=(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

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

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

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

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

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

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

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

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

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 !

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

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

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

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

Vậy \(P=\dfrac{19}{100}\)

b) Ta có: \(S=\frac{2}{2\cdot4}+\frac{2}{4\cdot6}+\frac{2}{6\cdot8}+...+\frac{2}{298\cdot300}\)

\(=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{298}-\frac{1}{300}\)

\(=\frac{1}{2}-\frac{1}{300}=\frac{149}{300}< \frac{200}{300}=\frac{2}{3}\)

hay \(S< \frac{2}{3}\)(1)

Ta có: \(\frac{1}{101}>\frac{1}{102}>\frac{1}{103}>...>\frac{1}{300}\)

nên \(\left(\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{200}\right)+\left(\frac{1}{201}+\frac{1}{202}+\frac{1}{203}+...+\frac{1}{300}\right)>\left(\frac{1}{200}+\frac{1}{200}+\frac{1}{200}+...+\frac{1}{200}\right)+\left(\frac{1}{300}+\frac{1}{300}+\frac{1}{300}+...+\frac{1}{300}\right)\)(vì mỗi ngoặc trên đều có 100 phân số có tử là 1)

\(\Leftrightarrow\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{300}>\frac{1}{200}\cdot100+\frac{1}{300}\cdot100\)

\(\Leftrightarrow Q>\frac{1}{2}+\frac{1}{3}=\frac{5}{6}\)

mà \(\frac{5}{6}>\frac{4}{6}=\frac{2}{3}\)

nên \(Q>\frac{2}{3}\)

hay \(\frac{2}{3}< Q\)(2)

Từ (1) và (2) suy ra S<Q

\(\dfrac{1}{5.10}+\dfrac{1}{10.15}+...+\dfrac{1}{395.400}\\ =\dfrac{1}{5}\left(\dfrac{5}{5.10}+\dfrac{5}{10.15}+...+\dfrac{5}{395.400}\right)\\ =\dfrac{1}{5}\left(\dfrac{1}{5}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{15}+...+\dfrac{1}{395}-\dfrac{1}{400}\right)\\ =\dfrac{1}{5}\left(\dfrac{1}{5}-\dfrac{1}{400}\right)\\ =\dfrac{1}{5}.\dfrac{79}{400}\\ =\dfrac{79}{2000}\)

=79/400

\(\dfrac{2}{5.10}+\dfrac{2}{10.15}+...+\dfrac{2}{995.1000}\\ =2\left(\dfrac{1}{5.10}+\dfrac{1}{10.15}+...+\dfrac{1}{995.1000}\right)\\ =\dfrac{2}{5}\left(\dfrac{5}{5.10}+\dfrac{5}{10.15}+...+\dfrac{5}{995.1000}\right)\\ =\dfrac{2}{5}\left(\dfrac{1}{5}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{15}+...+\dfrac{1}{995}-\dfrac{1}{1000}\right)\\ =\dfrac{2}{5}\left(\dfrac{1}{5}-\dfrac{1}{1000}\right)\)

\(=\dfrac{2}{5}.\dfrac{199}{1000}\\ =\dfrac{199}{2500}\)

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

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

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

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

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

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

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

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

a) Ta có: \(\dfrac{1}{2022}-\dfrac{5}{2\cdot4}-\dfrac{5}{4\cdot6}-\dfrac{5}{6\cdot8}-...-\dfrac{5}{2020\cdot2022}\)

\(=\dfrac{1}{2022}-5\left(\dfrac{1}{2\cdot4}+\dfrac{1}{4\cdot6}+\dfrac{1}{6\cdot8}+...+\dfrac{1}{2020\cdot2022}\right)\)

\(=\dfrac{1}{2022}-\dfrac{5}{2}\left(\dfrac{2}{2\cdot4}+\dfrac{2}{4\cdot6}+\dfrac{2}{6\cdot8}+...+\dfrac{2}{2020\cdot2022}\right)\)

\(=\dfrac{1}{2022}-\dfrac{5}{2}\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{2020}-\dfrac{1}{2022}\right)\)

\(=\dfrac{1}{2022}-\dfrac{5}{2}\left(\dfrac{1}{2}-\dfrac{1}{2022}\right)\)

\(=\dfrac{1}{2022}-\dfrac{5}{2}\cdot\dfrac{1010}{2022}\)

\(=\dfrac{1}{2022}-\dfrac{2025}{2022}=\dfrac{-1262}{1011}\)

b) Ta có: \(\dfrac{2^2}{1\cdot3}+\dfrac{2^2}{3\cdot5}+...+\dfrac{2^2}{197\cdot199}\)

\(=2\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{197\cdot199}\right)\)

\(=2\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{197}-\dfrac{1}{199}\right)\)

\(=2\left(1-\dfrac{1}{199}\right)\)

\(=2\cdot\dfrac{198}{199}=\dfrac{396}{199}\)