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

=(x² -4x²)-((x²-2x)(x+2))

=(x²-4x²)-(x³+2x²-2x²-4x)

=x²-4x²-x³+4x

=-x³-3x²+4x=-x(x²+3x-4)

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

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

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

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

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

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

\(=x\left(x-1\right)\left(x-4\right)\)

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

Lời giải:

$x^4-x^3-2x-4=(x^4+x^3)-(2x^3+2x^2)+(2x^2+2x)-(4x+4)$

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

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

$=(x+1)[x^2(x-2)+2(x-2)]=(x+1)(x-2)(x^2+2)$

56.B

57.B

58.B

56B

57B

58B

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

= (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

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

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

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

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

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