Theo bài ra ta cs 

\(2x=3y\Rightarrow\frac{x}{3}=\frac{y}{2}\)

\(4y=5z\Rightarrow\frac{y}{5}=\frac{z}{4}\)

T lại cs 

\(\frac{x}{3}=\frac{y}{2}\Rightarrow\frac{x}{15}=\frac{y}{10}\left(1\right)\)

\(\frac{y}{5}=\frac{z}{4}\Rightarrow\frac{x}{10}=\frac{z}{8}\left(2\right)\)

Từ (1);(2) \(\Rightarrow\frac{x}{15}=\frac{y}{10}=\frac{z}{8}\)

ADTC dãy tỉ số bằng nhau ta cs 

\(\frac{x}{15}=\frac{y}{10}=\frac{z}{8}=\frac{2x+3y-4z}{2.15+3.10-4.8}=\frac{56}{28}=2\)

\(\Rightarrow\hept{\begin{cases}\frac{x}{15}=2\\\frac{y}{10}=2\\\frac{z}{8}=2\end{cases}\Rightarrow\hept{\begin{cases}x=30\\y=20\\z=16\end{cases}}}\)

\(2x=3y;4y=5z\) => \(8x=12y;12y=15z\)

=>  \(\frac{8x}{120}=\frac{12y}{120}=\frac{15z}{120}\)=> \(\frac{x}{15}=\frac{y}{10}=\frac{z}{8}\)

=>   \(\frac{2x}{30}=\frac{3y}{30}=\frac{4z}{32}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\frac{2x}{30}=\frac{3y}{30}=\frac{4z}{32}=\frac{2x+3y-4z}{30+30-32}=\frac{56}{28}\)

=> \(\frac{2x}{30}=2=>2x=60=>x=30\)

\(\frac{3y}{30}=2=>3y=60=>y=20\)

\(\frac{4z}{32}=2=>4z=64=>z=16\)

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

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

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

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

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

a

Đặt \(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}=k\)

\(\Rightarrow x=2k+1;y=3k+2;z=4k+3\)

Thay vào,ta được:

\(2\left(2k+1\right)+3\left(3k+2\right)-\left(4k+3\right)=50\)

\(\Leftrightarrow4k+2+9k+6-4k-3=50\)

\(\Leftrightarrow9k+5=50\)

\(\Leftrightarrow9k=45\)

\(\Leftrightarrow k=5\)

\(\frac{x-1}{2}=\frac{y+3}{4}=\frac{z-5}{6}=\frac{5x-5}{10}=\frac{3y+9}{12}=\frac{4z-20}{24}\)

\(=\frac{5x-5-3y-9-4z+20}{10-12-24}=\frac{\left(5x-3y-4z\right)+\left(20-5-9\right)}{26}=\frac{46+6}{26}=2\)

\(\Rightarrow x=2\cdot2+1=5\)

\(y=4\cdot2-3=5\)

\(z=2\cdot6+5=17\)

Câu c tương tự như câu 1

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

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

bạn ơi cái này là tìm về cái gì?

ý bạn là \(x-y-z=-33?\)

Ta có \(2x=3y=5z\Rightarrow\dfrac{2x}{30}=\dfrac{3y}{30}=\dfrac{5z}{30}\Rightarrow\dfrac{x}{15}=\dfrac{y}{10}=\dfrac{z}{6}\)

Áp dụng t/c dtsbn:

\(\dfrac{x}{15}=\dfrac{y}{10}=\dfrac{z}{6}=\dfrac{x-y-z}{15-10-6}=\dfrac{-33}{-1}=33\\ \Rightarrow\left\{{}\begin{matrix}x=33\cdot15=495\\y=33\cdot10=330\\z=33\cdot6=198\end{matrix}\right.\)

Ngân đầu tiên né

=> x/15=y/20=z/24

đặt ....................... =k (k#0)

=> x=15k y=20k z=24k

=> M=2x+3y+4z/3x+4y+5z =.......=186k/245k=186/245

vậy M=186/245

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