Đặt \(\frac{x}{3}=\frac{y}{5}=\frac{z}{7}=k\Rightarrow x=3k;y=5k;z=7k\)
\(xy+yz+zx=3k.5k+5k.7k+7k.3k=k^2\left(15+35+21\right)=71k^2;xyz=3k.5k.7k=105k^3\)
Ta có : \(xyz\left(xz+yz+xy+xz+yz+xy\right)=477120\)
\(\Rightarrow xyz\left(xz+yz+xy\right)=238560\)\(\Rightarrow105k^3.71k^2=238560\Rightarrow k^5=32=2^5\Rightarrow k=2\)
Vậy : x= 6 ; y = 10 ; z = 14