Giải thích các bước giải:
Ta có :
$x^2-2xy+4x-3y+1=0$
$\to x^2+4x+1=2xy+3y$
$\to x^2+4x+1=y(2x+3)$
$\to x^2+4x+1\quad\vdots\quad 2x+3$
$\to 4x^2+16x+4\quad\vdots\quad 2x+3$
$\to (4x^2+6x)+(10x+15)-11\quad\vdots\quad 2x+3$
$\to 2x(2x+3)+5(2x+3)-11\quad\vdots\quad 2x+3$
$\to 11\quad\vdots\quad 2x+3$
$\to 2x+3\in\{1,11,-1,-11\}$
$\to x\in\{-1,4, -2, -7\}$
$\to y\in\{-2,3,3,-2\}$
