Đáp án:
`x/3 - 1/y = 1/3`
`-> 1/y = x/3 - 1/3`
`-> 1/y = (x - 1)/3`
`-> 3 = y (x - 1)`
`-> (x - 1)y = ±1 ×±3 = ± 3 × ±1`
Ta có bảng :
$\begin{array}{|c|c|c|c|c|c|c|}\hline x-1& 1 & 3 & -1 & -3 \\\hline y& 3 & 1 & -3 & -1 \\\hline x & 2&4 &0 &-2 \\\hline y & 3& 1&-3 &-1 \\\hline\end{array}$
Vậy `(x;y) = (2;3), (4;1), (0;-3), (-2;-1)`