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Question Number 100330 by bemath last updated on 26/Jun/20

Commented by bobhans last updated on 26/Jun/20

$$\left(1\right)+\left(2\right)+\left(3\right)\Rightarrow {\mathrm{x}}^2+{\mathrm{y}}^2+{\mathrm{z}}^2+2(\mathrm{xy}+\mathrm{xz}+\mathrm{yz})=16$$$${(\mathrm{x}+\mathrm{y}+\mathrm{z})}^2=16\Rightarrow \{\begin{array}{l}\mathrm{x}+\mathrm{y}+\mathrm{z}=4\\ \mathrm{x}+\mathrm{y}+\mathrm{z}=-4\end{array}$$$$\mathrm{case}:1$$$$\bullet \mathrm{x}(\mathrm{x}+\mathrm{y}+\mathrm{z})=1\Rightarrow \mathrm{x}=\frac{1}{4}$$$$\bullet \mathrm{y}(\mathrm{x}+\mathrm{y}+\mathrm{z})=6\Rightarrow \mathrm{y}=\frac{3}{2}$$$$\bullet \mathrm{z}(\mathrm{x}+\mathrm{y}+\mathrm{z})=9\Rightarrow \mathrm{z}=\frac{9}{4}$$

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