`x-3=y(x+2)`
`<=>x-3-y(x+2)=0`
`<=>(x+2)-y(x+2)-5=0`
`<=>(x+2)(1-y)=5`
Ta có `5=1.5=(-1).(-5)`
\begin{array}{|c|c|c|c|}\hline x+2&1&5&-1&-5\\\hline 1-y&5&1&-5&-1\\\hline x&-1&3&-3&-7\\\hline y&-4&0&6&2\\\hline x;y \in \mathbb{N}&(loại)&(nhận)&(loại)&(loại)\\\hline\end{array}