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

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

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

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

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

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

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

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

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

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

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

ủng hộ nha

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

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

A=1-1/100=99/100

KO CHẮC NHƯNG NHỚ K NHÉ BN

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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 = \(\frac{2}{5.10}+\frac{2}{10.15}+...+\frac{2}{35.40}\)

   = \(\frac{2}{5}.\left(\frac{5}{5.10}+\frac{5}{10.15}+...+\frac{5}{35.40}\right)\)

   = \(\frac{2}{5}.\left(\frac{1}{5}-\frac{1}{10}+\frac{1}{10}-\frac{1}{15}+...+\frac{1}{35}-\frac{1}{40}\right)\)

   = \(\frac{2}{5}.\left(\frac{1}{5}-\frac{1}{40}\right)\)

   = \(\frac{2}{5}.\frac{7}{40}\)

   = \(\frac{7}{100}\)

E=\(\frac{2}{5\cdot10}+\frac{2}{10\cdot15}+.....+\frac{2}{35\cdot40}\)

E=\(2\cdot\left(\frac{1}{5\cdot10}+\frac{1}{10\cdot15}+...+\frac{1}{35\cdot45}\right)\)

5E=\(2\cdot5\cdot\left(\frac{1}{5\cdot10}+\frac{1}{10\cdot15}+...+\frac{1}{35\cdot45}\right)\)

5E=\(2\cdot\left(\frac{5}{5\cdot10}+\frac{5}{10\cdot15}+...+\frac{5}{35\cdot45}\right)\)

5E=\(2\cdot\left(\frac{1}{5}-\frac{1}{10}+\frac{1}{10}-\frac{1}{15}+...+\frac{1}{35}-\frac{1}{45}\right)\)

5E=\(2\cdot\left(\frac{1}{5}-\frac{1}{40}\right)\)

5E=\(2\cdot\frac{7}{40}\)

5E=\(\frac{7}{20}\)

E=\(\frac{7}{20}:5\)

E=\(\frac{7}{100}\)

\(1-\frac{1}{5\cdot10}-\frac{1}{10\cdot15}-\frac{1}{15\cdot20}-...-\frac{1}{95\cdot100}\)

\(=1-\left(\frac{1}{5\cdot10}+\frac{1}{10\cdot15}+...+\frac{1}{95\cdot100}\right)\)

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

\(=1-\frac{1}{5}\left(\frac{1}{5}-\frac{1}{100}\right)=1-\frac{19}{500}=\frac{481}{500}\)