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

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

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

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

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

<=> \(3x-\frac{9}{10}=\frac{21}{10}\)

<=> \(3x=3\)

<=> \(x=1\)

_Sai đề

Cái phân số cuối cùng phải là 15/92.95

15/5.8+15/8.11+15/11.14+...+15/92.95

=5.(3/5.8+3/8.11+3/11.14+...+3/92.95)

=5.(1/5-1/8+1/8-1/11+1/11-1/14+...+1/92-1/95)

=5.(1/5-1/95)

=5.(19/95-1/95)

=5.18/95=18/19

`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`

đối với câu a thì bạn phân tích ra nha:

ta có:

 A = \(\frac{-7}{10^{2005}}+\frac{-15}{10^{2006}}=\frac{-7}{10^{2005}}+\frac{-7}{10^{2006}}+\frac{-8}{10^{2006}}\)

 B = \(\frac{-15}{10^{2005}}+\frac{-7}{10^{2006}}=\frac{-8}{10^{2005}}+\frac{-7}{10^{2005}}+\frac{-7}{10^{2006}}\)

vì \(\frac{8}{10^{2005}}>\frac{8}{10^{2006}}=>\frac{-8}{10^{2005}}< \frac{-8}{10^{2006}}\)

=> A > B

CÂU b mk làm phân số hơi mất thời gian nên bn thông cảm cho mk nha:

1/5*8 + 1/8*11 + 1/11*14 +...+ 1/x(x+3) = 101/1540

=> 1/5 - 1/8 + 1/8 - 1/11 + 1/11 -...+ (1/x) - (1/ x+3) = 101/1540

=>1/5 - 1/x+3 = 101/1540

=> 1/x+3 = 1/5 - 101/1540

=> 1/x+3 = 1/308

=> 308*1 = (x+3)*1

=> 308 = x+3

=> x = 308 - 3

=> x = 305

Chúc bn học tốt !

b.Đặt S=\(\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+...+\frac{1}{x.\left(x+3\right)}=\frac{101}{1540}\)  ta có

\(S=\frac{1}{3}.\left(\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+...+\frac{1}{x.\left(x+3\right)}\right)=\frac{101}{1540}\)

    \(\frac{1}{3}.\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{x}-\frac{1}{x+3}\right)=\frac{101}{1540}\)                  

    \(\frac{1}{3}.\left(\frac{1}{5}-\frac{1}{x+3}\right)=\frac{101}{1540}\)              

    \(\frac{1}{5}-\frac{1}{x+3}=\frac{101}{1540}:\frac{1}{3}\)          

    \(\frac{1}{5}-\frac{1}{x+3}=\frac{303}{1540}\)            

    \(\frac{1}{x+3}=\frac{1}{5}-\frac{303}{1540}\)                

    \(\frac{1}{x+3}=\frac{308}{1540}-\frac{303}{1540}\)               

    \(\frac{1}{x+3}=\frac{1}{308}\)                 

\(\Rightarrow x+3=308\)              

     \(x=308-3\)

     \(x=305\)

a,1/5-1/608=......

b tự tính nha

Ta có: 3S = 3/2.5 + 3/5.8 + ... + 3/47.50

           3S = 1/2 - 1/5 + 1/5 - 1/8 + ... +1/47 - 1/50

           3S = 1/2 - 1/50

           3S = 12/25

           => S = 12/25 : 3 = 4/25 

k, đây là dạng toán sai phân hữu hạn. 
----------- 
số hạng tổng quát là 1/[n.(n+3)] = (1/3).[(n+3)-n]/[n.(n+3)] = (1/3). [1/n - 1/(n+3)] 
=> 
A = (1/3).[(1/2 - 1/5) + (1/5 - 1/8) + (1/8 - 1/11) +...+(1/44 - 1/47) + (1/47 - 1/50)] 
= (1/3).[1/2 - 1/50] 
= (1/3). (24/50) = (1/3).(12/25) = 4/25 
vậy A = 4/25 
--------- 
good luck!

\(B=\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{2018.2021}\)

\(B=\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{2018}-\frac{1}{2021}\)

\(B=\frac{1}{5}-\frac{1}{2021}\)

\(B=\frac{2016}{10105}\)

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!!!!