Giải thích các bước giải:
$\left \{ {{x²-2y²=2x+y} \atop {y²-2x²=2y+x}} \right.$
<=>$\left \{ {{3x²-3y²=x-y} \atop {x²-2y²=2x+y}} \right.$
<=>$\left \{ {{3(x-y)(x+y)-(x-y)=0} \atop {x²-2y²=2x+y}} \right.$
<=>$\left \{ {{(x-y)(3x+3y-1)=0 (1)} \atop {x²-2y²=2x+y (2)}} \right.$
(1)<=>\(\left[ \begin{array}{l}x-y=0\\3x+3y-1=0\end{array} \right.\)
+) x-y=0<=>x=y
=>(2)<=>x²-2x²=2x+x
<=>x²+3x=0
<=>\(\left[ \begin{array}{l}x=0\\x=-3\end{array} \right.\)
=>\(\left[ \begin{array}{l}x=y=0\\x=y=-3\end{array} \right.\)
+) 3x+3y-1=0<=> x=$\frac{1}{3}$-y
=>(2)<=> ($\frac{1}{3}$-y)²-2y²=2y+($\frac{1}{3}$-y)
<=>\(\left[ \begin{array}{l}y=\frac{-5+\sqrt[]{17}}{6}\\y=\frac{-5-\sqrt[]{17}}{6}\end{array} \right.\)
=>$\left \{ {{y=\frac{-5+\sqrt[]{17}}{6}} \atop {x=\frac{-5-\sqrt[]{17}}{6}}} \right.$ hoặc $\left \{ {{y=\frac{-5-\sqrt[]{17}}{6}} \atop {x=\frac{-5+\sqrt[]{17}}{6}}} \right.$
