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

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

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

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

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

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

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

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

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

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

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

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

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

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) Mình ko ghi lại đề nhé!

= \(\frac{1}{2}\) - ( \(\frac{1}{3.7}\) + \(\frac{1}{7.11}\) + ... + \(\frac{1}{23.27}\) )

= \(\frac{1}{2}\) - \(\frac{1}{4}\) . ( \(\frac{1}{3}\) -  \(\frac{1}{7}\) + \(\frac{1}{7}\) - .... - \(\frac{1}{27}\) )

= \(\frac{1}{2}\) - \(\frac{1}{4}\) . ( \(\frac{1}{3}\) - \(\frac{1}{27}\) )

= \(\frac{1}{2}\) - \(\frac{1}{4}\) . \(\frac{8}{27}\)

= \(\frac{1}{2}\) - \(\frac{2}{27}\) = \(\frac{23}{54}\)

b) ..............................................................................

= \(\frac{1}{5}\)  . ( \(\frac{5}{5.10}\) - \(\frac{5}{10.15}\) - ... - \(\frac{5}{95.100}\) )

= \(\frac{1}{5}\) . ( \(\frac{1}{5}\) - \(\frac{1}{10}\) + \(\frac{1}{10}\) - ... - \(\frac{1}{100}\) )

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

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

= \(\frac{19}{500}\) 

k mình nha! Chúc bạn học tốt và được nhiều k!

\(B=\frac{5}{5\cdot10}+\frac{5}{10\cdot15}+...+\frac{5}{95\cdot100}\)

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

\(B=\frac{1}{5}-\frac{1}{100}\)

\(B=\frac{19}{100}\)

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

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

\(B=\frac{1}{5}-\frac{1}{100}\)

\(B=\frac{19}{100}\)

b=\(\frac{5}{5.10}+\frac{5}{10.15}+...+\frac{5}{95.100}\)

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

b=\(\frac{1}{5}-\frac{1}{100}\)

b=\(\frac{19}{100}\)

 hoc tot nha bn !

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