Đặt z = 1 - x.
Ta có: \(\left(1-z\right)^3-\left(1-z\right)^2+\left(1-z\right)-5=0\)
\(\Leftrightarrow-z^3+2z^2-2z-4=0\)
Mà \(y^3-2y^2+2y+4=0\)
Nên \(\left(y^3-z^3\right)-2\left(y^2-z^2\right)+2\left(y-z\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}y-z=0\\y^2+yz+z^2-2y-2z+2=0\end{matrix}\right.\).
Mặt khác ta có: \(y^2+yz+z^2-2y-2z+2=0\Leftrightarrow4y^2+4yz+4z^2-8y-8z+8=0\Leftrightarrow\left(2y+z-2\right)^2+z^2+\left(z-2\right)^2=0\) (vô lí).
Do đó y = z \(\Leftrightarrow y=1-x\Leftrightarrow x+y=1\).
