Đáp án + Giải thích các bước giải:

`xy^2 + 2xy + x - 3y = 0`

`=> x(y^2 + 2y + 1) - 3y = 0`

`=> x (y+1)^2 - 3y = 0`

`=> x (y+1)^2 -3y - 3 = -3`

`=> x(y+1)^2 - 3(y+1) = -3`

`=> (y+1)(xy + x - 3) = -3`

Vì `x,y in ZZ` nên ta có bảng:

\begin{array}{|c|c|c|}\hline \text{y+1}&\text{-1}&\text{3}&\text{1}&\text{-3}\\\hline \text{xy+x-3}&\text{3}&\text{-1}&\text{-3}&\text{1}\\\hline \text{y}&\text{-2}&\text{2}&\text{0}&\text{-4}\\\hline \text{xy+x}&\text{6}&\text{2}&\text{0}&\text{4}\\\hline \text{x}&\text{-6}&\text{$\frac{2}{3}$ (Loại)}&\text{0}&\text{$\frac{-4}{3}$ (Loại)}\\\hline\end{array}

`=> (x;y) = {(-6;-2);(0;0)}`

Vậy `(x;y) = {(-6;-2);(0;0)}`