giải hệ x^2+y^2=1+xy (x/y+1)^2 + (y/x+1)^2=1

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1. 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 $|\dfrac{x}y+\dfrac{y}x|\ge 2$

   $\to x^2+y^2+xy=-\sqrt{2}xy$

   $\to 1+xy+xy=-\sqrt{2}xy$

   $\to xy=-\dfrac{2-\sqrt{2}}{2}$

   $\to (x+y)^2=\dfrac{-4+3\sqrt{2}}{2}\to x+y=..\to x,y$

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