Áp dụng bđt |a|+|b| \(\ge\) |a+b| ta có:
|y - 1| + |y - 3| = |y - 1| + |3 - y| \(\ge\) |y - 1 + 3 - y| = |2| = 2
\(\Leftrightarrow\dfrac{14}{\left|y-1\right|+\left|y-3\right|}\le\dfrac{14}{2}=7\)
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 => \(\left[{}\begin{matrix}\left(x;y\right)=\left(1;1\right)\\\left(x;y\right)=\left(0;2\right)\\\left(x;y\right)=\left(-1;3\right)\end{matrix}\right.\)
Vậy ...
