\(\left\{{}\begin{matrix}x.\left(x+y+z\right)=-5\left(1\right)\\y.\left(x+y+z\right)=9\left(2\right)\\z.\left(x+y+z\right)=5\left(3\right)\end{matrix}\right.\)
Cộng theo vế của \(\left(1\right);\left(2\right)và\left(3\right)\) ta được:
\(\left(x+y+z\right)^2=9.\)
\(\Rightarrow x+y+z=\pm3\)
Xét \(x+y+z=3\)
\(\Rightarrow\left\{{}\begin{matrix}x.3=-5\\y.3=9\\z.3=5\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-\frac{5}{3}\\y=3\\z=\frac{5}{3}\end{matrix}\right.\)
Xét \(x+y+z=-3\)
\(\Rightarrow\left\{{}\begin{matrix}x.\left(-3\right)=-5\\y.\left(-3\right)=9\\z.\left(-3\right)=5\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\frac{5}{3}\\y=-3\\z=-\frac{5}{3}\end{matrix}\right.\)
Vậy........
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