Đặt \(\frac{x}{3}=\frac{y}{4}=\frac{z}{5}=k\left(k\ne0\right)\)\(\Rightarrow\begin{cases}x=3k\\y=4k\\z=5k\end{cases}\)
Ta có: \(b=\frac{x+y-z}{x+2y-z}=\frac{3k+4k-5k}{3k+2.4k-5k}=\frac{2k}{3k+8k-5k}=\frac{2k}{6k}=\frac{1}{3}\)
Giải:
Đặt \(\frac{x}{3}=\frac{y}{4}=\frac{z}{5}=k\)
\(\Rightarrow x=3k,y=4k,z=5k\)
Ta có:
\(B=\frac{x+y-z}{x+2y-z}=\frac{3k+4k-5k}{3k+8k-5k}=\frac{\left(3+4-5\right)k}{\left(3+8-5\right)k}=\frac{2k}{6k}=\frac{1}{3}\)
Vậy \(B=\frac{1}{3}\)