\(\dfrac{2}{2.4}+\dfrac{2}{4.6}+\dfrac{2}{6.8}+...+\dfrac{2}{98.100}\\ =\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+...+\dfrac{1}{98}-\dfrac{1}{100}\\ =\dfrac{1}{2}-\dfrac{1}{100}\\ =\dfrac{49}{100}\)

\(\dfrac{2}{2.4}+\dfrac{2}{4.6}+\dfrac{2}{6.8}+....+\dfrac{2}{98.100}\)\(=\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+....+\dfrac{1}{98}-\dfrac{1}{100}\)

                                                   \(=\dfrac{1}{2}-\dfrac{1}{100}=\dfrac{49}{100}\)

\(=\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{50}-\dfrac{1}{52}=\dfrac{1}{2}-\dfrac{1}{52}=\dfrac{25}{52}\)

Trả lời hô mình

Sửa đề: \(\dfrac{4}{2\cdot4}+\dfrac{4}{4\cdot6}+\dfrac{4}{6\cdot8}+...+\dfrac{4}{2018\cdot2020}+\dfrac{4}{2020\cdot2022}\)

Ta có: \(\dfrac{4}{2\cdot4}+\dfrac{4}{4\cdot6}+\dfrac{4}{6\cdot8}+...+\dfrac{4}{2018\cdot2020}+\dfrac{4}{2020\cdot2022}\)

\(=2\left(\dfrac{2}{2\cdot4}+\dfrac{2}{4\cdot6}+\dfrac{2}{6\cdot8}+...+\dfrac{2}{2018\cdot2020}+\dfrac{2}{2020\cdot2022}\right)\)

\(=2\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+...+\dfrac{1}{2018}-\dfrac{1}{2020}+\dfrac{1}{2020}-\dfrac{1}{2022}\right)\)

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

\(=2\cdot\dfrac{505}{1011}\)

\(=\dfrac{1010}{1011}\)

a. \(\dfrac{1}{2.4}\) + \(\dfrac{1}{4.6}\) + \(\dfrac{1}{6.8}\) + ...... + \(\dfrac{1}{20.22}\)

= 1/2 ( 1/2 - 1/4 + 1/4 - 1/6 + 1/6 - 1/8 + ..... + 1/20 - 1/22)

=1/2 ( 1/2 - 1/22)

= 1/2 . 5/11

= 5/22

b. 1+ 2/3 + 2/6 + 2/10 +...+ 2/45

=>1/2.(1+2/3+2/6+....+2/45)=1/2+2/6+2/12+...+2/90

=1/2+2/2.3+2/3.4+...+2/9.10
=2.(1/4+3-2/2.3+4-3/3.4+...+10-9/9.10)

=2. ( 1/4+1/2-1/3+1/3-1/4+.....+1/9-1/10)
= 2.( 1/4-1/10)=2.3/20=3/10

=> vì 1/2.*=3/10

=> *=3/10:1/2=3/5

tick mình nhé

B = 1 + \(\dfrac{2}{3}\) + \(\dfrac{2}{6}\) +\(\dfrac{2}{10}\) + \(\dfrac{2}{15}\)+...+ \(\dfrac{2}{45}\)

B = 1 + 2.(\(\dfrac{1}{3}\) + \(\dfrac{1}{6}\) + \(\dfrac{1}{10}\) + \(\dfrac{1}{15}\)+...+ \(\dfrac{1}{45}\))

B = 1 + \(\dfrac{4}{2}\).(\(\dfrac{1}{3}\) + \(\dfrac{1}{6}\) +\(\dfrac{1}{10}\) + \(\dfrac{1}{15}\) + ...+ \(\dfrac{1}{45}\))

B = 1 + 4.( \(\dfrac{1}{6}\) +\(\dfrac{1}{12}\)+ \(\dfrac{1}{20}\)+ \(\dfrac{1}{30}\)+...+ \(\dfrac{1}{90}\))

B = 1 + 4.(\(\dfrac{1}{2.3}\) + \(\dfrac{1}{3.4}\)+ \(\dfrac{1}{4.5}\) + \(\dfrac{1}{5.6}\)+...+\(\dfrac{1}{9.10}\))

B = 1 + 4 .( \(\dfrac{1}{2}\) - \(\dfrac{1}{3}\) + \(\dfrac{1}{3}\) - \(\dfrac{1}{4}\) + \(\dfrac{1}{4}\) - \(\dfrac{1}{5}\) + \(\dfrac{1}{5}\) - \(\dfrac{1}{6}\)+...+ \(\dfrac{1}{9}\) - \(\dfrac{1}{10}\))

B = 1 + 4.( \(\dfrac{1}{2}\) - \(\dfrac{1}{10}\))

B = 1 + 4. \(\dfrac{2}{5}\)

B = \(\dfrac{13}{5}\)

A = \(\dfrac{1}{2.4}\) + \(\dfrac{1}{4.6}\) + \(\dfrac{1}{6.8}\) +...+ \(\dfrac{1}{20.22}\)

A = \(\dfrac{2}{2}\).( \(\dfrac{1}{2.4}\) + \(\dfrac{1}{4.6}\) + \(\dfrac{1}{6.8}\)+...+ \(\dfrac{1}{20.22}\))

A = \(\dfrac{1}{2}\).( \(\dfrac{2}{2.4}\) + \(\dfrac{2}{4.6}\) + \(\dfrac{2}{6.8}\)+...+ \(\dfrac{1}{20.22}\))

A = \(\dfrac{1}{2}\).( \(\dfrac{1}{2}\) - \(\dfrac{1}{4}\) + \(\dfrac{1}{4}\) - \(\dfrac{1}{6}\) + \(\dfrac{1}{6}\) - \(\dfrac{1}{8}\) +...+ \(\dfrac{1}{20}\) - \(\dfrac{1}{22}\))

A = \(\dfrac{1}{2}\).( \(\dfrac{1}{2}\) - \(\dfrac{1}{22}\))

A = \(\dfrac{1}{2}\) . \(\dfrac{5}{11}\)

A = \(\dfrac{5}{22}\) 

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

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

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

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

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

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

=\(\frac{1}{2}-\frac{1}{8}\)

=\(\frac{3}{8}\)

Ta có  \(A=\dfrac{2}{1.3}-\dfrac{2}{2.4}+\dfrac{2}{3.5}-\dfrac{2}{4.6}+\dfrac{2}{5.7}-\dfrac{2}{6.8}+\dfrac{2}{7.9}-\dfrac{2}{8.10}+\dfrac{2}{9.11}-\dfrac{2}{10.12}\) 

\(\Rightarrow A=\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+\dfrac{2}{9.11}\right)-\left(\dfrac{2}{2.4}+\dfrac{2}{4.6}+\dfrac{2}{6.8}+\dfrac{2}{8.10}+\dfrac{2}{10.12}\right)\) \(\Rightarrow A=\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}\right)-\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{12}\right)\) \(\Rightarrow A=\left(1-\dfrac{1}{11}\right)-\left(\dfrac{1}{2}-\dfrac{1}{12}\right)\) 

\(\Rightarrow A=1-\dfrac{1}{11}-\dfrac{1}{2}+\dfrac{1}{12}\) 

\(\Rightarrow A=\dfrac{9}{22}+\dfrac{1}{12}\) 

\(\Rightarrow A=\dfrac{65}{132}\) 

Mà \(\dfrac{65}{132}< 1\) \(\Rightarrow A< 1\) 

Vậy \(A< 1\)