Ta có: \(15x=10y=6z\Leftrightarrow\dfrac{x}{\dfrac{1}{15}}=\dfrac{y}{\dfrac{1}{10}}=\dfrac{z}{\dfrac{1}{6}}\)
Đặt \(\dfrac{x}{\dfrac{1}{15}}=\dfrac{y}{\dfrac{1}{10}}=\dfrac{z}{\dfrac{1}{6}}=k\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{15}k\\y=\dfrac{1}{10}k\\z=\dfrac{1}{6}k\end{matrix}\right.\)
Mà \(xyz=240\)
\(\Rightarrow\dfrac{1}{15}k.\dfrac{1}{10}k.\dfrac{1}{6}k=240\)
\(\Rightarrow\dfrac{1}{900}k^3=240\)
\(\Rightarrow k^3=\dfrac{240}{\dfrac{1}{900}}=216000\)
\(\Rightarrow k=\sqrt[3]{216000}=60\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{15}.60=4\\y=\dfrac{1}{10}.60=6\\z=\dfrac{1}{6}.60=10\end{matrix}\right.\)
Vậy \(x=4\\ y=6\\ z=10\)
Từ 15x = 10y = 6z ⇒ \(\dfrac{15x}{30}=\dfrac{10y}{30}=\dfrac{6z}{30}\)
⇒ \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}\)
Đặt \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}=k\Rightarrow\left\{{}\begin{matrix}x=2k\\y=3k\\z=5k\end{matrix}\right.\Rightarrow2k.3k.5k=240\)
Từ 2k.3k.5k ta có:
2k.3k.5k=240
(2.3.5) . \(k^3\)=240
\(k^3\)=240:30
\(k^3\) =8
k=2
⇒x=2.2=4
⇒y=3.2=6
⇒z=5.2=10
