Ta có: \(13⋮\left(x-3\right)\)
\(\Rightarrow x-3\inƯ\left(13\right)=\left\{\pm1;\pm13\right\}\)
TH1: \(x-3=\pm1\Leftrightarrow\left[{}\begin{matrix}x-3=1\\x-3=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=2\end{matrix}\right.\)
TH2: \(x-3=\pm13\Leftrightarrow\left[{}\begin{matrix}x-3=13\\x-3=-13\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=16\\x=-10\end{matrix}\right.\)
Vậy \(x\in\left\{-10;2;4;16\right\}\)