Ta có :
\(x\left(y+2\right)+y=1\)
\(\Leftrightarrow x\left(y+2\right)+\left(y+2\right)=1+2\)
\(\Leftrightarrow\left(y+2\right)\left(x+1\right)=3\)
Vì \(x;y\in N\Leftrightarrow y+2;x+1\in N;x+1;y+2\inƯ\left(3\right)\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x+1=1\\y+2=3\end{matrix}\right.\\\left[{}\begin{matrix}x+1=3\\y+2=1\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x=0\\y=1\end{matrix}\right.\\\left[{}\begin{matrix}x=2\\y=-1\end{matrix}\right.\end{matrix}\right.\)
Vì \(x,y\in N\Leftrightarrow\left[{}\begin{matrix}x=0\\y=1\end{matrix}\right.\)
Vậy \(\left(x,y\right)=\left(0,1\right)\) là giá trị cần tìm
