Ta có:
\(PT\Leftrightarrow x\left(y+2\right)+2\left(y+2\right)=5\)
\(\Leftrightarrow\left(x+2\right)\left(y+2\right)=5\)
Ta có: -TH1:\(\left\{{}\begin{matrix}x+2=5\\y+2=1\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=-1\end{matrix}\right.\left(TM\right)\)
-TH2:\(\left\{{}\begin{matrix}x+2=1\\y+2=5\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=3\end{matrix}\right.\)\(\left(TM\right)\)
-TH3:\(\left\{{}\begin{matrix}x+2=-1\\y+2=-5\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=-3\\y=-7\end{matrix}\right.\)\(\left(TM\right)\)
-TH4:\(\left\{{}\begin{matrix}x+2=-5\\y+2=-1\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=-7\\y=-3\end{matrix}\right.\)\(\left(TM\right)\)
Vậy \(\left(x;y\right)\in\left\{\left(3:-1\right);\left(-1;3\right);\left(-3;-7\right);\left(-7;-3\right)\right\}\)
