ĐKXĐ: x<>2
Để \(\dfrac{x}{2-x}=\left|y\right|\) và x,y nguyên thì \(\left\{{}\begin{matrix}\dfrac{x}{2-x}>=0\\x⋮2-x\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{x}{x-2}< =0\\x⋮x-2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}0< =x< 2\\x-2+2⋮x-2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}0< =x< 2\\x-2\inƯ\left(2\right)\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}0< =x< 2\\x-2\in\left\{1;-1;2;-2\right\}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}0< =x< 2\\x\in\left\{3;1;4;0\right\}\end{matrix}\right.\)
=>\(x\in\left\{0;1\right\}\)
Khi x=1 thì \(\left|y\right|=\dfrac{1}{2-1}=1\)
=>y=1 hoặc y=-1
Khi x=0 thì \(\left|y\right|=\dfrac{0}{2-0}=0\)
=>y=0
