C=4/2.4+4/4.6+4/6.8+...+4/2008.2010
 C = 2 ( 2 / 2.4 + 2/4.6 + 2/6.8 + ...+2/2008.2010)
 C = 2 ( 1 - 1/4 + 1/4 - 1/6+1/6 - 1/8 +....+1/2008 - 1/2010 )
 C = 2 ( 1 - 1 / 2010 )
 C = 2 . 2009/2010 
 C = 2009 / 1005
Chúc bạn học tốt !

bạn tách ra từng bài một mình sẽ giúp 

Ta có :

\(B=\frac{1}{25.27}+\frac{1}{27.29}+\frac{1}{29.31}+...+\frac{1}{73.75}\)

\(2B=\frac{2}{25.27}+\frac{1}{27.29}+\frac{2}{29.31}+...+\frac{2}{73.75}\)

\(2B=\frac{1}{25}-\frac{1}{27}+\frac{1}{27}-\frac{1}{29}+....+\frac{1}{73}-\frac{1}{75}\)

\(2B=\frac{1}{25}-\frac{1}{75}\)

\(2B=\frac{2}{75}\)

\(\Rightarrow B=\frac{1}{75}\)

Vậy B = \(\frac{1}{75}\)

\(F=\frac{4}{2.3}+\frac{4}{4.6}+\frac{4}{6.8}+...+\frac{4}{2008.2010}\)

\(\Rightarrow F=2\left(\frac{2}{2.4}+\frac{2}{4.6}+...+\frac{2}{2008.2010}\right)\)

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

\(\Rightarrow F=2\left(\frac{1}{2}-\frac{1}{2010}\right)\)

\(\Rightarrow F=2.\frac{502}{1005}=\frac{1004}{1005}\)

Vậy F = \(\frac{1004}{1005}\)

\(\text{Ta có:}\) \(C=\frac{4}{2.4}+\frac{4}{4.6}+\frac{4}{6.8}+...+\frac{4}{2008.2010}\)

\(\Rightarrow\frac{1}{2}C=\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+....+\frac{2}{2008.2010}\)

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

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

\(\Rightarrow\frac{1}{2}C=\frac{502}{1005}\)

\(\Rightarrow C=\frac{502}{1005}:\frac{1}{2}=\frac{1004}{1005}\)

Ta có: \(B=\frac{1}{25.27}+\frac{1}{27.29}+...+\frac{1}{73.75}\)

\(\Rightarrow2B=\frac{2}{25.27}+\frac{2}{27.29}+...+\frac{2}{73.75}\)

\(\Rightarrow2B=\frac{1}{25}-\frac{1}{27}+\frac{1}{27}-\frac{1}{29}+....+\frac{1}{73}-\frac{1}{75}\)

\(\Rightarrow B=\left(\frac{1}{25}-\frac{1}{75}\right):2\)

\(\Rightarrow B=\frac{1}{75}\)

Vậy \(B=\frac{1}{75}\)

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

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

\(\Rightarrow\frac{2}{4}C=\frac{1}{2}-\frac{1}{2010}=\frac{502}{1005}\)

\(\Rightarrow C=\frac{502}{1005}:\frac{2}{4}=\frac{1004}{1005}\)

Vậy \(C=\frac{1004}{1005}\)

Ủng hộ tớ nha m.n ^_^

A = \(\frac{7}{10.11}+\frac{7}{11.12}+\frac{7}{12.13}+...+\frac{7}{69.70}\)

=\(7\left(\frac{1}{10.11}+\frac{1}{11.12}+\frac{1}{12.13}+...+\frac{1}{69.70}\right)\)

=\(7\left(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+\frac{1}{12}-\frac{1}{13}+...+\frac{1}{69}-\frac{1}{70}\right)\)

=\(7\left(\frac{1}{10}-\frac{1}{70}\right)\)

=\(7.\frac{3}{35}\)

=\(\frac{3}{5}\)

B=\(\frac{1}{25.27}+\frac{1}{27.29}+\frac{1}{29.31}+...+\frac{1}{73.75}\)

=\(\frac{1}{2}\left(\frac{2}{25.27}+\frac{2}{27.29}+\frac{2}{29.31}+...+\frac{2}{73.75}\right)\)

=\(\frac{1}{2}\left(\frac{1}{25}-\frac{1}{27}+\frac{1}{27}-\frac{1}{29}+\frac{1}{29}-\frac{1}{31}+...+\frac{1}{73}-\frac{1}{75}\right)\)

=\(\frac{1}{2}\left(\frac{1}{25}-\frac{1}{75}\right)\)

=\(\frac{1}{2}.\frac{2}{75}\)

=\(\frac{1}{75}\)

C : có ở bên dưới rồi, còn A và B thôi

a, 

suy ra A = 7. (1/10.11+1/11.12+1/12.13+.......+1/69.70)

suy ra A = 7. ( 1/10 - 1/11+ 1/11 - 1/12 + 1/12 - 1/13+ ............. + 1/69 - 1/70)

suy ra A = 7. ( 1/ 10 - 1/70) 

suy ra  A= 7. 3/35

suy ra A= 3/5

mấy câu kia tương tự bạn nhá

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 

A=4/2.4+4/4.6+4/6.8+...+4/2008.2010

=2.(2/2.4+2/4.6+2/6.8+...+2/2008.2010)

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

=2.(1/2-1/2010)

=2.502/1005

=1004/1005

Vậy A=1004/1005

100% giải đúng đầu tiên:

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

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

                      \(=2\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+..+\frac{2}{2008.2010}\right)\)

                      \(=2\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{2008}-\frac{1}{2010}\right)\)

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

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

                       \(=1-\frac{1}{1005}=\frac{1004}{1005}\)

\(A=\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)

\(2A=\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{99.100}\)

\(2A=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)

\(2A=\frac{1}{1}-\frac{1}{100}\)

\(2A=\frac{99}{100}\Rightarrow A=\frac{99}{100}:2\Rightarrow A=\frac{99}{200}\)

Câu B và C làm tương tự.

bạn Nhi làm sai rồi

\(\frac{2}{2\cdot3}\) sao có thể bằng \(\frac{1}{2}-\frac{1}{3}\) được

\(\frac{1}{2\cdot3}\) mới bằng \(\frac{1}{2}-\frac{1}{3}\)

kết quả là : \(\frac{49}{100}\)

\(\frac{2}{5}:\frac{1}{3}-\frac{2}{15}:\frac{1}{5}+\frac{3}{5}.\frac{1}{3}\)

\(=\frac{6}{5}+\frac{-2}{3}+\frac{1}{5}\)

\(=\frac{11}{15}\)

~ Hok tốt ~

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

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

\(=4.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{2008}-\frac{1}{2010}\right)\)

\(=4.\left[\frac{1}{2}+\left(\frac{1}{4}-\frac{1}{4}\right)+\left(\frac{1}{6}-\frac{1}{6}\right)+\left(\frac{1}{8}-\frac{1}{8}\right)+...+\left(\frac{1}{2008}-\frac{1}{2008}\right)-\frac{1}{2010}\right]\)

\(=4.\left[\frac{1}{2}-\frac{1}{2010}\right]\)

\(=4.\frac{502}{1005}=\frac{2008}{1005}\)

~ Hok tốt ~