Ta có:
\(\dfrac{2x}{3}=\dfrac{3y}{4}\Rightarrow y=\dfrac{4}{3}.\dfrac{2x}{3}=\dfrac{8x}{9}\)
\(\dfrac{2x}{3}=\dfrac{4z}{5}\Rightarrow z=\dfrac{5}{4}.\dfrac{2x}{3}=\dfrac{10x}{12}=\dfrac{5x}{6}\)
\(\Rightarrow x+y+z=x+\dfrac{8x}{9}+\dfrac{5x}{6}=49\)
Hay \(\left(18+16+15\right).\dfrac{x}{18}=49\).

tức là $x = 18 $
\(\Rightarrow y=16\)
và \(z=15\)

áp dụng tính chất dảy tỉ số bằng nhau

ta có : \(\dfrac{2\left(x-1\right)+3\left(y-2\right)-\left(z-3\right)}{\left(2.2\right)+\left(3.3\right)-4}=\dfrac{2x-2+3y-6-z+3}{4+9-4}\)

\(=\dfrac{\left(2x+3y-z\right)-5}{9}=\dfrac{50-5}{9}=\dfrac{45}{9}=5\)

suy ra ta có : \(\left\{{}\begin{matrix}\dfrac{x-1}{2}=5\\\dfrac{y-2}{3}=5\\\dfrac{z-3}{4}=5\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x-1=2.5\\y-2=3.5\\z-3=4.5\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x-1=10\\y-2=15\\z-3=20\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=10+1\\y=15+2\\z=20+3\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=11\\y=17\\z=23\end{matrix}\right.\) vậy \(x=11;y=17;z=23\)

2x = 3y = 6

=> 2x = 6 => x  = 3 

     3y = 6 => y = 2

vậy x = 3

      y = 2

2x=6

  => x=6:2=3

3y=6

  =>y=6:3=2

Vậy x=3, y=2

2x=3y=6

=> 2x=6

      x=6:2  

       x=3

\(5x\times\left(2x-\frac{1}{2}\right)+2\times\left(2x-\frac{1}{2}\right)=0\)

\(\Rightarrow\left(2x-\frac{1}{2}\right)\left(5x+2\right)=0\)

\(\Rightarrow\orbr{\begin{cases}2x-\frac{1}{2}=0\\5x+2=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}\frac{1}{4}\\x=\frac{-2}{5}\end{cases}}\)

Ta có 5x.(2x-1/2)+2.(2x-1/2)=0

        = (2x-1/2)(5x+2)=0

       Suy ra 2x-1/2=0 hoặc 5x+2=0

      Vậy x=1/4 hoặc x=-2/5

............................//............................

\(\Leftrightarrow x\left(2x^2+10x-x-5\right)-\left(2x^3+9x^2+x+4.5\right)=3.5\)

\(\Leftrightarrow2x^3+9x^2-5x-2x^3-9x^2-x-4.5=3.5\)

=>-6x=8

hay x=-4/3