\(\left(8x-16\right)\left(x-5\right)=0\\ \Leftrightarrow8\left(x-2\right)\left(x-5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-2=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\)
(8\(x\) - 16).(\(x\) - 5) = 0
\(\left[{}\begin{matrix}8x-16=0\\x-5=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}8x=16\\x-5=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\)
Vậy \(x\) \(\in\){2; 5}