\(\left(x^2-9\right)\left(-2x+8\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-9=0\\-2x+8=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2=9\\-2x=-8\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\pm3\\x=4\end{matrix}\right.\)
\(S=\left\{\pm3,4\right\}\)
(x2 - 9)(-2x + 8) = 0
<=> (x - 3) (x + 3) (-2x + 8) = 0
<=. \(\left[{}\begin{matrix}x-3=0\\x+3=0\\\text{-2x + 8 = 0}\end{matrix}\right.\) <=> \(\left[{}\begin{matrix}x-3=0\\x+3=0\\\text{-2x + 8 = 0}\end{matrix}\right.\) <=> \(\left[{}\begin{matrix}x=3\\x=-3\\\text{}x=4\end{matrix}\right.\)
Vậy x \(\in\left\{3;-3;4\right\}\)