\(\left\{{}\begin{matrix}x^3=2x+y\left(1\right)\\y^3=2y+x\left(2\right)\end{matrix}\right.\)
\(\Leftrightarrow x^3-y^3=2x+y-2y-x\)
\(\Leftrightarrow\left(x-y\right)\left(x^2+xy+y^2\right)=x-y\)
\(\Leftrightarrow\left(x-y\right)\left(x^2+xy+y^2\right)-\left(x-y\right)=0\)
\(\Leftrightarrow\left(x-y\right)\left(x^2+xy+y^2-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=y\left(3\right)\\x^2+xy+y^2=1\left(4\right)\end{matrix}\right.\)
từ (1)(3) ta có hệ \(\Leftrightarrow\left\{{}\begin{matrix}x=y\\x^3=2x+y\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=y\\x^3=2x+x\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=y\\x^3-3x=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=y\\\left[{}\begin{matrix}x=0\\x=\sqrt{3}\\x=-\sqrt{3}\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=0\\y=0\end{matrix}\right.\\\left\{{}\begin{matrix}x=\sqrt{3}\\y=\sqrt{3}\end{matrix}\right.\\\left\{{}\begin{matrix}x=-\sqrt{3}\\y=-\sqrt{3}\end{matrix}\right.\end{matrix}\right.\)
từ (1)(4) ta có hệ \(\left\{{}\begin{matrix}x^3=2x+y\\x^2+xy+y^2=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=x^3-2x\\x^2+xy+y^2=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=x^3-2x\\x^2+x\left(x^3-2x\right)+\left(x^3-2x\right)^2=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=x^3-2x\\x^2+x^4-2x^2+x^6-4x^4+4x^2=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=x^3-2x\\x^6-3x^4+3x^2-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=x^3-2x\\\left(x^2-1\right)^3=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=x^3-2x\\\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=1\\y=x^3-2x\end{matrix}\right.\\\left\{{}\begin{matrix}x=-1\\y=x^3-2x\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\\\left\{{}\begin{matrix}x=-1\\y=1\end{matrix}\right.\end{matrix}\right.\)
vậy \(S=\left\{\left(0;0\right),\left(\sqrt{3};\sqrt{3}\right),\left(-\sqrt{3};-\sqrt{3}\right),\left(1;-1\right),\left(-1;1\right)\right\}\)
