\(x\) \((\)\(\dfrac{3}{2.5}\) \(+ \) \(\dfrac{3}{5.8}\) \(+\) \(\dfrac{3}{8.11}\) \(+\) \(\dfrac{3}{11.14}\)\()\) \(=\) \(\dfrac{1}{21}\)
\(x\) \((\)\(\dfrac{1}{2}\) \(-\) \(\dfrac{1}{5}\) \(+\) \(\dfrac{1}{5}\) \(-\) \(\dfrac{1}{8}\) \(+\) \(\dfrac{1}{8}\) \(-\) \(\dfrac{1}{11}\) \(+\) \(\dfrac{1}{11}\) \(-\) \(\dfrac{1}{14}\)\()\) \(=\) \(\dfrac{1}{21}\)
\(x\) \((\)\(\dfrac{1}{2}\) \(-\) \(\dfrac{1}{14}\)\()\) \(=\) \(\dfrac{1}{21}\)
\(x\) x \(\dfrac{3}{7}\) \(=\) \(\dfrac{1}{21}\)
\(x\)        \(=\) \(\dfrac{1}{21}\) \(:\) \(\dfrac{3}{7}\) 
\(x\)        \(=\) \(\dfrac{1}{9}\)

`3x-15/(5*8)-15/(8*11)-15/(11*14)-...-15/(47*50)=2 1/10`

`3x-(15/(5*8)+15/(8*11)+15/(11*14)+...+15/(47*50))=21/10`

`3x-5(3/(5*8)+3/(8*11)+3/(11*14)+...+3/(47*50))=21/10`

`3x-5(1/5-1/8+1/8-1/11+1/11-1/14+...+1/47-1/50)=21/10`

`3x-5(1/5-1/50)=21/10`

`3x-5*9/50=21/10`

`3x-9/10=21/10`

`3x=21/10+9/10`

`3x=3`

`x=1`

a, dễ, tự làm

b, \(\dfrac{3x}{2.5}+\dfrac{3x}{5.8}+.........+\dfrac{3x}{11.14}=\dfrac{1}{21}\)

\(\Leftrightarrow x\left(\dfrac{3}{2.5}+\dfrac{3}{5.8}+.........+\dfrac{3}{11.14}\right)=\dfrac{1}{21}\)

\(\Leftrightarrow x\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+.....+\dfrac{1}{11}-\dfrac{1}{14}\right)=\dfrac{1}{21}\)

\(\Leftrightarrow x\left(\dfrac{1}{2}-\dfrac{1}{14}\right)=\dfrac{1}{21}\)

\(\Leftrightarrow x.\dfrac{3}{7}=\dfrac{1}{21}\)

\(\Leftrightarrow x=\dfrac{1}{9}\)

Vậy ...

a) (x-2)³ = (x-2)²

<=> (x-2)³-(x-2)² = 0

<=> (x-2)²(x-2-1) = 0

<=> \(\left\{{}\begin{matrix}\left(x-2\right)^2=0\\x-3=0\end{matrix}\right.\)

<=> \(\left\{{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)

b) \(\dfrac{3x}{2.5}+\dfrac{3x}{5.8}+...+\dfrac{3x}{11.14}=\dfrac{1}{21}\)

<=> \(x\left(\dfrac{3}{2.5}+\dfrac{3}{5.8}+...+\dfrac{3}{11.14}\right)=\dfrac{1}{21}\)

<=> \(x\left(\dfrac{1}{2}-\dfrac{1}{14}\right)=\dfrac{1}{21}\)

<=> \(x=\dfrac{1}{21}:\dfrac{3}{7}\)

<=> \(x=\dfrac{1}{9}\)

a) \(\left(x-2\right)^3=\left(x-2\right)^2\)

=> x - 2 = 0 hoặc x - 2 = 1

Với x - 2 = 0

x = 0+2

x = 2

Với x - 2 = 1

x = 1 + 2

x = 3

Vậy x \(\in\){2;3}

b) \(\dfrac{3x}{2.5}+\dfrac{3x}{5.8}+...+\dfrac{3x}{11.14}=\dfrac{1}{21}\)

x .\(\left(\dfrac{3}{2.5}+\dfrac{3}{5.8}+....+\dfrac{3}{11.14}\right)=\dfrac{1}{21}\)

x . \(\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+....+\dfrac{1}{11}-\dfrac{1}{14}\right)=\dfrac{1}{21}\)

x . \(\left(\dfrac{1}{2}-\dfrac{1}{14}\right)=\dfrac{1}{21}\)

x . \(\dfrac{3}{7}=\dfrac{1}{21}\)

x \(=\dfrac{1}{21}:\dfrac{3}{7}=\dfrac{1}{9}\)

sao 0 có kết quả                                                                          

3x/2.5 + 3x/5.8+3x/8.11+3x/11.14 = 1/21

 x(1/2-1/5+1/5-1/8+1/8-1/11+1/11-1/14) = 1/21

 x(1/2-1/14) = 1/21

x . 3/7 = 1/21

=> x = 1/21 : 3/7 

=> x = 1/9

Hihi mình giải zầy mk hk bik đúng hay sai

\(\frac{3x}{2.5}+\frac{3x}{5.8}+\frac{3x}{8.11}+\frac{3x}{11.14}=\frac{1}{21}\)

\(x.\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}\right)\)\(=\frac{1}{21}\)

\(x.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}\right)\)\(=\frac{1}{21}\)

\(x.\left(\frac{1}{2}-\frac{1}{4}\right)\)\(=\frac{1}{21}\)

\(x.\frac{3}{7}=\frac{1}{21}\)

\(x=\frac{1}{21}:\frac{3}{7}\)

\(x=\frac{1}{9}\)

Ta có : \(\frac{3x}{2\times5}+\frac{3x}{5\times8}+\frac{3x}{8\times11}+\frac{3x}{11\times14}=\frac{1}{21}\)

    \(\Rightarrow x\times\left(\frac{3}{2\times5}+\frac{3}{5\times8}+\frac{3}{8\times11}+\frac{3}{11\times14}\right)=\frac{1}{21}\)

    \(\Rightarrow x\times\left(\frac{1}{2\times5}+\frac{1}{5\times8}+\frac{1}{8\times11}+\frac{1}{11\times14}\right)=\frac{1}{21}\)

         \(x\times\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}\right)=\frac{1}{21}\) 

         \(x\times\left(\frac{1}{2}-\frac{1}{14}\right)\)                                          \(=\frac{1}{21}\)

        \(x\times\frac{3}{7}\)                                                             \(=\frac{1}{21}\)

        \(x\)                                                                        \(=\frac{1}{21}\div\frac{3}{7}=\frac{1}{21}\times\frac{7}{3}\)

        \(\Rightarrow x=\frac{1}{9}\)

Ta có 3x/2.5+3x/5.8+3x/8.11+3x/11.14=1/21

=>x(3/2.5+3/5.8+3/8.11+3/11.14)=1/21

=>3x(1/2.5+1/5.8+1/8.11+1/11.14)=1/21

=>3x(1/2-1/14)=1/21

=>3x.3/7=1/21

=>3x=1/21:3/7

=>3x=1

=>x=1:3=1/3

ta có : 3x / 2.5 + 3x / 5.8 + 3x / 8.11 + 3x / 11.14 = 1 / 21

            => 3x ( 1 /2.5 + 1 / 5.8 +1/8.11 +  1/11.14) = 1/21

            => 3x ( 1/2 - 1/5 + 1/5 - 1/8 + 1/8 - 1/11 + 1/11 - 1/14 ) = 1/21

            3x ( 1/2 - 1/14 ) = 1/21

            3x . 3/7 = 1/21

            3x = 1/21 : 3/7

            3x = 1/9

            x =  1/9 : 3

            x = 1/27

\(\dfrac{3}{2\cdot5}+\dfrac{3}{5\cdot8}+\dfrac{3}{8\cdot11}+\dfrac{3}{11\cdot14}+\dfrac{3}{14\cdot17}\)

= \(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{17}\)

\(=\dfrac{1}{2}-\dfrac{1}{17}\)

\(=\dfrac{15}{34}\)

Vì \(\dfrac{15}{34}< \dfrac{1}{2}=>\dfrac{3}{2\cdot5}+\dfrac{3}{5\cdot8}+\dfrac{3}{8\cdot11}+\dfrac{3}{11\cdot14}+\dfrac{3}{14\cdot27}< \dfrac{1}{2}\)

`(3x)/(2.5)+(3x)/(5.8)+(3x)/(8.11)+(3x)/(11.14)=1/21`
`=>x(3(2.5)+3/(5.8)+3/(8.11)+3/(11.14))=1/21`
`=>x(1/2-1/5+1/5-1/8+1/8-1/11-1/14)=1/21`
`=>x*(1/2-1/14)=1/21`
`=>x*3/7=1/21`
`=>x=1/21:3/7=1/9`
Vậy `x=1/9`

Vậy 

\(\dfrac{3x}{2.5}+\dfrac{3x}{5.8}+\dfrac{3x}{8.11}+\dfrac{3x}{11.14}=\dfrac{1}{21}\) 

\(x.\left(\dfrac{3}{2.5}+\dfrac{3}{5.8}+\dfrac{3}{8.11}+\dfrac{3}{11.14}\right)=\dfrac{1}{21}\) 

\(x.\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{14}\right)=\dfrac{1}{21}\) 

                       \(x.\left(\dfrac{1}{2}-\dfrac{1}{14}\right)=\dfrac{1}{21}\) 

                                      \(x.\dfrac{3}{7}=\dfrac{1}{21}\) 

                                          \(x=\dfrac{1}{21}:\dfrac{3}{7}\) 

                                          \(x=\dfrac{1}{9}\)

Đề bài là tìm x nhé các bạn, các bạn giải giúp mik nhé!!

Bài làm

\(\frac{3x}{2.5}+\frac{3x}{5.8}+\frac{3x}{8.11}+\frac{3x}{11.14}=\frac{1}{21}\)

\(\Leftrightarrow\frac{x}{2}-\frac{x}{5}+\frac{x}{5}-\frac{x}{8}+\frac{x}{8}-\frac{x}{11}+\frac{x}{11}-\frac{x}{14}=\frac{1}{21}\)

\(\Leftrightarrow\frac{x}{2}-\frac{x}{14}=\frac{1}{21}\)

\(\Leftrightarrow\frac{7x}{14}-\frac{x}{14}=\frac{1}{21}\)

\(\Leftrightarrow\frac{6x}{14}=\frac{1}{21}\)

\(\Leftrightarrow126x=14\)

\(\Leftrightarrow x=\frac{1}{9}\)

Học tôt 

Bài làm:
Ta có: \(\frac{3x}{2.5}+\frac{3x}{5.8}+\frac{3x}{8.11}+\frac{3x}{11.14}=\frac{1}{21}\)

\(\Leftrightarrow x\left(\frac{5-2}{2.5}+\frac{8-5}{5.8}+\frac{11-8}{8.11}+\frac{14-11}{11.14}\right)=\frac{1}{21}\)

\(\Leftrightarrow x\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}\right)=\frac{1}{21}\)

\(\Leftrightarrow x\left(\frac{1}{2}-\frac{1}{14}\right)=\frac{1}{21}\)

\(\Leftrightarrow\frac{3}{7}x=\frac{1}{21}\)

\(\Leftrightarrow x=\frac{1}{9}\)

Học tốt!!!!
 

\(\dfrac{1}{5\cdot8}+\dfrac{1}{8\cdot11}+\dfrac{1}{11\cdot14}+...+\dfrac{1}{x\left(x+3\right)}=\dfrac{101}{1540}\)

\(\Leftrightarrow\dfrac{1}{3}\left(\dfrac{3}{5\cdot8}+\dfrac{3}{8\cdot11}+\dfrac{3}{11\cdot14}+...+\dfrac{3}{x\left(x+3\right)}\right)=\dfrac{101}{1540}\)

\(\Leftrightarrow\dfrac{1}{3}\left(\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{x}-\dfrac{1}{x+3}\right)=\dfrac{101}{1540}\)

\(\Leftrightarrow\dfrac{1}{3}\left(\dfrac{1}{5}-\dfrac{1}{x+3}\right)=\dfrac{101}{1540}\)\(\Leftrightarrow\dfrac{1}{5}-\dfrac{1}{x+3}=\dfrac{303}{1540}\)

\(\Leftrightarrow\dfrac{1}{x+3}=\dfrac{1}{308}\)\(\Rightarrow x+3=308\Rightarrow x=305\)