Question

tim x y z

\(x\left(x+y+z\right)=13;y\left(x+y+z\right)=7;z\left(x+y+z\right)=-4\)

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Answer 1

\(\left\{{}\begin{matrix}x\left(x+y+z\right)=13\\y\left(x+y+z\right)=7\\z\left(x+y+z\right)=-4\end{matrix}\right.\) \(\Leftrightarrow x\left(x+y+z\right)+y\left(x+y+z\right)+z\left(x+y+z\right)=13+7-4\)

\(\Rightarrow\left(x+y+z\right)\left(x+y+z\right)=16\)

\(\Rightarrow\left(x+y+z\right)^2=16\)

\(\Rightarrow\left[{}\begin{matrix}x+y+z=4\\x+y+z=-4\end{matrix}\right.\)

Với \(x+y+z=4\)

\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{13}{4}\\y=\dfrac{7}{4}\\z=-1\end{matrix}\right.\)

Với \(x+y+z=-4\)

\(\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{13}{4}\\y=-\dfrac{7}{4}\\z=1\end{matrix}\right.\)