Đặt \(\dfrac{a}{2}=\dfrac{b}{5}=\dfrac{c}{7}=k\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=2k\\b=5k\\c=7k\end{matrix}\right.\)
Ta có: \(A=\dfrac{a-b+c}{a+2b-c}\)
\(=\dfrac{2k-5k+7k}{2k+2\cdot5k-7k}=\dfrac{4k}{5k}=\dfrac{4}{5}\)
Vậy: \(A=\dfrac{4}{5}\)