\(\left\{{}\begin{matrix}3xy^2=x^2+2\left(1\right)\\3x^2y=y^2+2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3xy^2\left(y^2+2\right)=\left(x^2+2\right)\left(y^2+2\right)\left(2\right)\\3x^2y\left(x^2+2\right)=\left(y^2+2\right)\left(x^2+2\right)\left(3\right)\end{matrix}\right.\)
Trừ vế theo vế \(\left(2\right)\) cho \(\left(3\right)\) ta được
\(3xy^2\left(y^2+2\right)-3x^2y\left(x^2+2\right)=0\)
\(\Leftrightarrow3xy\left(y-x\right)\left(x^2+y^2+xy+2\right)=0\)
Do \(x^2+xy+y^2+2>0\forall x,y\) nên\(\left[{}\begin{matrix}x=0\\y=0\\x=y\end{matrix}\right.\)
Nếu \(x=0\Rightarrow\left\{{}\begin{matrix}0=0+2\\0=y^2+2\end{matrix}\right.\left(VN\right)\)
Nếu \(y=0\Rightarrow\left\{{}\begin{matrix}0=x^2+2\\0=0+2\end{matrix}\right.\left(VN\right)\)
Nếu \(x=y\), \(\left(1\right)\Leftrightarrow3x^3-x^2-2=0\)
\(\Leftrightarrow\left(x-1\right)\left(3x^2+2x+2\right)=0\)
\(\Leftrightarrow x=y=1\)
Vậy hệ phương trình có nghiệm \(x=y=1\)
