Giải:
Đặt \(\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{5}=k\Rightarrow\left\{{}\begin{matrix}x=3k\\y=4k\\z=5k\end{matrix}\right.\)
Ta có: \(xyz=60\)
\(\Rightarrow3k.4k.5k=60\)
\(\Rightarrow k^3.60=60\)
\(\Rightarrow k^3=1\)
\(\Rightarrow k=1\)
\(\Rightarrow x=3,y=4,z=5\)
Vậy bộ số \(\left(x;y;z\right)\) là \(\left(3;4;5\right)\)
\(\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{5}=\dfrac{x.y.z}{3.4.5}=\dfrac{60}{60}=1\)
với \(\dfrac{x}{3}=1\Rightarrow x=3\)
\(\dfrac{y}{4}=1\Rightarrow y=4\)
\(\dfrac{z}{5}=1\Rightarrow z=5\)
\(\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{5}\) và \(xyz=60\) (2)
Đặt \(\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{5}=k\Rightarrow\left\{{}\begin{matrix}x=3k\\y=4k\\z=5k\end{matrix}\right.\)(1)
Thay (1) vào (2) ta có:
\(3k.4k.5k=60\Rightarrow60k^3=60\Rightarrow k^3=1\Rightarrow k=1\)
\(\Rightarrow\left\{{}\begin{matrix}x=3\\y=4\\z=5\end{matrix}\right.\)
Vậy (x;y;z)=(3;4;5)
Chúc bạn học tốt nha!!!
