\(1-\frac{2}{5.10}-\frac{2}{10.15}-.....-\frac{2}{95.100}\)
Tinh tong
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\(1-\frac{1}{5.10}-\frac{1}{10.15}-\frac{1}{15.20}-...-\frac{1}{95.100}\)
\(=1-\left(\frac{1}{5.10}+\frac{1}{10.15}+\frac{1}{15.20}+...+\frac{1}{95.100}\right)\)
\(=1-\frac{1}{5}.\left(\frac{1}{5}-\frac{1}{10}+\frac{1}{10}-\frac{1}{15}+\frac{1}{15}-\frac{1}{20}+...+\frac{1}{95}-\frac{1}{100}\right)\)
\(=1-\frac{1}{5}.\left(\frac{1}{5}-\frac{1}{100}\right)\)
\(=1-\frac{1}{5}.\frac{19}{100}\)
\(=1-\frac{19}{500}\)
\(=\frac{481}{500}\)
\(1-\frac{1}{5.10}-\frac{1}{10.15}-\frac{1}{15.20}-.....-\frac{1}{95.100}\)
\(=1-\left(\frac{1}{5.10}+\frac{1}{10.15}+\frac{1}{15.20}+...+\frac{1}{95.100}\right)\)
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
ta có B = 1- 1/5.10 - 1/10.15 -.......- 1/95 .100
=> 5B = 5 -( 5/5.10+5/10.15 +....+ 5/95.100
= > 5B = 5 - ( 1/5 -1/100 )
=> 5B= 481/100
=> B = 481/500
\(A=\frac{1}{1.5}+\frac{1}{5.10}+\frac{1}{10.15}+...+\frac{1}{95.100}\)
\(\Rightarrow\)\(5A=1+\frac{5}{5.10}+\frac{5}{10.15}+...+\frac{5}{95.100}\)
\(=1+\frac{1}{5}-\frac{1}{10}+\frac{1}{10}-\frac{1}{15}+...+\frac{1}{95}-\frac{1}{100}\)
\(=1+\frac{1}{5}-\frac{1}{100}=\frac{119}{100}\)
\(\Rightarrow\)\(A=\frac{119}{500}\)
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
\(\frac{2}{5.10}+\frac{2}{10.15}+\frac{2}{15.20}+...+\frac{2}{2015.2020}\)
\(=2.\left(\frac{1}{5.10}+\frac{1}{10.15}+\frac{1}{15.20}+...+\frac{1}{2015.2020}\right)\)
\(=2.\left(\frac{1}{5}-\frac{1}{10}+\frac{1}{10}-\frac{1}{15}+\frac{1}{15}-\frac{1}{20}+...+\frac{1}{2015}-\frac{1}{2020}\right)\)
\(=2.\left(\frac{1}{5}-\frac{1}{2020}\right)\)
\(=2.\frac{403}{2020}=\frac{403}{1010}\)
\(\frac{2}{5.10}+\frac{2}{10.15}+\frac{2}{15.20}+...+\frac{2}{2015.2020}\)
=\(\frac{2}{5}\left(\frac{5}{5.10}+\frac{5}{10.15}+\frac{5}{15.20}+...+\frac{5}{2015.2020}\right)\)
=\(\frac{2}{5}\left(\frac{1}{5}-\frac{1}{10}+\frac{1}{10}-\frac{1}{15}+\frac{1}{15}-\frac{1}{20}+...+\frac{1}{2015}-\frac{1}{2016}\right)\)
=\(\frac{2}{5}.\left(\frac{1}{5}-\frac{1}{2020}\right)\)
=\(\frac{2}{5}.\frac{403}{2020}\)
=\(\frac{403}{5005}\)
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
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
E = \(\frac{2}{5.10}+\frac{2}{10.15}+...+\frac{2}{35.40}\)
= \(\frac{2}{5}.\left(\frac{5}{5.10}+\frac{5}{10.15}+...+\frac{5}{35.40}\right)\)
= \(\frac{2}{5}.\left(\frac{1}{5}-\frac{1}{10}+\frac{1}{10}-\frac{1}{15}+...+\frac{1}{35}-\frac{1}{40}\right)\)
= \(\frac{2}{5}.\left(\frac{1}{5}-\frac{1}{40}\right)\)
= \(\frac{2}{5}.\frac{7}{40}\)
= \(\frac{7}{100}\)
\(1-\frac{1}{5\cdot10}-\frac{1}{10\cdot15}-\frac{1}{15\cdot20}-...-\frac{1}{95\cdot100}\)
\(=1-\left(\frac{1}{5\cdot10}+\frac{1}{10\cdot15}+...+\frac{1}{95\cdot100}\right)\)
\(=1-\frac{1}{5}\left(\frac{1}{5}-\frac{1}{10}+\frac{1}{10}-\frac{1}{15}+\frac{1}{15}-...+\frac{1}{95}-\frac{1}{100}\right)\)
\(=1-\frac{1}{5}\left(\frac{1}{5}-\frac{1}{100}\right)=1-\frac{19}{500}=\frac{481}{500}\)
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