Giải:
Ta có: \(\frac{x}{3}=\frac{y}{4}\Rightarrow\frac{x}{15}=\frac{y}{20}\)
\(\frac{y}{5}=\frac{z}{6}\Rightarrow\frac{y}{20}=\frac{z}{24}\)
\(\Rightarrow\frac{x}{15}=\frac{y}{20}=\frac{z}{24}\)
Đặt \(\frac{x}{15}=\frac{y}{20}=\frac{z}{24}=k\)
\(\Rightarrow\hept{\begin{cases}x=15k\\y=20k\\z=24k\end{cases}}\)
\(\Rightarrow M=\frac{2x+3y+4z}{3x+4y+5z}=\frac{30k+60k+96k}{45k+80k+120k}=\frac{\left(30+60+96\right)k}{\left(45+80+120\right)k}\)
bạn tự tính nốt nhé
Mình giải tiếp cho:
\(M=\frac{\left(30+60+96\right)k}{\left(45+80+120\right)k}=\frac{186k}{245k}=\frac{186}{245}\)
Vậy \(M=\frac{186}{245}\)
Ta có :
\(\frac{x}{3}=\frac{y}{4}\) \(\Rightarrow\)\(\frac{x}{15}=\frac{y}{20}\)
\(\frac{y}{5}=\frac{z}{6}\)\(\Rightarrow\)\(\frac{y}{20}=\frac{z}{24}\)
\(\Rightarrow\)\(\frac{x}{15}=\frac{y}{20}=\frac{z}{24}\)
Đặt \(\frac{x}{15}=\frac{y}{20}=\frac{z}{24}=k\)\(\Rightarrow\hept{\begin{cases}x=15k\\y=20k\\z=24k\end{cases}}\)
\(\Rightarrow M=\frac{2x+3y+4z}{3x+4y+5z}=\frac{30k+60k+96k}{45k+80k+120k}=\frac{k.\left(30+60+96\right)}{k.\left(45+80+120\right)}=\frac{186k}{245k}=\frac{186}{245}\)
Vậy \(M=\frac{186}{245}\)
ta có
\(\frac{x}{3}=\frac{y}{4},\frac{y}{5}=\frac{z}{6}\Rightarrow\hept{\begin{cases}\frac{x}{15}=\frac{y}{20}\\\frac{y}{20}=\frac{z}{24}\end{cases}\Rightarrow}\frac{x}{15}=\frac{y}{20}=\)\(\frac{z}{24}\)
đặt \(\frac{x}{15}=\frac{y}{20}=\frac{z}{24}=k\left(k\in Z\right)\)
=>x=15k, y=20k, z=24k
thay vào biểu thức ta có
M= \(\frac{2.15k+3.20k+4.24k}{3.15k+4.20k+5.24k}=\frac{30k+60k+96k}{45k+80k+120k}=\)\(\frac{186k}{245k}\)=......
bạn tự rút gọn và kết luận nhé
k cho mk nh bn
*****Chúc bạn học giỏi*****
Giải
ta có:
\(\frac{x}{3}=\frac{y}{4}\Rightarrow\frac{x}{3}.\frac{1}{5}=\frac{y}{4}.\frac{1}{5}\Rightarrow\frac{x}{15}=\frac{y}{20}\)(1)
\(\frac{y}{5}=\frac{z}{6}\Rightarrow\frac{y}{5}.\frac{1}{4}=\frac{z}{6}.\frac{1}{4}\Rightarrow\frac{y}{20}=\frac{z}{24}\)(2)
Từ (1) và (2) \(\Rightarrow\frac{x}{15}=\frac{y}{20}=\frac{z}{24}\)
Đặt \(\frac{x}{15}=\frac{y}{20}=\frac{z}{24}=k\)
\(\Rightarrow\hept{\begin{cases}x=15k\\y=20k\\z=24k\end{cases}}\)
\(\Rightarrow M=\frac{2x+3y+4z}{3x+4y+5z}=\frac{30k+60k+96k}{45k+80k+120k}=\frac{\left(30+60+96\right)k}{\left(45+80+120\right)k}=\frac{186k}{245k}=\frac{186}{245}\)
Vậy \(M=\frac{186}{245}\)
