Đáp án:
$(x;y)=\left\{(-8;6),(-2;12),(0;-2),(6;4)\right\}$
Giải thích các bước giải:
$y = \dfrac{5x - 2}{x + 1}=\dfrac{5(x +1) - 7}{x +1}=5 - \dfrac{7}{x +1}$
$y \in \Bbb Z \Leftrightarrow \dfrac{7}{x + 1}\in\Bbb Z$
$\Leftrightarrow x + 1 \in Ư(7)=\left\{-7;-1;1;7\right\}$
$\Leftrightarrow x = \left\{-8;-2;0;6\right\}$
$\Rightarrow y = \left\{6;12;-2;4\right\}$
Vậy $(x;y)=\left\{(-8;6),(-2;12),(0;-2),(6;4)\right\}$
