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. 
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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 
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good luck!

a) \(\frac{6}{2.5}+\frac{6}{5.8}+\frac{6}{8.11}+.......+\frac{6}{44.47}+\frac{6}{47.50}\)

\(=2\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+......+\frac{3}{44.47}+\frac{3}{47.50}\right)\)

\(=2\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+......+\frac{1}{44}-\frac{1}{47}+\frac{1}{47}-\frac{1}{50}\right)\)

\(=2\left(\frac{1}{2}-\frac{1}{50}\right)\)

\(=1-\frac{1}{25}\)

\(=\frac{24}{25}\)

đặt \(A=\frac{1}{9.11}+\frac{1}{11.13}+........+\frac{1}{41.43}+\frac{1}{43.45}\)

\(2A=\frac{2}{9.11}+\frac{2}{11.13}+.......+\frac{2}{41.43}+\frac{2}{43.45}\)

\(2A=\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}+......+\frac{1}{41}-\frac{1}{43}+\frac{1}{43}-\frac{1}{45}\)

\(2A=\frac{1}{9}-\frac{1}{45}\)

\(2A=\frac{4}{45}\)

\(A=\frac{4}{45}\div2\)

\(A=\frac{2}{45}\)

\(A=\frac{5}{2.5}+\frac{5}{5.8}+\frac{5}{8.11}+...+\frac{5}{47.50}\)

\(=\frac{5}{3}\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{47.50}\right)\)

\(=\frac{5}{3}\left(\frac{5-2}{2.5}+\frac{8-5}{5.8}+\frac{11-8}{8.11}+...+\frac{50-47}{47.50}\right)\)

\(=\frac{5}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{47}-\frac{1}{50}\right)\)

\(=\frac{5}{3}\left(\frac{1}{2}-\frac{1}{50}\right)\)

\(=\frac{4}{5}\)

A = \(\dfrac{4}{2.5}\) + \(\dfrac{4}{5.8}\)+...+ \(\dfrac{4}{47.50}\)

A = \(\dfrac{4}{3}\).( \(\dfrac{3}{2.5}+\dfrac{3}{5.8}+...+\dfrac{3}{47.50}\))

A = \(\dfrac{4}{3}\).(\(\dfrac{1}{2}\) - \(\dfrac{1}{5}\) + \(\dfrac{1}{5}\) - \(\dfrac{1}{8}\)+...+ \(\dfrac{1}{47}\) - \(\dfrac{1}{50}\))

A = \(\dfrac{4}{3}\).( \(\dfrac{1}{2}\) - \(\dfrac{1}{50}\)) 

A = \(\dfrac{4}{3}\). \(\dfrac{24}{50}\)

A = \(\dfrac{16}{25}\)

S = 1/2.5 +1/5.8 +1/8.11+1/11.14+1/14.17+1/17.20

S=1/3.(1/2-1/5+1/5-1/8+1/8-1/11+1/11-1/14+1/14-1/17+1/17-1/20)

S=1/3.(1/2-1/20)

S=1/3.(10/20-1/20)

S=1/3.9/20

S= 3/20

k nha

3S=3/2.5+3/5.8+3/8.11+...+3/101.104

3S=1/2-1/5+1/5-1/8+1/8-1/11+...+1/101-1/104

3S=1/2-1/104

S=51/104:3

S=17/104

Vậy S=17/104

         \(S=\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+........+\frac{1}{101.104}\)

\(\Rightarrow3S=3\left(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+.......+\frac{1}{101.104}\right)\)

           \(=\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+........+\frac{3}{101.104}\)

           \(=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+.........+\frac{1}{101}-\frac{1}{104}\)

           \(=\frac{1}{2}-\frac{1}{104}\)

           \(=\frac{51}{104}\)

           \(\Rightarrow S=\frac{51}{104}:3=\frac{51}{104}.\frac{1}{3}\)

                     \(=\frac{7}{104}\)

                VẬY   \(S=\frac{7}{104}\)

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

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

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

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

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

<=> \(3x-\frac{9}{10}=\frac{21}{10}\)

<=> \(3x=3\)

<=> \(x=1\)

`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`