Đặt \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}=k\)
\(\Rightarrow\left\{{}\begin{matrix}x=2k\\y=3k\\z=4k\end{matrix}\right.\)(1)
Thay (1) vào biểu thức M ta được:
\(M=\dfrac{y+z-x}{x-y+z}=\dfrac{3k+4k-2k}{2k-3k+4k}=\dfrac{5k}{3k}=\dfrac{5}{3}\)
Vậy.................
Chúc bạn học tốt!!!
\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\)
\(\Rightarrow\left\{{}\begin{matrix}x=2k\\y=3k\\z=4k\end{matrix}\right.\)
\(\Rightarrow\dfrac{y+z-x}{x-y+z}=\dfrac{3k+4k-2k}{2k-3k+4k}=\dfrac{5k}{3k}=\dfrac{5}{3}\)