`x(x+y+z)+y(x+y+z)+z(x+y+z)=-5+9+5`
`to (x+y+z)(x+y+z)=9`
`to (x+y+z)^2=9`
`to` \(\left[ \begin{array}{l}x+y+z=3\\x+y+z=-3\end{array} \right.\)
`+)` Với `x+y+z=3`
`x.3=-5 to x=-5/3`
`y.3=9 to y=3`
`z.3=5 to z=5/3`
`+)` Với `x+y+z=-3`
`x.(-3)=-5 to x=5/3`
`y.(-3)=9 to y=-3`
`z.(-3)=5 to z=-5/3`
Vậy `(x;y;z)=(-5/3;3;5/3);(5/3;-3;-5/3)`