Đáp án: $(x,y)\in\{(0,2), (2,2), (-2,-6), (-6,-4)\}$
Giải thích các bước giải:
Ta có:
$x^2-xy=y-2$
$\to x^2+2=xy+y$
$\to x^2+2=y(x+1)$
$\to x^2+2\quad\vdots\quad x+1$
$\to x^2-1+3\quad\vdots\quad x+1$
$\to (x-1)(x+1)+3\quad\vdots\quad x+1$
$\to 3\quad\vdots\quad x+1$
$\to x+1\in\{1,3,-1,-3\}$
$\to x\in\{0, 2, -2, -4\}$
Lại có $x^2+2=y(x+1)\to y=\dfrac{x^2+2}{x+1}$
$\to y\in\{2, 2, -6, -6\}$
$\to (x,y)\in\{(0,2), (2,2), (-2,-6), (-6,-4)\}$
