Đáp án:

`(x - 6) (x^2 - 1) = 0`

\(\left[ \begin{array}{l}x - 6 = 0\\x^2 - 1 = 0\end{array} \right.\) 

\(\left[ \begin{array}{l}x=0+6\\x^2=0+1\end{array} \right.\) 

\(\left[ \begin{array}{l}x=6\\x^2=1\end{array} \right.\) 

\(\left[ \begin{array}{l}x=6\\x^2=1^2\end{array} \right.\) 

\(\left[ \begin{array}{l}x=6\\x=\pm1\end{array} \right.\) 

Vậy `x in {6;+-1}`