=(x² -4x²)-((x²-2x)(x+2))

=(x²-4x²)-(x³+2x²-2x²-4x)

=x²-4x²-x³+4x

=-x³-3x²+4x=-x(x²+3x-4)

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

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

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

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

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

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

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

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

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

Ta có: \(\left(x-1\right)^2\ge0\forall x\)

\(\Rightarrow\left(x-1\right)^2+1\ge1\forall x\)

\(\Rightarrow\left(x-1\right)^2+1>0\forall x\)

\(\Rightarrow\)đa thức \(x^2-2x+2\)vô nghiệm

\(\Rightarrow\)đa thức \(x^2-2x+2\)không phân tích được thành nhân tử

Cái kia tương tự

Tham khảo nhé~

`(x-1)^2-2(x-1)(2x+1)+(2x+1)^2`

`=(x-1-2x-1)^2`

`=(-x-2)^2`

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

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

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

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