Ta có: \(\frac{x}{6}-\frac{1}{y}=\frac{1}{2}\)
\(\Rightarrow\frac{1}{y}=\frac{1}{2}+\frac{x}{6}\)
\(\Rightarrow\frac{1}{y}=\frac{3+x}{6}\)
\(\Rightarrow\left(3+x\right)y=6\)
\(\Rightarrow3+x\inƯ\left(6\right);y\inƯ\left(6\right)\)
mà \(Ư\left(6\right)=\left\{\pm1;\pm2;\pm3;\pm6\right\}\)
\(\Rightarrow3+x,y\in\left\{\pm1;\pm2;\pm3;\pm6\right\}\)
| \(3+x\) | \(1\) | \(-1\) | 2 | \(-2\) | 3 | \(-3\) | \(6\) | \(-6\) |
| \(y\) | 6 | \(-6\) | 3 | \(-3\) | 2 | \(-2\) | 1 | \(-1\) |
| \(x\) | \(-2\) | \(-4\) | \(-1\) | \(-5\) | 0 | \(-6\) | 3 | \(-9\) |
Vậy \(\left(x,y\right)\in\left\{\left(-4,-6\right);\left(-2,6\right);\left(-1,3\right);\left(-5,-3\right);\left(0,2\right);\left(-6,-2\right);\left(3,1\right);\left(-9,-1\right)\right\}\)
