Đặt \(\dfrac{x}{2}=\dfrac{y}{5}=\dfrac{z}{7}=k\)
\(\Rightarrow\left\{{}\begin{matrix}x=2k\\y=5k\\z=7k\end{matrix}\right.\)
Thay vào A ta có :
A=\(\dfrac{x-y+z}{x+2y-z}=\dfrac{2k-5k+7k}{2k+2.5k-7k}=\dfrac{4k}{5k}=\dfrac{4}{5}\)
Ta có: \(\dfrac{x}{2}=\dfrac{y}{5}=\dfrac{z}{7}\)
Đặt \(\dfrac{x}{2}=\dfrac{y}{5}=\dfrac{z}{7}=k\)
\(\Rightarrow x=2k;y=5k;z=7k\)
Theo đề ta có:
\(A=\dfrac{x-y+z}{x+2y-z}=\dfrac{2k-5k+7k}{2k+2\left(5k\right)-7k}\)
\(A=\dfrac{\left(2-5+7\right)k}{2k+10k-7k}=\dfrac{\left(2-5+7\right)k}{\left(2+10-7\right)k}\)
\(A=\dfrac{4k}{5k}=\dfrac{4}{5}\)
Vậy \(A=\dfrac{4}{5}\)
