\(\left\{{}\begin{matrix}x^3+1=2y\left(1\right)\\y^3+1=2x\left(2\right)\end{matrix}\right.\)
Lấy \(\left(1\right)-\left(2\right)\Leftrightarrow x^3-y^2=2\left(y-x\right)\)
\(\Rightarrow\left(x-y\right)\left(x^2+xy+y^2\right)+2\left(x-y\right)=0\Leftrightarrow\left(x-y\right)\left(x^2+xy+y^2+2\right)=0\)
mà \(x^2+xy+y^2+2=x^2+2.x.\dfrac{1}{2}y+\dfrac{1}{4}y^2+\dfrac{3}{4}y^2+2\)
\(=\left(x+\dfrac{1}{2}y\right)^2+\dfrac{3}{4}y^2+2>0\Rightarrow x=y\)
Thế vào (1),ta được: \(x^3+1=2x\Rightarrow x^3-2x+1=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2+x-1\right)=0\Rightarrow\left[{}\begin{matrix}x=1\\x^2+x-1=0\end{matrix}\right.\)
\(\Delta=1^2-4.\left(-1\right)=5\Rightarrow\left[{}\begin{matrix}x=\dfrac{-b-\sqrt{\Delta}}{2a}=\dfrac{-1-\sqrt{5}}{2}\\x=\dfrac{-b+\sqrt{\Delta}}{2}=\dfrac{-1+\sqrt{5}}{2}\end{matrix}\right.\)
Vậy hệ có bộ nghiệm (x,y) là:\(\left(1,1\right);\left(\dfrac{-1-\sqrt{5}}{2};\dfrac{-1-\sqrt{5}}{2}\right);\left(\dfrac{-1+\sqrt{5}}{2};\dfrac{-1+\sqrt{5}}{2}\right)\)
