Lại thấy: \(\left(x+y-2\right)^2+7\ge7\forall x;y\)
Do đó, \(\left(x+y-2\right)^2+7=\dfrac{14}{\left|y-1\right|+\left|y-3\right|}=7\) \(\Rightarrow\left\{{}\begin{matrix}y-1\ge0\\y-3\le0\\x+y-2=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y\ge1\\y\le3\\x+y=2\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}1\le y\le3\\x+y=2\end{matrix}\right.\)
Mà x;y nguyên => \(\left[{}\begin{matrix}\left(x;y\right)=\left(1;1\right)\\\left(x;y\right)=\left(0;2\right)\\\left(x;y\right)=\left(-1;3\right)\end{matrix}\right.\)