Ta có: \(\left(x+y+z\right)^2=x^2+y^2+z^2+2\left(xy+yz+zx\right)\)
\(=x^2+y^2+z^2+2.1=x^2+y^2+z^2+2\left(2y^2-3z^2\right)\)\(=x^2+5y^2-5z^2\)
\(\Leftrightarrow\left(x+y+z\right)^2-x^2+5\left(z-y\right)\left(z+y\right)=0\)
\(\Leftrightarrow\left(2x+y+z\right)\left(y+z\right)+5\left(z-y\right)\left(z+y\right)=0\)
\(\Leftrightarrow\left(y+z\right)\left(2x+y+z+5z-5y\right)=0\)
\(\Leftrightarrow\left(y+z\right)\left(2x-4y+6z\right)=0\)
\(\Leftrightarrow\left(y+z\right)\left(x-2y+3z\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}y=-z\\x-2y+3z=0\end{cases}}\)
Với y=-z ta có: \(2y^2-3z^2=1\Rightarrow2y^2-3y^2=1\Leftrightarrow-y^2=1\)( do \(y^2\ge0\)) => pt vô nghiệm