\(\left\{\begin{matrix}x^3-3x-2=2-y\\y^3-3y-2=2-z\\z^3-3z-2=2-x\end{matrix}\right.\)
ta có: x3-3x-2=x3-2x2+2x2-4x+x-2
=x2(x-2)+2x(x-2)+(x-2)=(x-2)(x+1)2
tương tự : y3-3y-2=(y-2)(y+1)2; z3-3z-2=(z-2)(z+1)2
ta có hệ pt:\(\left\{\begin{matrix}\left(x-2\right)\left(x+1\right)^2=2-y\\\left(y-2\right)\left(y+1\right)^2=2-z\\\left(z-2\right)\left(z+1\right)^2=2-x\end{matrix}\right.\)
nhân vế vs vế 3 pt trên :\(\left(x-2\right)\left(y-2\right)\left(z-2\right)\left(x+1\right)^2\left(y+1\right)^2\left(z+1\right)^2=\left(2-x\right)\left(2-y\right)\left(2-z\right)\)
\(\Leftrightarrow\left(x+1\right)^2\left(y+1\right)^2\left(z+1\right)^2=-1\)
vt >= 0, vF <0 vậy hệ pt vô nghiệm
