Ta có \(y=1-2z^2;x=3-y-z=2z^2-z+2\)
\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{3}\Rightarrow\frac{3\left(yz+xz+xy\right)}{3xyz}=\frac{xyz}{3xyz}\)
\(\Rightarrow3z\left(1-2z^2\right)+3z\left(2z^2-z+2\right)+3\left(1-2z^2\right)\left(2z^2-z+2\right)\)
\(=z\left(1-2z^2\right)\left(2z^2-z+2\right)\)
\(\Leftrightarrow4z^5-14z^4+8z^3-8z^2+4z+6=0\)
\(\Leftrightarrow z=1\vee z=3\vee z=-\frac{1}{2}\)
Với z = 1, ta có y = -1, x = 3
Với z = 3, x = 17, y = -17
Với \(z=-\frac{1}{2},x=3,y=\frac{1}{2}\)
Tóm lại hệ có 3 nghiệm \(\left(3;-1;1\right),\left(17;-17;3\right),\left(3;\frac{1}{2};-\frac{1}{2}\right)\)