Question

Tìm x ; y ; z :

\(x.y=z\) ; \(y.z=4x\) ; \(z.x=9y\)

Answer 1

Giải:

Theo đề ra, ta có:

$x.y=z$ ; $y.z=4x$ ; $z.x=9y$

$\mathrm{\backslash (\backslash Leftrightarrow\backslash left(x.y\backslash right)\backslash left(y.z\backslash right)\backslash left(z.x\backslash right)=\backslash left(xyz\backslash right)^2\backslash )}$

$\mathrm{\backslash (\backslash Leftrightarrow\backslash left(xyz\backslash right)^2=z.4x.9y\backslash )}$

\(\Leftrightarrow\left(xyz\right)^2=36xyz\)

\(\Leftrightarrow\left(xyz\right)^2-36xyz=0\)

\(\Leftrightarrow xyz\left(xyz-36\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}xyz=0\\xyz-36=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=y=z=0\\xyz=36\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=y=z=0\\\left\{{}\begin{matrix}x=3\\y=2\\z=6\end{matrix}\right.\end{matrix}\right.\)

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