Ta có: \(\dfrac{x}{4}-\dfrac{2}{y}\: =\dfrac{3}{2}\: \: \: \Leftrightarrow\dfrac{xy}{4y}\: -\dfrac{8}{4y}=\dfrac{6y}{4y}\Rightarrow xy-8=6y\)\(xy-6y=8\Rightarrow y.\left(x-6\right)=8\)
Vì x,y \(\in Z\Rightarrow x-6\in Z\)
\(\Rightarrow y.\left(x-6\right)=1.8=-1.\left(-8\right)=2.4=-2.\left(-4\right)\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-6=1\\y=8\end{matrix}\right.\\\left\{{}\begin{matrix}x-6=8\\y=1\end{matrix}\right.\end{matrix}\right.hoặc\left[{}\begin{matrix}\left\{{}\begin{matrix}x-6=-1\\y=-8\end{matrix}\right.\\\left\{{}\begin{matrix}x-6=-8\\y=-1\end{matrix}\right.\end{matrix}\right.hoặc\left[{}\begin{matrix}\left\{{}\begin{matrix}x-6=2\\y=4\end{matrix}\right.\\\left\{{}\begin{matrix}x-6=4\\y=2\end{matrix}\right.\end{matrix}\right.hoặc\left[{}\begin{matrix}\left\{{}\begin{matrix}x-6=-2\\y=-4\end{matrix}\right.\\\left\{{}\begin{matrix}x-6=-4\\y=-2\end{matrix}\right.\end{matrix}\right.\)
Rồi làm như tìm x,y bình thường nha bạn......chúc bạn học tốt 
