`2xy - 10x + y = 8`

`y(2x + 1) - 10x = 8`

`y(2x + 1) - (10x + 5) = 8 - 5 = 3`

`y(2x + 1) - 5(2x + 1) = 3`

`(y - 5)(2x + 1) = 3`

Vì `x, y ∈ ZZ ⇒ (y - 5), (2x + 1) ∈ ZZ`

`⇒ (y - 5), (2x + 1) ∈ Ư(3) = {± 1, ± 3}`

Ta có bảng:

$\begin{array}{|c|c|c|}\hline\text{2x + 1}&\text{-1}&\text{1}&\text{-3}&\text{3}\\\hline\text{y - 5}&\text{-3}&\text{3}&\text{-1}&\text{1}\\\hline \text{x}&\text{-1}&\text{0}&\text{-2}&\text{1}\\\hline \text{y}&\text{2}&\text{8}&\text{4}&\text{6}\\\hline\end{array}$

Vậy `(x, y) ∈ {(-1, 2); (0, 8); (-2, 4); (1, 6)}`