a) Ta có: \(3xy+x-3y=6\)
\(\Rightarrow3xy-3y+x-1+1=6\)
\(\Rightarrow3y\left(x-1\right)+\left(x-1\right)=5\)
\(\Rightarrow\left(x-1\right)\left(3y+1\right)=5\)
\(\Rightarrow x-1\inƯ\left(5\right);3y+1\inƯ\left(5\right)\)
mà \(Ư\left(5\right)=\left\{\pm1;\pm5\right\}\)
\(\Rightarrow x-1\) và \(3y+1\in\left\{\pm1;\pm5\right\}\)
Ta có bảng sau:
| \(x-1\) | \(1\) | \(-1\) | 5 | \(-5\) |
| \(3y+1\) | \(5\) | \(-5\) | 1 | \(-1\) |
| \(x\) | 0 | 0 | 6 | \(-4\) |
| \(y\) | \(\dfrac{4}{3}\) | \(-2\) | 0 | \(\dfrac{-2}{3}\) |
Vậy \(\left(x,y\right)\in\left\{\left(0,\dfrac{4}{3}\right);\left(0,-2\right);\left(6,0\right);\left(-4,\dfrac{-2}{3}\right)\right\}\).
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