\(\frac{x}{3}=\frac{y}{4}\Rightarrow\frac{x}{15}=\frac{y}{20}\left(1\right)\)

\(\frac{y}{5}=\frac{z}{6}\Rightarrow\frac{y}{20}=\frac{z}{24}\left(2\right)\)

từ (1) và (2) => \(\frac{x}{15}=\frac{y}{20}=\frac{z}{24}\)

đặt \(\frac{x}{15}=\frac{y}{20}=\frac{z}{24}=k\Rightarrow x=15k,y=20k,z=24k\)

thay x=15k, y=20k, z=24k vào M ta có:

\(M=\frac{2.15k+3.20k+4.24k}{3.15k+4.20k+5.24k}=\frac{30k+60k+96k}{45k+80k+120k}=\frac{186k}{245k}=\frac{186}{245}\)

vậy M=\(\frac{186}{245}\)

Ta có:\(\hept{\begin{cases}\frac{x}{3}=\frac{y}{4}\Rightarrow\frac{x}{15}=\frac{y}{20}\\\frac{y}{5}=\frac{z}{6}\Rightarrow\frac{y}{20}=\frac{z}{24}\end{cases}}\)

\(\Rightarrow\frac{x}{15}=\frac{y}{20}=\frac{z}{24}\)\(\Leftrightarrow\frac{2x}{30}=\frac{3y}{60}=\frac{4z}{96}=\frac{2x+3y+4z}{30+60+96}=\frac{2x+3y+4z}{186}\)(theo tính chất dãy tỉ số bằng nhau).(1)

= \(\frac{3x}{45}=\frac{4y}{80}=\frac{5z}{120}=\frac{3x+4y+5z}{45+80+120}=\frac{3x+4y+5z}{245}\)(theo tính chất dãy tỉ số bằng nhau). (2)

Từ (1) và (2) \(\Rightarrow\frac{2x+3y+4z}{186}=\frac{3x+4y+5z}{245}\Rightarrow\frac{2x+3y+4z}{3x+4y+5z}=\frac{186}{245}\)

Ta có: \(\frac{x}{3}=\frac{y}{4}\Rightarrow\frac{x}{15}=\frac{y}{20}\left(1\right)\)

\(\frac{y}{5}=\frac{z}{6}\Rightarrow\frac{y}{20}=\frac{z}{24}\left(2\right)\)

Từ (1) và (2) \(\Rightarrow\frac{x}{15}=\frac{y}{20}=\frac{z}{24}\)

Đặt \(\frac{x}{15}=\frac{y}{20}=\frac{z}{24}=k\left(k\ne0\right)\)\(\Rightarrow\begin{cases}x=15k\\y=20k\\z=24k\end{cases}\)

\(A=\frac{2x+3y+4z}{3x+4y+5z}=\frac{2.15k+3.20k+4.24k}{3.15k+4.20k+5.24k}=\frac{30k+60k+96k}{45k+80k+120k}=\frac{186k}{245k}=\frac{186}{245}\)

Vậy \(A=\frac{186}{245}\)

\(\frac{x}{3}=\frac{y}{4}\Rightarrow\frac{x}{15}=\frac{y}{20}\left(1\right)\\ \frac{y}{5}=\frac{z}{6}\Rightarrow\frac{y}{20}=\frac{z}{24}\left(2\right)\)

Từ (1) và (2 ) suy ra :\(\frac{x}{15}=\frac{y}{20}=\frac{z}{24}\)

Đặt :\(\frac{x}{15}=\frac{y}{20}=\frac{z}{24}=k\)

\(\Rightarrow\)x=15k; y=20k và z=24k (3)

Thay (3) vào A ta được:

A=\(\frac{2.15k+3.20k+4.24k}{3.15k+4.20k+5.24k}=\frac{186k}{245k}=\frac{186}{245}\)

Vậy A=\(\frac{186}{245}\)

Bài lm của mk có j thiếu sót thì bn tự bổ xung nha

Vì \(\frac{x}{3}=\frac{y}{4}=>x=\frac{3}{4}y\)

\(\frac{y}{5}=\frac{z}{6}=>z=\frac{6}{5}y\)

Ta có

\(M=\frac{2x+3y+4z}{3x+4y+5z}=\frac{2.\frac{3}{4}y+3y+4.\frac{6}{5}y}{3.\frac{3}{4}y+4y+5.\frac{6}{5}y}=\frac{\frac{93}{10}y}{\frac{49}{4}y}=\frac{93}{10}:\frac{49}{4}=\frac{186}{245}\)

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 \(M=\frac{186}{245}\)

Vì \(\frac{x}{3}\) = \(\frac{y}{4}\) => \(\frac{x}{15}\) = \(\frac{y}{20}\)

\(\frac{y}{5}\) = \(\frac{z}{6}\) => \(\frac{y}{20}\) = \(\frac{z}{24}\)

nên \(\frac{x}{15}\) = \(\frac{y}{20}\) = \(\frac{z}{24}\)

Đặt \(\frac{x}{15}\) = \(\frac{y}{20}\) = \(\frac{z}{24}\) = k

=> x = 15k; y = 20k và z = 24k

Thay vào M ta đc:

M = \(\frac{2.15k+3.20k+4.24k}{3.15k+4.20k+5.24k}\)

= \(\frac{30k+60k+96k}{45k+80k+120k}\)

= \(\frac{\left(30+60+96\right)k}{\left(45+80+120\right)k}\)

= \(\frac{186k}{245k}\) = \(\frac{186}{245}\)

Vậy M = \(\frac{186}{245}\).

Ta có \(\frac{3}{x+3}=\frac{4}{y+4}\Rightarrow3\left(y+4\right)=4\left(x+3\right)\Rightarrow3y=4x\Rightarrow\frac{x}{3}=\frac{y}{4}\)

Tương tự \(\frac{y+10}{5}=\frac{z+12}{6}\Rightarrow6\left(y+10\right)=5\left(z+12\right)\Rightarrow6y=5z\Rightarrow\frac{y}{5}=\frac{z}{6}\)

Từ đó ta có \(\frac{x}{15}=\frac{y}{20}=\frac{z}{16}\) 

Đặt \(\frac{x}{15}=\frac{y}{20}=\frac{z}{16}=t\Rightarrow\hept{\begin{cases}x=15t\\y=20t\\z=16t\end{cases}}\)

Vậy thì \(M=\frac{2.15t+3.20t+4.16t}{3.15t+4.20t+5.16t}=\frac{154t}{205t}=\frac{154}{205}.\)

M=104/205

\(\frac{x}{3}=\frac{y}{4}\Rightarrow\frac{x}{15}=\frac{y}{20}\)

\(\frac{y}{5}=\frac{z}{6}\Rightarrow\frac{y}{20}=\frac{z}{24}\)

\(\Rightarrow\frac{x}{15}=\frac{y}{20}=\frac{z}{24}=k\)

\(\Rightarrow x=15k;y=20k;z=24k\)

\(\frac{2x+3y+4z}{3x+4y+5z}=\frac{2.15k+3.20k+4.24k}{3.15k+4.20k+5.24k}=\frac{30k+60k+96k}{45k+80k+120k}=\frac{186k}{245k}=\frac{186}{245}\)

Ngân đầu tiên né