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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
\(\frac{5}{5.10}+\frac{5}{10.15}+.............+\frac{5}{2015.2020}\)
\(=\frac{1}{5}-\frac{1}{10}+\frac{1}{10}-\frac{1}{15}+..............+\frac{1}{2015}-\frac{1}{2020}\)
\(=\frac{1}{5}-\frac{1}{2020}\)
\(=\frac{403}{2020}\)
CT
1
14 tháng 8 2015
\(a=3\left(\frac{1}{5.10}+\frac{1}{10.15}+...+\frac{1}{45.50}\right)\)
\(a=\frac{3}{5}\left(\frac{5}{5.10}+\frac{5}{10.15}+...+\frac{5}{45.50}\right)\)
\(a=\frac{3}{5}\left(\frac{1}{5}-\frac{1}{10}+\frac{1}{10}-\frac{1}{15}+...+\frac{1}{45}-\frac{1}{50}\right)\)
\(a=\frac{3}{5}.\left(\frac{1}{5}-\frac{1}{50}\right)\)
\(a=\frac{3}{5}\cdot\frac{9}{50}\)
\(a=\frac{27}{250}\)
24 tháng 2 2022
\(\dfrac{2}{5.10}+\dfrac{2}{10.15}+...+\dfrac{2}{995.1000}\\ =2\left(\dfrac{1}{5.10}+\dfrac{1}{10.15}+...+\dfrac{1}{995.1000}\right)\\ =\dfrac{2}{5}\left(\dfrac{5}{5.10}+\dfrac{5}{10.15}+...+\dfrac{5}{995.1000}\right)\\ =\dfrac{2}{5}\left(\dfrac{1}{5}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{15}+...+\dfrac{1}{995}-\dfrac{1}{1000}\right)\\ =\dfrac{2}{5}\left(\dfrac{1}{5}-\dfrac{1}{1000}\right)\)
\(=\dfrac{2}{5}.\dfrac{199}{1000}\\ =\dfrac{199}{2500}\)
NT
7
15 tháng 7 2019
E = 2/5.10 + 2/10.15 + ... + 2/35.40
E = 2/5.(1/5 - 1/10 + 1/10 - 1/15 + ... + 1/35 - 1/40)
E = 2/5.(1/5 - 1/40)
E = 2/5.7/40
E = 7/100
XO
15 tháng 7 2019
E = \(\frac{2}{5.10}+\frac{2}{10.15}+...+\frac{2}{35.40}\)
= \(\frac{2}{5}.\left(\frac{5}{5.10}+\frac{5}{10.15}+...+\frac{5}{35.40}\right)\)
= \(\frac{2}{5}.\left(\frac{1}{5}-\frac{1}{10}+\frac{1}{10}-\frac{1}{15}+...+\frac{1}{35}-\frac{1}{40}\right)\)
= \(\frac{2}{5}.\left(\frac{1}{5}-\frac{1}{40}\right)\)
= \(\frac{2}{5}.\frac{7}{40}\)
= \(\frac{7}{100}\)
KT
17 tháng 7 2018
\(A=\frac{1}{1.5}+\frac{1}{5.10}+\frac{1}{10.15}+...+\frac{1}{95.100}\)
\(\Rightarrow\)\(5A=1+\frac{5}{5.10}+\frac{5}{10.15}+...+\frac{5}{95.100}\)
\(=1+\frac{1}{5}-\frac{1}{10}+\frac{1}{10}-\frac{1}{15}+...+\frac{1}{95}-\frac{1}{100}\)
\(=1+\frac{1}{5}-\frac{1}{100}=\frac{119}{100}\)
\(\Rightarrow\)\(A=\frac{119}{500}\)
BO
17 tháng 7 2018
A=1/1.5+1/5.10+....+1/95.100
=(5/1.5+5/5.10+...+5/95.100):5
=(1-1/5+1/5-1/10+...+1/95-1/100):5
=(1-1/100):5
=99/100:5
=99/500
11 tháng 5 2016
mk chỉ làm thôi nhé
Ta tách biểu thức trên thành hai phần A và B
\(A=\frac{12-\frac{12}{7}-\frac{12}{289}-\frac{12}{85}}{4-\frac{4}{7}-\frac{4}{289}-\frac{12}{85}}\)
\(A=\frac{12.\left(1-\frac{1}{7}-\frac{1}{289}-\frac{1}{85}\right)}{4.\left(1-\frac{1}{7}-\frac{1}{289}-\frac{1}{85}\right)}\)
\(A=3\)
\(B=\frac{3+\frac{3}{13}+\frac{3}{169}+\frac{3}{91}}{7+\frac{7}{13}+\frac{7}{169}+\frac{7}{91}}\)
\(B=\frac{3.\left(1+\frac{1}{13}+\frac{1}{169}+\frac{1}{91}\right)}{7.\left(1+\frac{1}{13}+\frac{1}{169}+\frac{1}{91}\right)}\)
\(B=\frac{3}{7}\)
=> A:B=7
=>b=7
AF
1
31 tháng 8 2016
\(\frac{1.2+3.6+5.10+7.14}{2.3+6.9+10.15+14.21}\)
\(=\frac{1.2+3.6+5.10+7.14}{1.2.3+3.6.3+5.10.3+7.14.3}\)
\(=\frac{1.2+3.6+5.10+7.14}{3.\left(1.2+3.6+5.10+7.14\right)}\)
\(=\frac{1}{3}\)
\(\frac{2}{5.10}+\frac{2}{10.15}+\frac{2}{15.20}+...+\frac{2}{2015.2020}\)
\(=2.\left(\frac{1}{5.10}+\frac{1}{10.15}+\frac{1}{15.20}+...+\frac{1}{2015.2020}\right)\)
\(=2.\left(\frac{1}{5}-\frac{1}{10}+\frac{1}{10}-\frac{1}{15}+\frac{1}{15}-\frac{1}{20}+...+\frac{1}{2015}-\frac{1}{2020}\right)\)
\(=2.\left(\frac{1}{5}-\frac{1}{2020}\right)\)
\(=2.\frac{403}{2020}=\frac{403}{1010}\)
\(\frac{2}{5.10}+\frac{2}{10.15}+\frac{2}{15.20}+...+\frac{2}{2015.2020}\)
=\(\frac{2}{5}\left(\frac{5}{5.10}+\frac{5}{10.15}+\frac{5}{15.20}+...+\frac{5}{2015.2020}\right)\)
=\(\frac{2}{5}\left(\frac{1}{5}-\frac{1}{10}+\frac{1}{10}-\frac{1}{15}+\frac{1}{15}-\frac{1}{20}+...+\frac{1}{2015}-\frac{1}{2016}\right)\)
=\(\frac{2}{5}.\left(\frac{1}{5}-\frac{1}{2020}\right)\)
=\(\frac{2}{5}.\frac{403}{2020}\)
=\(\frac{403}{5005}\)