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25 tháng 2 2022

\(\dfrac{1}{5.10}+\dfrac{1}{10.15}+...+\dfrac{1}{395.400}\\ =\dfrac{1}{5}\left(\dfrac{5}{5.10}+\dfrac{5}{10.15}+...+\dfrac{5}{395.400}\right)\\ =\dfrac{1}{5}\left(\dfrac{1}{5}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{15}+...+\dfrac{1}{395}-\dfrac{1}{400}\right)\\ =\dfrac{1}{5}\left(\dfrac{1}{5}-\dfrac{1}{400}\right)\\ =\dfrac{1}{5}.\dfrac{79}{400}\\ =\dfrac{79}{2000}\)

25 tháng 2 2022

Cảm ơn

 

11 tháng 3 2017

C=1/5.10+1/10.15+...+1/95.100

   = 5/5.10+5/10.15+...+5/95.100

   = 1/5-1/10+1/10-1/15+...+1/95-1/100

   = 1/5-1/100

   = 19/100

11 tháng 3 2017

\(C=5\times\left(1+\frac{1}{5}-\frac{1}{10}+\frac{1}{10}-\frac{1}{15}+..+\frac{1}{95}-\frac{1}{100}\right)\)

\(C=5\times\left(1-\frac{1}{100}\right)\)

\(C=5\times\frac{99}{100}\)

\(C=\frac{99}{20}\)

27 tháng 6 2018

\(1-\frac{1}{5.10}-\frac{1}{10.15}-\frac{1}{15.20}-...-\frac{1}{95.100}\)

\(=1-\left(\frac{1}{5.10}+\frac{1}{10.15}+\frac{1}{15.20}+...+\frac{1}{95.100}\right)\)

\(=1-\frac{1}{5}.\left(\frac{1}{5}-\frac{1}{10}+\frac{1}{10}-\frac{1}{15}+\frac{1}{15}-\frac{1}{20}+...+\frac{1}{95}-\frac{1}{100}\right)\)

\(=1-\frac{1}{5}.\left(\frac{1}{5}-\frac{1}{100}\right)\)

\(=1-\frac{1}{5}.\frac{19}{100}\)

\(=1-\frac{19}{500}\)

\(=\frac{481}{500}\)

27 tháng 6 2018

\(1-\frac{1}{5.10}-\frac{1}{10.15}-\frac{1}{15.20}-.....-\frac{1}{95.100}\)

\(=1-\left(\frac{1}{5.10}+\frac{1}{10.15}+\frac{1}{15.20}+...+\frac{1}{95.100}\right)\)

Đặt \(C=\frac{1}{5.10}+\frac{1}{10.15}+\frac{1}{15.20}+....+\frac{1}{95.100}\)

\(\Rightarrow C=\frac{1}{5}.\left(\frac{5}{5.10}+\frac{5}{10.15}+\frac{5}{15.20}+....+\frac{5}{95.100}\right)\)

           \(=\frac{1}{5}.\left(\frac{1}{5}-\frac{1}{10}+\frac{1}{10}-\frac{1}{15}+....+\frac{1}{95}-\frac{1}{100}\right)\)

             \(=\frac{1}{5}.\left(\frac{1}{5}-\frac{1}{100}\right)=\frac{1}{5}.\frac{19}{100}=\frac{19}{500}\)

\(\Rightarrow1-C=1-\frac{19}{500}=\frac{481}{500}\)

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24 tháng 2 2022

\(\dfrac{2}{5.10}+\dfrac{2}{10.15}+...+\dfrac{2}{995.1000}\\ =2\left(\dfrac{1}{5.10}+\dfrac{1}{10.15}+...+\dfrac{1}{995.1000}\right)\\ =\dfrac{2}{5}\left(\dfrac{5}{5.10}+\dfrac{5}{10.15}+...+\dfrac{5}{995.1000}\right)\\ =\dfrac{2}{5}\left(\dfrac{1}{5}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{15}+...+\dfrac{1}{995}-\dfrac{1}{1000}\right)\\ =\dfrac{2}{5}\left(\dfrac{1}{5}-\dfrac{1}{1000}\right)\)

\(=\dfrac{2}{5}.\dfrac{199}{1000}\\ =\dfrac{199}{2500}\)

31 tháng 3 2016

=(5/5-5/10+5/10-5/15+.........+5/2015-5/2020)

=(1/5-1/10+1/10-1/20+.......+1/2015-1/2020)

=1/5-1/2020

=403/2020

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31 tháng 3 2016

\(\frac{5}{5.10}+\frac{5}{10.15}+.............+\frac{5}{2015.2020}\)

\(=\frac{1}{5}-\frac{1}{10}+\frac{1}{10}-\frac{1}{15}+..............+\frac{1}{2015}-\frac{1}{2020}\)

\(=\frac{1}{5}-\frac{1}{2020}\)

\(=\frac{403}{2020}\)

8 tháng 8 2018

\(B=\frac{5}{5\cdot10}+\frac{5}{10\cdot15}+...+\frac{5}{95\cdot100}\)

\(B=\frac{1}{5}-\frac{1}{10}+\frac{1}{10}-\frac{1}{15}+...+\frac{1}{95}-\frac{1}{100}\)

\(B=\frac{1}{5}-\frac{1}{100}\)

\(B=\frac{19}{100}\)

8 tháng 8 2018

\(B=\frac{5}{5.10}+\frac{5}{10.15}+...+\frac{5}{95.100}\)

\(B=\frac{1}{5}-\frac{1}{10}+\frac{1}{10}-\frac{1}{15}+...+\frac{1}{95}-\frac{1}{100}\)

\(B=\frac{1}{5}-\frac{1}{100}\)

\(B=\frac{19}{100}\)

6 tháng 7 2016

\(\frac{2}{5.10}+\frac{2}{10.15}+\frac{2}{15.20}+...+\frac{2}{2015.2020}\)

\(=2.\left(\frac{1}{5.10}+\frac{1}{10.15}+\frac{1}{15.20}+...+\frac{1}{2015.2020}\right)\)

\(=2.\left(\frac{1}{5}-\frac{1}{10}+\frac{1}{10}-\frac{1}{15}+\frac{1}{15}-\frac{1}{20}+...+\frac{1}{2015}-\frac{1}{2020}\right)\)

\(=2.\left(\frac{1}{5}-\frac{1}{2020}\right)\)

\(=2.\frac{403}{2020}=\frac{403}{1010}\)

6 tháng 7 2016

\(\frac{2}{5.10}+\frac{2}{10.15}+\frac{2}{15.20}+...+\frac{2}{2015.2020}\)

=\(\frac{2}{5}\left(\frac{5}{5.10}+\frac{5}{10.15}+\frac{5}{15.20}+...+\frac{5}{2015.2020}\right)\)

=\(\frac{2}{5}\left(\frac{1}{5}-\frac{1}{10}+\frac{1}{10}-\frac{1}{15}+\frac{1}{15}-\frac{1}{20}+...+\frac{1}{2015}-\frac{1}{2016}\right)\)

=\(\frac{2}{5}.\left(\frac{1}{5}-\frac{1}{2020}\right)\)

=\(\frac{2}{5}.\frac{403}{2020}\)

=\(\frac{403}{5005}\)

14 tháng 8 2015

\(a=3\left(\frac{1}{5.10}+\frac{1}{10.15}+...+\frac{1}{45.50}\right)\)

\(a=\frac{3}{5}\left(\frac{5}{5.10}+\frac{5}{10.15}+...+\frac{5}{45.50}\right)\)

\(a=\frac{3}{5}\left(\frac{1}{5}-\frac{1}{10}+\frac{1}{10}-\frac{1}{15}+...+\frac{1}{45}-\frac{1}{50}\right)\)

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

\(a=\frac{3}{5}\cdot\frac{9}{50}\)

\(a=\frac{27}{250}\)

11 tháng 4 2018

Đặt A = \(\frac{10}{5.10}+\frac{10}{10.15}+...+\frac{10}{2015.2020}\)

\(=10\left(\frac{1}{5.10}+\frac{1}{10.15}+...+\frac{1}{2015.2020}\right)\)

\(=\frac{10}{5}\left(\frac{1}{5}-\frac{1}{10}+\frac{1}{10}-\frac{1}{15}+...+\frac{1}{2015}-\frac{1}{2020}\right)\)

\(=2\left(\frac{1}{5}-\frac{1}{2020}\right)\)\(=\frac{2}{5}-\frac{2}{2020}\)

\(=\frac{2}{5}-\frac{1}{1010}\)\(=\frac{404}{1010}-\frac{1}{1010}\)\(=\frac{403}{1010}\)

Vậy giá trị của biểu thức đã cho là 403/1010

11 tháng 4 2018

\(\frac{10}{5.10}+\frac{10}{10.15}+...+\frac{10}{2015.2020}\)

\(=2.\left(\frac{5}{5.10}+\frac{5}{10.15}+...+\frac{5}{2015.2020}\right)\)

\(=2.\left(\frac{1}{5}-\frac{1}{10}+\frac{1}{10}-\frac{1}{15}+...+\frac{1}{2015}-\frac{1}{2020}\right)\)

\(=2.\left(\frac{1}{5}-\frac{1}{2020}\right)\)

\(\frac{2}{5}-\frac{1}{1010}\)

Tính nốt nha