a) Ta có: xy+3x+y=8
\(\Leftrightarrow x\left(y+3\right)+y+3=11\)
\(\Leftrightarrow\left(y+3\right)\cdot\left(x+1\right)=11\)
\(\Leftrightarrow y+3;x+1\inƯ\left(11\right)\)
\(\Leftrightarrow y+3;x+1\in\left\{1;-1;11;-11\right\}\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}y+3=1\\x+1=11\end{matrix}\right.\\\left\{{}\begin{matrix}y+3=11\\x+1=1\end{matrix}\right.\\\left\{{}\begin{matrix}y+3=-1\\x+1=-11\end{matrix}\right.\\\left\{{}\begin{matrix}y+3=-11\\x+1=-1\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}y=-2\\x=10\end{matrix}\right.\\\left\{{}\begin{matrix}y=8\\x=0\end{matrix}\right.\\\left\{{}\begin{matrix}y=-4\\x=-12\end{matrix}\right.\\\left\{{}\begin{matrix}y=-14\\x=-2\end{matrix}\right.\end{matrix}\right.\)
Vậy: (x,y)={(10;-2);(0;8);(-12;-4);(-2;-14)}
