Ta có: \(x+7⋮x+2\)
\(\Rightarrow\left(x+2\right)+5⋮x+2\)
\(\Rightarrow5⋮x+2\)
\(\Rightarrow x+2U\left(5\right)=\left\{-1;1;-5;5\right\}\)
\(\Rightarrow\left\{{}\begin{matrix}x+2=-1\Rightarrow x=-3\\x+2=1\Rightarrow x=-1\\x+2=-5\Rightarrow x=-7\\x+2=5\Rightarrow x=3\end{matrix}\right.\)
Vậy \(x\in\left\{-3;-1;-7;3\right\}\)