\(\left\{{}\begin{matrix}x^3+1=2x\left(1\right)\\y^3+1=2y\left(2\right)\end{matrix}\right.\)
Lấy (1)-(2) có:
(1)-(2) có: \(x^3-y^3=2\left(y-x\right)\)
\(\Leftrightarrow\left(x-y\right)\left(x^2+xy+y^2\right)-2\left(y-x\right)=0\Leftrightarrow\left(x-y\right)\left(x^2+xy+y^2+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x-y=0\\x^2+xy+y^2+2=0\end{matrix}\right.\)
1, x - y =0 => x = y
Thay vào (1) có: \(y^3+1=2y\Leftrightarrow y^3+1-2y=0\Leftrightarrow y^3-y-y+1=0\Leftrightarrow y\left(y^2-1\right)-\left(y-1\right)=0\Leftrightarrow\left(y-1\right)\left(y^2+y-1\right)=0\Leftrightarrow\left[{}\begin{matrix}y=1\\y=\dfrac{-1-\sqrt{5}}{2}\\y=\dfrac{-1+\sqrt{5}}{2}\end{matrix}\right.\)2, \(x^2+xy+y^2+2=0\Leftrightarrow\left(x+\dfrac{1}{2}y\right)^2+\dfrac{3}{4}y^2+2=0\)(vô nghiêm)
Vậy hệ PT có cặp nghiệm (x, y)=....
