Vì x,y,z tỉ lệ với các số 2,3,4.
\(\Rightarrow\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}=k\)
\(\Rightarrow\left\{{}\begin{matrix}x=2k\\y=3k\\z=4k\end{matrix}\right.\)
Thay x = 2k; y = 3k ; z = 4k vào M, ta được:
\(M=\dfrac{5x+2y+z}{x+4y-3z}\)
\(M=\dfrac{5.\left(2k\right)+2.\left(3k\right)+4k}{2k+4.\left(3k\right)-3.\left(4k\right)}\)
\(M=\dfrac{10k+6k+4k}{2k+12k-12k}\)
\(M=\dfrac{20k}{2k}\)
\(M=\dfrac{20}{2}\)
\(M=10\)
Vậy M = 10.
Vì x,y,z tỉ lệ nghịch với các số 2;3;4 nên \(\dfrac{x}{\dfrac{1}{2}}=\dfrac{y}{\dfrac{1}{3}}=\dfrac{z}{\dfrac{1}{4}}\)
Đặt x = \(\dfrac{1}{2}k\), y = \(\dfrac{1}{3}k\), z = \(\dfrac{1}{4}k\)
\(\Rightarrow M=\dfrac{\dfrac{1}{2}k.5+\dfrac{1}{3}k.2+\dfrac{1}{4}k}{\dfrac{1}{2}k+\dfrac{1}{3}k.4-\dfrac{1}{4}k.3}\)
\(M=\dfrac{\dfrac{5}{2}k+\dfrac{2}{3}k+\dfrac{1}{4}k}{\dfrac{1}{2}k+\dfrac{4}{3}k-\dfrac{3}{4}k}\)
\(M=\dfrac{k\left(\dfrac{5}{2}+\dfrac{2}{3}+\dfrac{1}{4}\right)}{k\left(\dfrac{1}{2}+\dfrac{4}{3}-\dfrac{3}{4}\right)}\)
\(M=\dfrac{\dfrac{41}{12}}{\dfrac{13}{12}}\)
\(M=\dfrac{41}{13}\)
Vậy \(M=\dfrac{41}{13}\)
