$\begin{array}{l}\dfrac x9+\dfrac1y=\dfrac16\\\Leftrightarrow \dfrac{xy}{9y}+\dfrac{9}{9y}=\dfrac16\\\Leftrightarrow \dfrac{xy+9}{9y}=\dfrac16\\\Leftrightarrow 6(xy+9)=9y\\\Leftrightarrow 2(xy+9)=3y\\\Leftrightarrow 2xy+18=3y\\\Leftrightarrow 2xy-3y=-18\\\Leftrightarrow (2x-3).y=-18\\\text{mà $x,y\in\mathbb{Z}$}\\\to 2x-3,y\in Ư(-18)=\{\pm1;\pm2;\pm3;\pm6;\pm9;\pm18\}\\\text{- Do $2x-3\ \not\vdots2\to 2x-3\in\{\pm1;\pm3;\pm9\}$}\\\text{- Ta có bảng sau :}\\\begin{array}{|c|c|}\hline2x-3&-9&-3&-1&1&3&9\\\hline y&2&6&18&-18&-6&-2\\\hline x&-3&0&1&2&3&6 \\\hline\end{array}\\\text{- Vậy các cặp số $(x,y)$ thỏa mãn là :}\\ (-3,2);(0,6);(1,18);(2,-18);(3,-6);(6,-2) \end{array}$
