\right)\left(x^2+2x+7x+14\right)\)

\(=\left(x-3\right)\left[x\left(x+2\right)+7\left(x+2\right)\right]\)

\(=\left(x-3\right)\left(x+2\right)\left(x+7\right)\)