\(1,=x\left(x^2-2x+1-y^2\right)=x\left[\left(x-1\right)^2-y^2\right]=x\left(x-y-1\right)\left(x+y-1\right)\\ 2,=\left(x+y\right)^3\\ 3,=\left(2y-z\right)\left(4x+7y\right)\\ 4,=\left(x+2\right)^2\\ 5,Sửa:x\left(x-2\right)-x+2=0\\ \Leftrightarrow\left(x-2\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

\(=\left(x^2+x\right)^2-2\left(x^2+x\right)+1-16\\ =\left(x^2+x-1\right)^2-16\\ =\left(x^2+x-1-4\right)\left(x^2+x-1+4\right)\\ =\left(x^2+x-5\right)\left(x^2+x+3\right)\)

= (1-4x²)-x(x²-4)

=1-4x²-x³+4x

( 1 + 2x ) ( 1 - 2x ) - x ( x + 2 ) ( x - 2 )

= ( 1 - 4 x² ) - x ( x² - 4 )

= 1 - 4 x² - x³ + 4x

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

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

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

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

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

Lời giải:

$2x^2-2xy-4y^2=2(x^2-xy-2y^2)$

$=2[(x^2-2xy)+(xy-2y^2)]$

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

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

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$x^2-2x-4y^2-4y=(x^2-2x+1)-(4y^2+4y+1)$

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

$=(x-2y-2)(x+2y)$
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$x^2-4y^2-x-2y=(x^2-4y^2)-(x+2y)=(x-2y)(x+2y)-(x+2y)$

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

56.B

57.B

58.B

56B

57B

58B