\(\frac{1}{2}-\frac{1}{14}=\frac{3}{7}\). Mà ở đây kết quả là \(\frac{1}{21}\)nên phân số phải nhóm ra ngoài là: 

\(\frac{1}{21}:\frac{3}{7}=\frac{1}{9}\). Ta có:

         \(\frac{1}{9}.\left(\frac{3x}{2.5}+\frac{3x}{5.8}+\frac{3x}{8.11}+\frac{3x}{11.14}\right)=\frac{1}{21}\). Suy ra 3x=9. Vậy x=3

Nhầm nha: Vì có 3x nên phân số nhóm ra ngoài là \(\frac{1}{3}\). Ta có tương tự. Suy ra 3x=3. Vậy x=1

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

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= 1/21 : 3/7

x= 1/21 . 7/3

x= 1/9

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+1/11-1/14] = 1/21

=> x. [1/2-1/14] = 1/21 => x . 3/7= 1/21 

=> x = 1/9

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+1/11-1/14] = 1/21

=> x. [1/2-1/14] = 1/21 => x . 3/7= 1/21 

=> x = 1/9

mình nhé 

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+1/11-1/14] = 1/21

x.[1/2-1/14] = 1/21 => x . 3/7 = 1/21

x = 1/9

\(3x-\frac{15}{5\cdot8}-\frac{15}{8\cdot11}-\frac{15}{11\cdot14}-...-\frac{15}{47\cdot50}=2\frac{1}{10}\)

<=> \(3x-5\left(\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+\frac{3}{11\cdot14}+...+\frac{3}{47\cdot50}\right)=\frac{21}{10}\)

<=> \(3x-5\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{47}-\frac{1}{50}\right)=\frac{21}{10}\)

<=> \(3x-5\left(\frac{1}{5}-\frac{1}{50}\right)=\frac{21}{10}\)

<=> \(3x-5\cdot\frac{9}{50}=\frac{21}{10}\)

<=> \(3x-\frac{9}{10}=\frac{21}{10}\)

<=> \(3x=3\)

<=> \(x=1\)

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

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

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

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

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

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

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

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

\(\frac{3x}{2\cdot5}+\frac{3x}{5\cdot8}+.........+\frac{3x}{14\cdot17}=\frac{1}{21}\)

\(\Rightarrow3x\cdot\left(\frac{1}{2\cdot5}+\frac{1}{5\cdot7}+........+\frac{1}{14\cdot17}\right)=\frac{1}{21}\)

\(\Rightarrow3x\cdot\left[\frac{1}{3}\left(\frac{3}{2\cdot5}+\frac{3}{5\cdot7}+......+\frac{3}{14\cdot17}\right)\right]=\frac{1}{21}\)

\(\Rightarrow3x\cdot\left[\frac{1}{3}\cdot\left(\frac{1}{2}-\frac{1}{17}\right)\right]=\frac{1}{21}\)

\(\Rightarrow3x\cdot\left(\frac{1}{3}\cdot\frac{15}{34}\right)=\frac{1}{21}\)

\(\Rightarrow3x\cdot\frac{5}{34}=\frac{1}{21}\)\(\Rightarrow x=\frac{1}{21}:\frac{1}{34}:3\)

\(\Rightarrow x=\frac{34}{315}\)

ai k mh mh k lại

k cho mh nha

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

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

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

\(3x.\frac{1}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-...+\frac{1}{11}-\frac{1}{14}\right)=\frac{1}{21}\)

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

\(\frac{3x}{7}=\frac{1}{21}\Rightarrow x=\frac{1}{9}\)

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

\(\Leftrightarrow3x\left(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+\frac{3x}{11.14}\right)=\frac{1}{21}\)

\(\Leftrightarrow3x.\frac{1}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}\right)=\frac{1}{21}\)

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

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

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

\(\Leftrightarrow63x=7\Leftrightarrow x=\frac{1}{9}\)

\(\Leftrightarrow21.9x=7\)

\(\Leftrightarrow189x=7\)

\(\Leftrightarrow x=\frac{1}{27}\)

b, Gọi số ki-lo-gam dưa là a kg(a>0)

Theo bài ra ta có:

\(\frac{3}{4}a=3\frac{1}{2}\Leftrightarrow\frac{3}{4}a=\frac{7}{2}\Leftrightarrow a=\frac{11}{4}\left(kg\right)\)

Vậy số ki-lo-gam dưa là \(\frac{11}{4}\) kg