\(x+x.y+y+1=2\)
\(\Leftrightarrow x\left(y+1\right)+\left(y+1\right)=2\)
\(\Leftrightarrow\left(x+1\right)\left(y+1\right)=2\)
vì x,y thuộc Z nên (x+1) và (y+1) \(\inƯ\left(2\right)=\left\{-1;1;2;-2\right\}\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+1=1\\y+1=2\end{matrix}\right.\\\left\{{}\begin{matrix}x+1=2\\y+1=1\end{matrix}\right.\\\left\{{}\begin{matrix}x+1=-1\\y+1=-2\end{matrix}\right.\\\left\{{}\begin{matrix}x+1=-2\\y+1=-1\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=0\\y=1\end{matrix}\right.\\\left\{{}\begin{matrix}x=1\\y=0\end{matrix}\right.\\\left\{{}\begin{matrix}x=-2\\y=-3\end{matrix}\right.\\\left\{{}\begin{matrix}x=-3\\y=-2\end{matrix}\right.\end{matrix}\right.\)
vậy \(S=\left\{\left(0;1\right),\left(1;0\right),\left(-2;-3\right),\left(-3;-2\right)\right\}\)
