\(x^2+7⋮x-2\Leftrightarrow\)\(\dfrac{x^2+7}{x-2}\in Z\Leftrightarrow\dfrac{x^2-4+11}{x-2}\in Z\Leftrightarrow x+2+\dfrac{11}{x-2}\in Z\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=1\\x-2=-1\\x-2=11\\x-2=-11\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=1\\x=13\\x=-9\end{matrix}\right.\)
Kl: x=3, x=1, x=13, x=-9