Ta có: \(\frac{1}{x}+\frac{1}{y}=\frac{1}{3}\)
\(\Leftrightarrow\frac{y}{xy}+\frac{x}{xy}=\frac{1}{3}\)
\(\Leftrightarrow\frac{y+x}{xy}=\frac{1}{3}\)
\(\Leftrightarrow3\left(y+x\right)=xy\)
\(\Leftrightarrow3y+3x=xy\)
\(\Leftrightarrow3y+3x-xy=0\)
\(\Leftrightarrow3y+x\left(3-y\right)=0\)
\(\Leftrightarrow3y-9-x\left(y-3\right)=0-9\)
\(\Leftrightarrow3\left(y-3\right)-x\left(y-3\right)=-9\)
\(\Leftrightarrow\left(y-3\right)\left(3-x\right)=-9\)
\(\Leftrightarrow y-3;3-x\inƯ\left(-9\right)\)
\(\Leftrightarrow y-3;3-x\in\left\{1;-1;3;-3;9;-9\right\}\)
| y-3 | 1 | -1 | 3 | -3 | 9 | -9 |
| 3-x | -9 | 9 | -3 | 3 | -1 | 1 |
| x | 12 | -6 | 6 | 0 | 4 | 2 |
| y | 4 | 2 | 6 | 0 | 12 | -6 |
Vậy: \(\left(x,y\right)\in\left\{\left(12;4\right);\left(-6;2\right)\left(6;6\right);\left(0;0\right);\left(4;12\right);\left(2;-6\right)\right\}\)