JL
4
K
Khách
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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
MM
4
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 \(C=\frac{1}{5.10}+\frac{1}{10.15}+\frac{1}{15.20}+....+\frac{1}{95.100}\)
\(\Rightarrow C=\frac{1}{5}.\left(\frac{5}{5.10}+\frac{5}{10.15}+\frac{5}{15.20}+....+\frac{5}{95.100}\right)\)
\(=\frac{1}{5}.\left(\frac{1}{5}-\frac{1}{10}+\frac{1}{10}-\frac{1}{15}+....+\frac{1}{95}-\frac{1}{100}\right)\)
\(=\frac{1}{5}.\left(\frac{1}{5}-\frac{1}{100}\right)=\frac{1}{5}.\frac{19}{100}=\frac{19}{500}\)
\(\Rightarrow1-C=1-\frac{19}{500}=\frac{481}{500}\)
Chúc bạn học tốt
DT
3
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
ai tích mk mk vs
31 tháng 3 2016
\(\frac{5}{5.10}+\frac{5}{10.15}+.............+\frac{5}{2015.2020}\)
\(=\frac{1}{5}-\frac{1}{10}+\frac{1}{10}-\frac{1}{15}+..............+\frac{1}{2015}-\frac{1}{2020}\)
\(=\frac{1}{5}-\frac{1}{2020}\)
\(=\frac{403}{2020}\)
27 tháng 6 2017
\(A=\dfrac{3}{5\cdot10}+\dfrac{3}{10\cdot15}+...+\dfrac{3}{95\cdot100}\)
\(=\dfrac{3}{5}\left(\dfrac{5}{5\cdot10}+\dfrac{5}{10\cdot15}+...+\dfrac{5}{95\cdot100}\right)\)
\(=\dfrac{3}{5}\left(\dfrac{1}{5}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{15}+...+\dfrac{1}{95}-\dfrac{1}{100}\right)\)
\(=\dfrac{3}{5}\left(\dfrac{1}{5}-\dfrac{1}{100}\right)\)\(=\dfrac{3}{5}\cdot\dfrac{19}{100}=\dfrac{57}{500}\)
27 tháng 6 2017
\(A=\dfrac{3}{5.10}+\dfrac{3}{10.15}+.....+\dfrac{3}{95.100}\)
\(A=\dfrac{3}{5}\left(\dfrac{1}{5}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{15}+.....+\dfrac{1}{95}-\dfrac{1}{100}\right)\)
\(A=\dfrac{3}{5}\left(\dfrac{1}{5}-\dfrac{1}{100}\right)\)
\(=\dfrac{3}{5}.\dfrac{19}{100}=\dfrac{19}{500}\)
25 tháng 2 2022
\(\dfrac{1}{5.10}+\dfrac{1}{10.15}+...+\dfrac{1}{395.400}\\ =\dfrac{1}{5}\left(\dfrac{5}{5.10}+\dfrac{5}{10.15}+...+\dfrac{5}{395.400}\right)\\ =\dfrac{1}{5}\left(\dfrac{1}{5}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{15}+...+\dfrac{1}{395}-\dfrac{1}{400}\right)\\ =\dfrac{1}{5}\left(\dfrac{1}{5}-\dfrac{1}{400}\right)\\ =\dfrac{1}{5}.\dfrac{79}{400}\\ =\dfrac{79}{2000}\)
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 7 2019
5/5.10 + 5/10.15 + ... + 5/45.50
= 1/5 - 1/10 + 1/10 - 1/15 + ... + 1/45 - 1/50
= 1/5 - 1/50
= 9/50
EC
2
K
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
\(\frac{2}{5.10}+\frac{2}{10.15}+\frac{2}{15.20}+...+\frac{2}{2015.2020}\)
=\(\frac{2}{5}\left(\frac{5}{5.10}+\frac{5}{10.15}+\frac{5}{15.20}+...+\frac{5}{2015.2020}\right)\)
=\(\frac{2}{5}\left(\frac{1}{5}-\frac{1}{10}+\frac{1}{10}-\frac{1}{15}+\frac{1}{15}-\frac{1}{20}+...+\frac{1}{2015}-\frac{1}{2016}\right)\)
=\(\frac{2}{5}.\left(\frac{1}{5}-\frac{1}{2020}\right)\)
=\(\frac{2}{5}.\frac{403}{2020}\)
=\(\frac{403}{5005}\)
\(B=\frac{5}{5\cdot10}+\frac{5}{10\cdot15}+...+\frac{5}{95\cdot100}\)
\(B=\frac{1}{5}-\frac{1}{10}+\frac{1}{10}-\frac{1}{15}+...+\frac{1}{95}-\frac{1}{100}\)
\(B=\frac{1}{5}-\frac{1}{100}\)
\(B=\frac{19}{100}\)
\(B=\frac{5}{5.10}+\frac{5}{10.15}+...+\frac{5}{95.100}\)
\(B=\frac{1}{5}-\frac{1}{10}+\frac{1}{10}-\frac{1}{15}+...+\frac{1}{95}-\frac{1}{100}\)
\(B=\frac{1}{5}-\frac{1}{100}\)
\(B=\frac{19}{100}\)