Ta có:
Đặt \(\frac{x}{2}=\frac{2y}{5}=\frac{3z}{7}=k\)
\(\Rightarrow\left\{{}\begin{matrix}x=2k\\y=\frac{5}{2}k\\z=\frac{7}{3}k\end{matrix}\right.\)
\(\Rightarrow A=\frac{2x+5y-3z}{7x+y-5z}=\frac{2.2k+5.\frac{5}{2}k-3.\frac{7}{3}k}{7.2k+\frac{5}{2}k-5.\frac{7}{3}k}\)
=\(\frac{4k+\frac{25}{2}k-7k}{14k+\frac{5}{2}k-\frac{35}{3}k}=\frac{\frac{19}{2}k}{\frac{29}{6}k}=\frac{57}{29}\)