\(2a=3b=5c=k\)
=> \(\left\{{}\begin{matrix}\frac{k}{2}=a\left(1\right)\\\frac{k}{3}=b\left(2\right)\\\frac{k}{5}=c\left(3\right)\end{matrix}\right.\)
=> \(a+b-c=\frac{k}{2}+\frac{k}{3}-\frac{k}{5}=38\)
=> \(30.\frac{k}{2}+30.\frac{k}{3}-30.\frac{k}{5}=30.38\)
=> \(15k+10k-6k=1140\)
=> \(19k=1140\)
=> \(k=1140:19=60\left(4\right)\)
Thay (4) vào (3);(2)và(1) , ta có :
\(\left\{{}\begin{matrix}\frac{60}{2}=a=30\\\frac{60}{3}=b=20\\\frac{60}{5}=c=12\end{matrix}\right.\)
Vậy \(\left(a;b;c\right)=\left(30;20;12\right)\)