2x−3y/5=5y−2z/3=3z−5x/2=10x-15y/25=15y-6z/9=6z-10x/4=...+..+..../25+9+4=0/31=0

=> 2x=3y;  5y=2z ;  3z=5x => x/3=y/2; y/2=z/5

=> x/3=y/2 =z/5 = 12x/36=5y/10=3z/15= (12x+5y-3z)/31

      x/3 = 3y/6=2z/10 = (x-3y+2z)/7

=>  (12x+5y-3z)/ (x-3y+2z)=31/7

ngu

help me ai nhanh nhất mik tích cho

a) Ta có: \(\left(\dfrac{3}{4}\right)^{2021}>\left(\dfrac{3}{4}\right)^1=\dfrac{3}{4}\)

\(\Leftrightarrow\left(\dfrac{3}{4}\right)^{2021}+1>\dfrac{3}{4}+1\)

Ta có :\(\frac{2x-3y}{5}=\frac{5y-2z}{3}=\frac{3z-5x}{2}\)

=> \(\frac{10x-15y}{25}=\frac{15y-6z}{9}=\frac{6z-10x}{4}=\frac{10x-15y+15y-6z+6z-10x}{25+9+4}=\frac{0}{38}=0\)

=> \(\hept{\begin{cases}2x=3y\\5y=2z\\3z=5x\end{cases}}\Leftrightarrow\frac{x}{3}=\frac{y}{2}=\frac{z}{5}\)

Đặt \(\frac{x}{3}=\frac{y}{2}=\frac{z}{5}=k\Leftrightarrow\hept{\begin{cases}x=3k\\y=2k\\z=5k\end{cases}}\)

Khi đó B = \(\frac{12x+5y-3z}{x-3y+2z}=\frac{36k+10k-15k}{3k-6k+10k}=\frac{31k}{7k}=\frac{31}{7}\)

\(a,4x=5y\:\Rightarrow\frac{x}{5}=\frac{y}{4}\Rightarrow\frac{x}{15}=\frac{y}{12}\)

\(4y=6z\Rightarrow\frac{y}{6}=\frac{z}{4}\Rightarrow\frac{y}{12}=\frac{z}{8}\)

\(\Rightarrow\frac{x}{15}=\frac{y}{12}=\frac{z}{8}\)

\(\Rightarrow\frac{x}{15}=\frac{2y}{24}=\frac{3z}{24}\)

\(\Rightarrow\frac{x-2y+3z}{15-24+24}=\frac{x}{15}=\frac{y}{12}=\frac{z}{8}\)

\(\Rightarrow\frac{5}{15}=\frac{x}{15}=\frac{y}{12}=\frac{z}{8}\)

\(\Rightarrow\frac{1}{3}=\frac{x}{15}=\frac{y}{12}=\frac{z}{8}\)

\(\Rightarrow\hept{\begin{cases}x=\frac{1}{3}\cdot15=5\\y=\frac{1}{3}\cdot12=4\\z=\frac{1}{3}\cdot8=\frac{8}{3}\end{cases}}\)

mọi người giúp mk câu b, c, d còn lại nha

nhanh