1\()\)x²-2x=0

x\(\left(x-2\right)\)=0

\(\Rightarrow\)x=0 hoặc x=2

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

\(\Rightarrow x^2-1=0\)hoặc x-1=0

\(\Leftrightarrow\)x=1 hoặc x=1

.\(x^2-6x+5\)=\(x^2-2\times3x+5\)

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

=\(\left(3x-1\right)^2-\left(x-1\right)^2=0\)

=\(\left(3x-1+x-1\right)\left(3x-1-x+1\right)=0\)

=\(\left(4x-2\right)2x=0\)

=>\(\left[{}\begin{matrix}4x-2=0\\2x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{1}{2}\\x=0\end{matrix}\right.\)

(x+1)(x²-x+1)-(x³+27)

=(x³+1)-(x³⁺27)

=x³+1-x³-27

=-26

$x^3+3x^2-x-3$

$\mathrm{=(x3+3x2)-(x+3)}$

$\mathrm{=x2(x+3)-(x+3)}$

$\mathrm{=(x+3)(x2-1)}$

$\mathrm{=(x+3)(x-1)(x+1)}$

x³+2x²+2x+1

(x³+1)+(2x²+2x)

(x+1)(x²-x+1)+2x(x+1)

(x+1)(x² -x+1+2x)

(x+1)(x²+x+1)

\(\left(8x^4-4x^3+x^2\right):2x^2\)

\(\Leftrightarrow4x^2-2x+\frac{1}{2}\)

\(x^2+x-12\)

\(\Leftrightarrow\)\(x^2-3x+4x-12\)

\(\Leftrightarrow x\left(x-3\right)+4\left(x-3\right)\)

\(\Leftrightarrow\left(x-3\right)\left(x+4\right)\)