Giải thích các bước giải:
Ta có :
$x^2+y^2=1+xy\to (x+y)^2=1+3xy$
$(\dfrac{x}{y}+1)^2+(\dfrac{y}{x}+1)^2=1$
$\to \dfrac{x^2}{y^2}+2\dfrac{x}{y}+1+\dfrac{y^2}{x^2}+2\dfrac{y}{x}+1=1$
$\to \dfrac{x^2}{y^2}+\dfrac{y^2}{x^2}+2(\dfrac{x}{y}+\dfrac{y}{x})+1=0$
$\to \dfrac{x^2}{y^2}+2\dfrac{x}y.\dfrac{y}x+\dfrac{y^2}{x^2}+2(\dfrac{x}{y}+\dfrac{y}{x})+1=2$
$\to (\dfrac{x}y+\dfrac{y}x)^2+2(\dfrac{x}{y}+\dfrac{y}{x})+1=2$
$\to (\dfrac{x}y+\dfrac{y}x+1)^2=2$
$\to \dfrac{x}y+\dfrac{y}x+1=-\sqrt{2}$
Vì $|\dfrac{x}y+\dfrac{y}x|\ge 2$
$\to x^2+y^2+xy=-\sqrt{2}xy$
$\to 1+xy+xy=-\sqrt{2}xy$
$\to xy=-\dfrac{2-\sqrt{2}}{2}$
$\to (x+y)^2=\dfrac{-4+3\sqrt{2}}{2}\to x+y=..\to x,y$
