Ta có HPT : \(\hept{\begin{cases}2x+y=x^2\\2y+x=y^2\end{cases}}\)
\(\Leftrightarrow x^2-y^2=2x+y-2y-x\)
\(\Leftrightarrow\left(x-y\right)\left(x+y\right)=x-y\)
\(\Leftrightarrow\left(x-y\right)\left(x+y-1\right)=0\)
TH1 : \(x-y=0\)
\(\Leftrightarrow x=y\)
\(\Leftrightarrow2x+x=x^2\)
\(\Leftrightarrow\orbr{\begin{cases}x=y=0\\x=y=3\end{cases}}\)
TH2 : \(x+y-1=0\)
\(\Leftrightarrow2\left(1-y\right)+y=\left(1-y\right)^2\)
\(\Leftrightarrow2-2y+y=1-2y+y^2\)
\(\Leftrightarrow y^2-y-1=0\)
\(\Leftrightarrow\orbr{\begin{cases}y=\frac{1+\sqrt{5}}{2}\Leftrightarrow x=\frac{1-\sqrt{5}}{2}\\y=\frac{1-\sqrt{5}}{2}\Leftrightarrow x=\frac{1+\sqrt{5}}{2}\end{cases}}\)
Vậy \(\left(x;y\right)\in\left\{\left(0;0\right);\left(3;3\right);\left(\frac{1-\sqrt{5}}{2};\frac{1+\sqrt{5}}{2}\right);\left(\frac{1+\sqrt{5}}{2};\frac{1-\sqrt{5}}{2}\right)\right\}\)
