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

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

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

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

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

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

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

\(\frac{5}{5\times10}+\frac{5}{10\times15}+...+\frac{5}{45\times50}\)

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

\(=\frac{1}{5}-\frac{1}{50}\)

\(=\frac{9}{50}\)

~Study well~

#Thạc_Trân

\(\frac{5}{5\cdot10}+\frac{5}{10\cdot15}+...+\frac{5}{45\cdot50}\)

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

\(=\frac{1}{5}-\frac{1}{50}=\frac{10-1}{50}=\frac{9}{50}\)

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

bai nay de ma hinh nhu day khnog phai cua lop 7 dau ban sai roi

\(\frac{1.2+3.6+5.10+7.14}{2.3+6.9+10.15+14.21}\)

\(=\frac{1.2+3.6+5.10+7.14}{1.2.3+3.6.3+5.10.3+7.14.3}\)

\(=\frac{1.2+3.6+5.10+7.14}{3.\left(1.2+3.6+5.10+7.14\right)}\)

\(=\frac{1}{3}\)