$5x+2 \quad\vdots\quad x+1$
$→ x+1 \quad\vdots\quad x+1$
$→ 5x-5 \quad\vdots\quad x+1$
$→ (5x+2)-(5x-5) \quad\vdots\quad x+1$
$→ 3 \quad\vdots\quad x+1$
$→ x+1 \in Ư(3)=\{±1;±3\}$
Ta có bảng sau:
\begin{array}{|c|c|}\hline x+1&-1&1&-3&3\\\hline x&-2&0&-4&2\\\hline\end{array}
Vậy $x=\{±2;-4;0\}$