a) Đặt \(\dfrac{x}{12}=\dfrac{y}{9}=\dfrac{z}{5}=k\Rightarrow\left\{{}\begin{matrix}x=12k\\y=9k\\z=5k\end{matrix}\right.\left(1\right)\)
Ta có: xyz = 20 => 12k . 9k . 5k = 20
=> \(k^3.540=20\)
=> \(k^3=\dfrac{1}{27}\)
=> k = \(\dfrac{1}{3}\)
Thay \(k=\dfrac{1}{3}\) vào (1) ta có: x = 4; y = 3; z = \(\dfrac{5}{3}\)