Đặt \(\frac{a}{2}=\frac{b}{5}=\frac{c}{7}=k\Rightarrow a=2k;b=5k;c=7k\)
Hay: \(\frac{a-b+c}{a+2b-c}=\frac{2k-5k+7k}{2k+2.5k-7k}=\frac{k\left(2-5+7\right)}{k\left(2+10-7\right)}=\frac{4k}{5k}=\frac{4}{5}\)
\(\Rightarrow A=\frac{4}{5}\)
Vậy \(A=\frac{4}{5}\)