\(\Rightarrow2A=\frac{2}{25.27}+\frac{2}{27.29}+\frac{2}{29.31}+...+\frac{2}{73.75}\)

\(\Rightarrow2A=\frac{1}{25}-\frac{1}{27}+\frac{1}{27}-\frac{1}{31}+...+\frac{1}{73}-\frac{1}{75}\)

\(\Rightarrow2A=\frac{1}{25}-\frac{1}{75}=\frac{3}{75}-\frac{1}{75}=\frac{2}{75}\)

\(\Rightarrow A=\frac{2}{75}\div2=\frac{1}{75}\)

B=1/25.27+1/27.29+1/29.31+.......+1/73.75

=1/25+1/75

=3/75+1/75

=4/75.

#)Giải :

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

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

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

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

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

\(B=\frac{2}{75}\div2\)

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

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

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

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

   = \(\frac{1}{2}.\frac{2}{75}\)

   = \(\frac{1}{75}\)

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

1) A=7(\(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-......+\frac{1}{69}-\frac{1}{70}\) )

 A=7 ( \(\frac{1}{10}-\frac{1}{70}\))

A=7x 6/70

A=3/5

`=>2B=(2)/(25.27)+(2)/(27.29)+(2)/(29.31)+....+(2)/(73.75)`

`=>2B=(1)/(25)-(1)/(27)+(1)/(27)-(1)/(29)+(1)/(29)-(1)/(31)+.....+(1)/(73)-(1)/(75)`

`=>2B=(1)/(25)-(1)/(75)`

`=>2B=(3)/(75)-(1)/(75)=(2)/(75)`

`=>B=(2)/(75):2`

`=>B=1/75`

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

\(\Rightarrow2B=\dfrac{2}{25.27}+\dfrac{2}{27.29}+...+\dfrac{2}{73.75}=\dfrac{1}{25}-\dfrac{1}{27}+\dfrac{1}{27}-\dfrac{1}{29}+...+\dfrac{1}{73}-\dfrac{1}{75}\)\(\Rightarrow2B=\dfrac{1}{25}-\dfrac{1}{75}=\dfrac{2}{75}\Rightarrow B=\dfrac{1}{75}\)

Ta có: \(B=\dfrac{1}{25\cdot27}+\dfrac{1}{27\cdot29}+\dfrac{1}{29\cdot31}+...+\dfrac{1}{73\cdot75}\)

\(=\dfrac{1}{2}\left(\dfrac{2}{25\cdot27}+\dfrac{2}{27\cdot29}+\dfrac{2}{29\cdot31}+...+\dfrac{2}{73\cdot75}\right)\)

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

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

\(=\dfrac{1}{2}\cdot\dfrac{2}{75}=\dfrac{1}{75}\)

Giải:

a) C = \(\frac{6}{15.18}+\frac{6}{18.21}+...+\frac{6}{87.90}\)

C = \(\frac{6}{3}.\left(\frac{3}{15.18}+\frac{3}{18.21}+...+\frac{3}{87.90}\right)\)

C = \(\frac{6}{3}.\left(\frac{1}{15}-\frac{1}{18}+\frac{1}{18}-\frac{1}{21}+...+\frac{1}{87}-\frac{1}{90}\right)\)

C = \(\frac{6}{3}.\left(\frac{1}{15}-\frac{1}{90}\right)\)

C = \(\frac{6}{3}.\frac{1}{18}\)

C = \(2.\frac{1}{18}\)

C = \(\frac{1}{9}\)

Vậy C = \(\frac{1}{9}\)

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

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

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

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

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

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

Vậy D = \(\frac{1}{75}\)

c) E = \(\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{38.41}\)

E = \(\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{38}-\frac{1}{41}\)

E = \(\frac{1}{8}-\frac{1}{41}\)

E = \(\frac{33}{328}\)

Vậy E = \(\frac{33}{328}\)

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

A=\(\frac{1}{7}\)x(\(\frac{1}{10}\)+\(\frac{1}{11}\)+\(\frac{1}{11}\)+\(\frac{1}{12}\)+\(\frac{1}{12}\)+\(\frac{1}{13}\)+......+\(\frac{1}{69}\)+\(\frac{1}{70}\))

A=\(\frac{1}{7}\)x(\(\frac{1}{10}\)+\(\frac{1}{70}\))

A=\(\frac{1}{7}\)x(\(\frac{7}{70}\)+\(\frac{1}{70}\))

A=\(\frac{1}{7}\)x\(\frac{4}{35}\)

A=\(\frac{4}{245}\)

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

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

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

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

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

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

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

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

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

\(=\frac{1}{2}.\left(\frac{3}{75}-\frac{1}{75}\right)\)

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

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

Câu dưới đặt 2 ra ngoài rồi làm bình thường.