Đặt \(\frac{x}{2}=\frac{y}{3}=\frac{z}{5}=k\)
\(\)\(\Rightarrow\left\{\begin{matrix}x=2k\\y=3k\\z=5k\end{matrix}\right.\)
\(\Rightarrow x.y.z=2k.3k.5k=30.k^3\)
\(\Rightarrow30.k^3=810\)
\(\Rightarrow k^3=27\)
\(\Rightarrow k=3\)
\(\Rightarrow\left\{\begin{matrix}x=6\\y=9\\z=15\end{matrix}\right.\)
Vậy x=6, y=9 và z=15
Đặt\(\frac{x}{2}\)=\(\frac{y}{3}\)=\(\frac{z}{5}\)=k
\(\Rightarrow\)x=2.k ; y=3.k ; z=5.k
Suy ra:x.y.z=2k.3k.5k=810
\(\Rightarrow\)2k.3k.5k=810
\(\Rightarrow\)(2.3.5)\(^{_k3}\)=810
\(\Rightarrow k^3\)=810:30
\(\Rightarrow k^3\)=27:\(3^3\)
\(\Rightarrow\)k=3
Suy ra:x=3.2=6
y=3.3=9
z=3.5=15
Vậy x=6;y=9;z=15
Ta có : \(\frac{x}{2}=\frac{y}{3}=\frac{z}{5}\) và \(x.y.z=810\)
Đặt \(\frac{x}{2}=\frac{y}{3}=\frac{z}{5}=k\)
\(\Rightarrow x=2k;y=3k;z=5k\)
Thay \(x=2k;y=3k;z=5k\) vào \(x.y.z=810\), ta được :
\(x.y.z=810\)
\(\Rightarrow\left(2k\right).\left(3k\right).\left(5k\right)=810\)
\(\Rightarrow30k^3=810\)
\(\Rightarrow k^3=27\)
\(\Rightarrow k^3=3^3\)
\(\Rightarrow k=3\)
+ Nếu \(k=3\)
\(\Rightarrow\left\{\begin{matrix}x=6\\y=9\\z=15\end{matrix}\right.\)
Vậy : \(\left\{\begin{matrix}x=6\\y=9\\z=15\end{matrix}\right.\)
