\(\dfrac{x}{12}=\dfrac{y}{9}=\dfrac{z}{5}\)
Đặt : \(\dfrac{x}{12}=k\Leftrightarrow x=12k\)
\(\dfrac{y}{9}=k\Leftrightarrow y=9k.\)
\(\dfrac{z}{5}=k\Leftrightarrow z=5k\)
Theo đề bài ta có : \(12k.9k.5k=20\)
\(540.k^3=20\)
\(\Leftrightarrow k=\dfrac{1}{3}\)
Thay vào ta sẽ có :
\(x=4,y=3,z=\dfrac{5}{3}\)
Giải:
Đặt \(\dfrac{x}{12}=\dfrac{y}{9}=\dfrac{z}{5}=k\Rightarrow\left\{{}\begin{matrix}x=12k\\y=9k\\z=5k\end{matrix}\right.\)
Ta có: \(xyz=20\)
\(\Rightarrow540k^3=20\)
\(\Rightarrow k^3=\dfrac{1}{27}\)
\(\Rightarrow k=\dfrac{1}{3}\)
\(\Rightarrow\left\{{}\begin{matrix}x=1\\y=\dfrac{3}{4}\\z=\dfrac{5}{12}\end{matrix}\right.\)
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