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

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

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

\(x^3-6x^2-x+30=x^3+2x^2-8x^2-16x+15x+30\)

                                       \(=x^2\left(x+2\right)-8x\left(x+2\right)+15\left(x+2\right)\)

                                       \(=\)  \(\left(x^2-8x+15\right)\left(x+2\right)\)

                                      \(=\left(x^2-5x-3x+15\right)\left(x+2\right)\)

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

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

\(-\left(x^2+2.x.\frac{1}{2}+\left(\frac{1}{2}\right)^2=0\right)\)

=> -(x+1/2)^2 =0

=> x + \(\frac{1}{2}\)=0

=> x = \(-\frac{1}{2}\)

x²-2x-1=0

<=> (x-1)²=1

<=> \(\orbr{\begin{cases}x-1=1\\x-1=-1\end{cases}}\)

<=>\(\orbr{\begin{cases}x=2\\x=0\end{cases}}\)

\(2x^3+3x^2+2x+3=0\)

\(\Leftrightarrow x^2\left(2x+3\right)+\left(2x+3\right)=0\)

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

\(\Leftrightarrow2x+3=0\Leftrightarrow x=-\frac{3}{2}\)

\(2x^3-12x^2+18x=0\)

\(x\left(2x^2-12x+18\right)=0\)

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

........

\(3\left(x+4\right)-x^2-4x=0\)

\(\Leftrightarrow3\left(x+4\right)-x\left(x+4\right)=0\)

\(\Leftrightarrow\left(x+4\right)\left(3-x\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+4=0\\3-x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-4\\x=3\end{cases}}}\)

Vậy x = -4 hoặc x =3

\(3x\left(x-2\right)+7\left(x-2\right)=0\)

\(\left(x-2\right)\left(3x+7\right)=0\)

\(\hept{\begin{cases}x-2=0\\3x+7=0\end{cases}}\)

\(\hept{\begin{cases}x=2\\x=-\frac{7}{3}\end{cases}}\)

(x - 2)(3x + 7) = 0

=> \(\orbr{\begin{cases}x+2=0\\3x+7=0\end{cases}=>\orbr{\begin{cases}x=-2\\x=\frac{-7}{3}\end{cases}}}\)

\(x^3-x=0\)

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

\(\Rightarrow\orbr{\begin{cases}x=0\\x^2-1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x^2=1\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=\pm1\end{cases}}\)

Vậy  \(x\in\left\{0;\pm1\right\}\)

_Chúc bạn học tốt_

\(x^3-x=0\)

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

\(\Leftrightarrow x.\left(x-1\right)\left(x+1\right)=0\)

Suy ra: x = 0 hoặc x - 1= 0 => x = 1

  Hoặc x + 1 = 0 => x = -1

Vậy x = 0 hoặc x = -1 hoặc x = 1