$2x-5y+5xy = 14$
$⇔ 2x+5xy - 5y = 14$
$⇔ x(2 + 5y) - 5y = 14$
$⇔ x(2+5y) - (2 + 5y) = 12$
$⇔ (x-1)(2+5y) = 12$
$⇒$ $x-1;2+5y$ $∈$ `Ư(12)={±1;±2;±3;±4;±6;±12}`
Mà $2+5y$ chia $5$ dư $2$, dư $-3$
$⇒$ $2+5y$ $∈$ `{-3;2;12}`
Ta có bảng:
$\left[\begin{array}{ccc}2+5y&-3&2&12\\x-1&-4&6&1\\y&-1&0&2\\x&-3&7&2\end{array}\right]$
Vậy `(x;y)=(-3;-1);(7;0);(2;2)`