Đáp án: $\begin{pmatrix}x=1,\:&y=1\\ x=3,\:&y=\dfrac{1}{3}\\ x=\dfrac{3+\sqrt{5}}{2},\:&y=\dfrac{\sqrt{5}-1}{2}\\ x=\dfrac{3-\sqrt{5}}{2},\:&y=-\dfrac{\sqrt{5}+1}{2}\end{pmatrix}$

Giải thích các bước giải:

Ta có:

$x^2y-2xy-xy^2-x+y+2=0$

$\to xy(x-2-y)-(x-y-2)=0$

$\to (xy-1)(x-y-2)=0$

$\to xy-1=0$ hoặc $x-y-2=0$

Giải $xy-1=0$

$\to xy=1$

Ta có:

$x(4-x-y)=2$

$\to 4x-x^2-xy=2$

$\to 4x-x^2-1=2$

$\to x^2-4x+3=0$

$\to (x-1)(x-3)=0$
$\to x\in\{1, 3\}$

$\to y\in\{1, \dfrac13\}$

Giải $x-y-2=0$

$\to y=x-2$

$\to x(4-x-(x-2))=2$

$\to -2x^2+6x=2$

$\to 2x^2-6x+2=0$

$\to x^2-3x+1=0$

$\to x=\dfrac{3\pm\sqrt5}2$

$\to y=\dfrac{-1\pm\sqrt5}2$