ĐK: y\(\ne0,y\ne\pm12\)
\(\left\{{}\begin{matrix}\dfrac{x}{y}-\dfrac{x}{y+12}=1\\\dfrac{x}{y-12}-\dfrac{x}{y}=2\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}\dfrac{xy+12x-xy}{y\left(y+12\right)}=1\\\dfrac{xy-xy+12x}{y\left(y-12\right)}=2\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}12x=y^2+12y\\12x=2y^2-24y\end{matrix}\right.\)\(\Leftrightarrow\)\(y^2+12y=2y^2-24y\Leftrightarrow y^2-36y=0\Leftrightarrow y\left(y-36\right)=0\Leftrightarrow\)\(\left[{}\begin{matrix}y=0\left(ktm\right)\\y=36\left(tm\right)\end{matrix}\right.\)\(\Leftrightarrow y=36\Leftrightarrow12x=36^2+12.36\Leftrightarrow x=144\)
Vậy (x;y)=(144;36)
