Ta có:

\(\dfrac{x}{8}-\dfrac{1}{y}=\dfrac{1}{4}\Leftrightarrow\dfrac{xy-8}{8y}=\dfrac{1}{4}\)

\(\Leftrightarrow4\left(xy-8\right)=8y\)

\(\Leftrightarrow4\left(xy-8\right)=4.2y\)

\(\Leftrightarrow xy-8=2y\)

\(\Leftrightarrow xy-2y=8\)

\(\Leftrightarrow y\left(x-2\right)8\)

(*)\(\left[{}\begin{matrix}\left\{{}\begin{matrix}y=1\\x-2=8\end{matrix}\right.\\\left\{{}\begin{matrix}y=-1\\x-2=-8\end{matrix}\right.\\\left\{{}\begin{matrix}y=2\\x-2=4\end{matrix}\right.\\\left\{{}\begin{matrix}y=-2\\x-2=-4\end{matrix}\right.\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}y=1\\x=10\end{matrix}\right.\\\left\{{}\begin{matrix}y=-1\\x=-6\end{matrix}\right.\\\left\{{}\begin{matrix}y=2\\x=6\end{matrix}\right.\\\left\{{}\begin{matrix}y=-2\\x=-2\end{matrix}\right.\end{matrix}\right.\)

(**)\(\left[{}\begin{matrix}\left\{{}\begin{matrix}y=8\\x-2=1\end{matrix}\right.\\\left\{{}\begin{matrix}y=-8\\x-2=-1\end{matrix}\right.\\\left\{{}\begin{matrix}y=4\\x-2=2\end{matrix}\right.\\\left\{{}\begin{matrix}y=-4\\x-2=-2\end{matrix}\right.\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}y=8\\x=3\end{matrix}\right.\\\left\{{}\begin{matrix}y=-8\\x=1\end{matrix}\right.\\\left\{{}\begin{matrix}y=4\\x=4\end{matrix}\right.\\\left\{{}\begin{matrix}y=-4\\x=0\end{matrix}\right.\end{matrix}\right.\)

Vậy ta tìm được các cặp x,y thỏa mãn từ (*), (**)