\(\left\{{}\begin{matrix}x^3+1=2y\\y^3+1=2x\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^3-y^3=2y-2x\\x^3+1=2y\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-y\right)\left(x^2+xy+y^2+2\right)=0\\x^3+1=2y\end{matrix}\right.\)
Do x2+xy+y2+2=(x+\(\dfrac{y}{2}\))2+\(\dfrac{3y^2}{4}+2>0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=y\\y^3+1=2y\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=y\\\left(y-1\right)\left(y^2+y-1\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=1\end{matrix}\right.hoặc\left\{{}\begin{matrix}y^2+y-1=0\\x=y\end{matrix}\right.\)
<=>\(\left\{{}\begin{matrix}x=1\\y=1\end{matrix}\right.\)hoặc\(\left\{{}\begin{matrix}x=\dfrac{-1+\sqrt{5}}{2}\\y=\dfrac{-1+\sqrt{5}}{2}\end{matrix}\right.\)hoặc\(\left\{{}\begin{matrix}x=\dfrac{-1-\sqrt{5}}{2}\\y=\dfrac{-1-\sqrt{5}}{2}\end{matrix}\right.\)
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