Đặt \(\dfrac{x}{12}=\dfrac{y}{9}=\dfrac{z}{5}=k\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=12k\\y=9k\\z=5k\end{matrix}\right.\)
Ta có: xyz=20
\(\Leftrightarrow k^3=\dfrac{1}{27}\)
\(\Leftrightarrow k=\dfrac{1}{3}\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=12k=4\\y=9k=3\\z=5k=\dfrac{5}{3}\end{matrix}\right.\)