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7 tháng 8 2021

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 = \(\frac{1}{75}\)

7 tháng 8 2021

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

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

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

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

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

Vậy F = \(\frac{1004}{1005}\)

10 tháng 7 2018

a, \(\frac{1}{25.27}+\frac{1}{27.29}+...+\frac{1}{73.75}\)

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

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

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

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

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

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

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

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

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

\(=2\left(\frac{1004}{2010}\right)\)

\(=2\left(\frac{502}{1005}\right)\)

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

Tk hộ =v

10 tháng 7 2018

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

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

\(\frac{4}{2.4}+\frac{4}{4.6}+\frac{4}{6.8}+...+\frac{4}{2008.2010}=2.\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{2008.2010}\right)\)

\(=2.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{2008}-\frac{1}{2010}\right)=2.\left(\frac{1}{2}-\frac{1}{2010}\right)=2.\frac{502}{1005}=\frac{1004}{1005}\)

29 tháng 6 2016

 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 !

29 tháng 6 2016

bạn tách ra từng bài một mình sẽ giúp 

11 tháng 7 2016

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

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

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

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

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

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

11 tháng 7 2016

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 ^_^

6 tháng 6 2015

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

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

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

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

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

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

6 tháng 6 2015

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

18 tháng 5 2018

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

18 tháng 5 2018

mấy câu kia tương tự bạn nhá

3 tháng 8 2016

\(A=\frac{3}{4\cdot7}+\frac{4}{7\cdot11}+\frac{4}{11\cdot15}+...+\frac{4}{100\cdot104}\)

\(A=\frac{7-4}{4\cdot7}+\frac{11-7}{7\cdot11}+\frac{15-11}{11\cdot15}+...+\frac{104-100}{100\cdot104}\)

\(A=\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...+\frac{1}{100}-\frac{1}{104}\)

\(A=\frac{1}{4}-\frac{1}{104}\)

\(A=\frac{25}{104}\)

3 tháng 8 2016

\(B=\frac{1}{25\cdot27}+\frac{1}{27\cdot29}+\frac{1}{29\cdot31}+...+\frac{1}{73\cdot75}\)

\(B\cdot2=\left(\frac{1}{25\cdot27}+\frac{1}{27\cdot29}+\frac{1}{29\cdot31}+...+\frac{1}{73\cdot75}\right)\cdot2\)

\(B\cdot2=\frac{2}{25\cdot27}+\frac{2}{27\cdot29}+\frac{2}{29\cdot31}+...+\frac{2}{73\cdot75}\)

\(B\cdot2=\frac{27-25}{25\cdot27}+\frac{29-27}{27\cdot29}+\frac{31-29}{29\cdot31}+...+\frac{75-73}{73\cdot75}\)

\(B\cdot2=\frac{1}{25}-\frac{1}{27}+\frac{1}{27}-\frac{1}{29}+\frac{1}{29}-\frac{1}{31}+...+\frac{1}{73}-\frac{1}{75}\)

\(B\cdot2=\frac{1}{25}-\frac{1}{75}\)

\(B\cdot2=\frac{2}{75}\)

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

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

8 tháng 5 2015

 

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

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

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

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

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

Câu B và C làm tương tự.

8 tháng 5 2015

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 : \(\frac{49}{100}\)

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

\(=7\left(\dfrac{1}{10}-\dfrac{1}{70}\right)=\dfrac{7.60}{700}=\dfrac{420}{700}=\dfrac{3}{5}\)

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

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