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\(1,=x\left(x^2-2x+1-y^2\right)=x\left[\left(x-1\right)^2-y^2\right]=x\left(x-y-1\right)\left(x+y-1\right)\\ 2,=\left(x+y\right)^3\\ 3,=\left(2y-z\right)\left(4x+7y\right)\\ 4,=\left(x+2\right)^2\\ 5,Sửa:x\left(x-2\right)-x+2=0\\ \Leftrightarrow\left(x-2\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
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a) Ta có: \(x^2-2xy+y^2-2x+2y\)
\(=\left(x-y\right)^2-2\left(x-y\right)\)
\(=\left(x-y\right)\left(x-y-2\right)\)
b) Ta có: \(x^2-4x+4-x^2y+2xy\)
\(=\left(x-2\right)^2-xy\left(x-2\right)\)
\(=\left(x-2\right)\left(x-2-xy\right)\)
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1 ) ( 2x - 1 ) ( 8x + 12 ) + x2( 2x - 1 ) + ( 1 - 2x ).( 2x - 3 )
= ( 2x - 1 ) . ( 8x + 12 ) + x2 ( 2x - 1 ) - ( 2x - 1 ) . ( 2x - 3 )
= ( 2x - 1 ) . ( 8x + 12 + x2 - 2x + 3 )
= ( 2x - 1 ) . ( x2 + 6x + 15 )
2 ) 3x ( x - y ) - 2y ( y - x ) - 4x + 4y
= 3x ( x - y ) + 2y ( x - y ) - 4. ( x - y )
= ( x - y ) ( 3x + 2y - 4 )
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\(1,x^2-xy-2x+2y\)
\(x\left(x-2\right)-y\left(x-2\right)\)
\(\left(x-2\right)\left(x-y\right)\)
\(2,x^2+4x+4-y^2\)
\(\left(x+2\right)^2-y^2\)
\(\left(x+2-y\right)\left(x+2+y\right)\)
\(3,x^2+x+y-y^2\)
\(\left(x-y\right)\left(x+y\right)+\left(x+y\right)\)
\(\left(x+y\right)\left(x-y+1\right)\)
\(4,x^3-x^2-4x+4\)
\(x^2\left(x-1\right)-4\left(x-1\right)\)
\(\left(x-1\right)\left(x^2-4\right)\)
\(\left(x-1\right)\left(x-2\right)\left(x+2\right)\)
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Lời giải:
$-4x^4+1+2x^2y-y=-(4x^4-1)+(2x^2y-y)$
$=-(2x^2-1)(2x^2+1)+y(2x^2-1)$
$=(2x^2-1)[-(2x^2+1)+y]$
$=(2x^2-1)(y-2x^2-1)$