\(P=\dfrac{5}{5.10}+\dfrac{5}{10.15}+...+\dfrac{5}{95.100}\)

\(\Rightarrow P=\dfrac{1}{5}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{15}+...+\dfrac{1}{95}-\dfrac{1}{100}\)

\(\Rightarrow P=\dfrac{1}{5}-\dfrac{1}{100}\)

\(\Rightarrow P=\dfrac{19}{100}\)

Vậy \(P=\dfrac{19}{100}\)

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

=(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

\(\frac{5}{5.10}+\frac{5}{10.15}+.............+\frac{5}{2015.2020}\)

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

\(=\frac{1}{5}-\frac{1}{2020}\)

\(=\frac{403}{2020}\)

5/5.10+5/10.15+5/15.20+...+5/2015.2020

=5(1/5.10+1/10.15+1/15.20+...+1/2015.2020) khoang cach tu 5-10;10-15;...;2015-2020 la 5 suy ra

=5/5(1/5-1/10+1/10-1/15+1/15-1/20+...+1/2015-1/2020)    ;   (-1/5+1/5;-1/10+1/10;-1/15+1/15;-1/20+1/20;... bang 0)

=1(1/5-1/2020)=2015/10100=403/2020

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

B=1+\(\dfrac{1}{5.10}\)+\(\dfrac{1}{10.15}\)+\(\dfrac{1}{15.20}\)+......+\(\dfrac{1}{95.100}\)

5B = 5 +\(\dfrac{5}{5.10}+\dfrac{5}{10.15}+\dfrac{5}{15.20}+........+\dfrac{5}{95.100}\)

5B=5+\(\dfrac{1}{5}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{15}+\dfrac{1}{15}-\dfrac{1}{20}+.........+\dfrac{1}{95}-\dfrac{1}{100}\)

5B=5+\(\dfrac{1}{5}-\dfrac{1}{100}\)

5B=\(\dfrac{519}{100}\)

=>B= \(\dfrac{519}{100}:5=\dfrac{519}{500}\)

A= \(\dfrac{1}{3.7}\) +\(\dfrac{1}{7.11}\)+\(\dfrac{1}{11.15}\)+\(\dfrac{1}{15.19}\)+\(\dfrac{1}{19.23}\)+\(\dfrac{1}{23.27}\)

A= 4.(\(\dfrac{1}{3.7}+\dfrac{1}{7.11}+\dfrac{1}{11.15}+\dfrac{1}{15.19}+\dfrac{1}{19.23}\)+\(\dfrac{1}{23.27}\)

A=4.\(\dfrac{1}{3.7}+4.\dfrac{1}{7.11}+4.\dfrac{1}{11.15}+4.\dfrac{1}{15.19}+4.\dfrac{1}{19.23}+4.\dfrac{1}{23.27}\)

A=\(4.(\dfrac{1}{3}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{15}+\dfrac{1}{15}-\dfrac{1}{19}+\dfrac{1}{19}-\dfrac{1}{23}+\dfrac{1}{23}-\dfrac{1}{25}+\dfrac{1}{25}-\dfrac{1}{29}+\dfrac{1}{29})\)

A= 4 (.\(\dfrac{1}{3}-\dfrac{1}{29}\))

A=\(\dfrac{104}{87}\)