a) \(\frac{1}{1x3}+\frac{1}{3x5}+\frac{1}{5x7}+...+\frac{1}{2007x2009}\)

\(=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2007}-\frac{1}{2009}\right)\)

\(=\frac{1}{2}.\left(1-\frac{1}{2009}\right)=\frac{1}{2}\cdot\frac{2008}{2009}=\frac{1004}{2009}\)

....

các bài cn lại bn lm tương tự nha

b, \(\dfrac{1}{18}+\dfrac{1}{54}+\dfrac{1}{108}+...+\dfrac{1}{990}\)

3A = \(\dfrac{1}{6}+\dfrac{1}{18}+...+\dfrac{1}{330}\)

3A-A = \(\dfrac{1}{6}-\dfrac{1}{990}\)

2A = 82/495

A =82/495 : 2 

A=41/495

K:2=2/2.4+2/4.6+2/6.8+...+2/2008.2010

     =1/2-1/4+1/4-1/6+1/6-1/8+...+1/2008-1/2010

     =1/2-1/2010

     =502/1005

  K=502/1005.2

    =1004/1005

F=1/3.6+1/6.9+1/9.12+...+1/30.33

3F=3/3.6+3/6.9+3/9.12+...+1/30.33

    =1/3-1/6+1/6-1/9+1/9-1/12+...+1/30-1/33

    =1/3-1-33

    =10/33

  F=10/33:3

    =10/99

Bn sai câu K = 4/2.4 + 4/4.6 + 4/6.8 +....+ 4/2008.2010 

\(=2\left(\frac{1}{2}-\frac{1}{2010}\right)=\frac{2.2004}{2010}=\frac{2004}{1005}\)

\(=\frac{2}{1\cdot2}+\frac{2}{2\cdot3}+\frac{2}{3\cdot4}+...+\frac{2}{1004\cdot1005}\)

\(=2\cdot\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{1004\cdot1005}\right)\)

\(=2\cdot\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{1004}-\frac{1}{1005}\right)\)

\(=2\cdot\left(1-\frac{1}{1005}\right)=2\cdot\frac{1004}{1005}=\frac{2008}{1005}\)

\(\frac{4}{2.4}+\frac{4}{4.6}+\frac{4}{6.8}+...+\frac{4}{2014.2016}=2.\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{2014.2016}\right)\)

\(=2.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{2014}-\frac{1}{2016}\right)\)

\(=2.\left(\frac{1}{2}-\frac{1}{2016}\right)=2.\left(\frac{1008}{2016}-\frac{1}{2016}\right)=2.\frac{1007}{2016}=\frac{1007}{1008}\)

\(\frac{4}{2.4}+\frac{4}{4.6}+\frac{4}{6.8}+\frac{4}{2014.2016}\)

\(=2.\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{2014.2016}\right)\)

\(=2.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{2014}-\frac{1}{2016}\right)\)

\(=2.\left(\frac{1}{2}-\frac{1}{2016}\right)\)

\(=2.\frac{1007}{2016}\)

\(=\frac{1007}{1008}\)

Đặt:A =  \(\frac{4}{2.4}+\frac{4}{4.6}+\frac{4}{6.8}+.....+\frac{4}{2008.2010}\)

=> A = 2.(\(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+.....+\frac{2}{2008.2010}\)

=> A = 2.(\(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+.....+\frac{1}{2008}-\frac{1}{2010}\)

=> A = 2.(\(\frac{1}{2}-\frac{1}{2010}\))

=> A = 2.\(\frac{502}{1005}\)

=> A = \(\frac{1004}{1005}\)

đặt A= \(\frac{4}{2.4}+\frac{4}{4.6}+...+\frac{4}{2008.2010}\) 

=> 1/2.A=\(\frac{2}{2.4}+\frac{2}{4.6}+...+\frac{2}{2008.2010}\) 

= \(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{2008}-\frac{1}{2010}\)

=\(\frac{1}{2}-\frac{1}{2010}\)

=\(\frac{502}{1005}\)

Vậy biểu thức cần tìm có giá trị là \(\frac{502}{1005}\)

\(\frac{4}{2.4}+\frac{4}{4.6}+\frac{4}{6.8}+....+\frac{4}{2014.2016}\)

\(=2.\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+....+\frac{2}{2014.2016}\right)\)

\(=2.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+.....+\frac{1}{2014}-\frac{1}{2016}\right)\)

\(=2.\left(\frac{1}{2}-\frac{1}{2016}\right)\)

\(=2.\frac{1007}{2016}=\frac{1007}{1008}\)

\(A=\frac{4}{2.4}+\frac{4}{4.6}+\frac{4}{6.8}+...+\frac{4}{2014.2016}\)

\(A=\frac{4}{2}\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{2014.2016}\right)\)

\(A=2\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{2014}-\frac{1}{2016}\right)\)

\(A=2\left(\frac{1}{2}-\frac{1}{2016}\right)=2.\frac{1007}{2016}=\frac{1007}{1008}\)

\(B=\dfrac{4}{2.4}+\dfrac{4}{4.6}+...+\dfrac{4}{98.100}\)

\(\Rightarrow5B=\dfrac{20}{2.4}+\dfrac{20}{4.6}+...+\dfrac{20}{98.100}\)

\(\Rightarrow5B=10\left(\dfrac{2}{2.4}+\dfrac{2}{4.6}+...+\dfrac{2}{98.100}\right)\)

\(\Rightarrow5B=10\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{98}-\dfrac{1}{100}\right)\)

\(\Rightarrow5B=10\left(\dfrac{1}{2}-\dfrac{1}{100}\right)\)

\(\Rightarrow5B=10.\dfrac{49}{100}\)

\(\Rightarrow5B=\dfrac{49}{10}\)

Vậy \(5B=\dfrac{49}{10}\)

Ta có: B = \(\dfrac{4}{2.4}\) + \(\dfrac{4}{4.6}\) + \(\dfrac{4}{6.8}\) + ... + \(\dfrac{4}{98.100}\).

=> \(\dfrac{B}{2}\) = \(\dfrac{2}{2.4}\) + \(\dfrac{2}{4.6}\) + \(\dfrac{2}{6.8}\) + ... + \(\dfrac{2}{98.100}\)

=\(\dfrac{1}{2}\) - \(\dfrac{1}{4}\) + \(\dfrac{1}{4}\) - \(\dfrac{1}{6}\) + \(\dfrac{1}{6}\) - \(\dfrac{1}{8}\) + ... + \(\dfrac{1}{98}\) - \(\dfrac{1}{100}\)

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

=> B = \(\dfrac{49}{200}\).

=> 5B = \(\dfrac{49}{200}\) . 5 = \(\dfrac{49}{40}\).

Vậy 5B = \(\dfrac{49}{40}\).

\(A=\frac{4}{2.4}+\frac{4}{4.6}+\frac{4}{6.8}+.....+\frac{4}{2008.2010}\)

\(\Rightarrow A=4\left(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+.....+\frac{1}{2008.2010}\right)\)

\(\Rightarrow A=4\left[\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+....+\frac{1}{2008}-\frac{1}{2010}\right)\right]\)

\(\Rightarrow A=4\left[\frac{1}{2}\left(\frac{1}{2}-\frac{1}{2010}\right)\right]\Rightarrow A=4\left(\frac{1}{2}.\frac{502}{1005}\right)\Rightarrow A=4.\frac{251}{1005}\Rightarrow A=\frac{1004}{1005}\)

\(B=\frac{1}{18}+\frac{1}{54}+\frac{1}{108}+....+\frac{1}{990}\)

\(\Rightarrow B=\frac{1}{3.6}+\frac{1}{6.9}+\frac{1}{9.12}+....+\frac{1}{30.33}\)

\(\Rightarrow B=\frac{1}{3}\left(\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+\frac{1}{9}-\frac{1}{12}+.....+\frac{1}{30}-\frac{1}{33}\right)\)

\(\Rightarrow B=\frac{1}{3}.\left(\frac{1}{3}-\frac{1}{33}\right)\Rightarrow B=\frac{1}{3}.\frac{10}{33}\Rightarrow B=\frac{10}{99}\)