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Uchiha Sasuke

26/12/2017 at 09:59
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What is the value of xyz where x,y and z are positive real numbers such that \(x\left(y+z\right)=32\) ; \(y\left(x+z\right)=27\) ; \(z\left(x+y\right)=35\)




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    Dao Trong Luan Coordinator 26/12/2017 at 10:19

    \(\left\{{}\begin{matrix}x\left(y+z\right)=32\\y\left(x+z\right)=27\\z\left(x+y\right)=35\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}xy+xz=32\\xy+yz=27\\xz+yz=35\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}xz-yz=5\\xz-xy=8\\yz-xy=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}z\left(x-y\right)=5\\x\left(z-y\right)=8\\y\left(z-x\right)=3\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}z\left(x+y-\left(x-y\right)\right)=30\\x\left(y+z-\left(z-y\right)\right)=24\\y\left(x+z-\left(z-x\right)\right)=24\end{matrix}\right.\)

    \(\Rightarrow\left\{{}\begin{matrix}2yz=30\\2zy=24\\2xz=24\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}yz=15\\zy=12\\xz=12\end{matrix}\right.\Rightarrow yzzyxz=15.12.12=2160\Leftrightarrow\left(xyz\right)^2=2160\Leftrightarrow xyz=\pm12\sqrt{15}\)

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