Đặt \(\dfrac{x}{5}=\dfrac{y}{3}=\dfrac{z}{2}=k\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=5k\\y=3k\\z=2k\end{matrix}\right.\)
Ta có: \(x^2-y^2-z^2=48\)
\(\Leftrightarrow25k^2-9k^2-4k^2=48\)
\(\Leftrightarrow k^2=4\)
TRường hợp 1: k=-2
\(\Leftrightarrow\left\{{}\begin{matrix}x=5k=-10\\y=3k=-6\\z=2k=-4\end{matrix}\right.\)
TRường hợp 2: k=2
\(\Leftrightarrow\left\{{}\begin{matrix}x=5k=10\\y=3k=6\\z=2k=4\end{matrix}\right.\)
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