Đặt \(B=\frac{2}{5\cdot9}+\frac{2}{9\cdot13}+\frac{2}{13\cdot17}+....+\frac{2}{97\cdot101}\)

\(\Rightarrow2B=\frac{4}{5\cdot9}+\frac{4}{9\cdot13}+\frac{4}{13\cdot17}+....+\frac{4}{97\cdot101}\)

\(\Leftrightarrow2B=\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}+....+\frac{1}{97}-\frac{1}{101}\)

\(\Leftrightarrow2B=\frac{1}{5}-\frac{1}{101}=\frac{96}{505}\)

\(\Leftrightarrow B=\frac{96}{505}:2\)

\(x+\frac{2}{5.9}+\frac{2}{9.13}+\frac{2}{13.17}+...+\frac{2}{41.45}=\frac{-37}{45}\)

\(\Leftrightarrow x+\left[\frac{2}{4}\cdot\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}+...+\frac{1}{41}-\frac{1}{45}\right)\right]=\frac{-37}{45}\)

\(\Leftrightarrow x+\left[\frac{1}{2}\cdot\left(\frac{1}{5}-\frac{1}{45}\right)\right]=\frac{-37}{45}\)

\(\Leftrightarrow x+\left[\frac{1}{2}\cdot\frac{8}{45}\right]=\frac{-37}{45}\)

\(\Leftrightarrow x+\frac{4}{45}=\frac{-37}{45}\)

\(\Leftrightarrow x=-\frac{41}{45}\)

\(4S=4.\left(\frac{1}{5.9}+\frac{1}{9.13}+...+\frac{1}{21.25}\right)\)

=\(\frac{4}{5.9}+\frac{4}{9.13}+....+\frac{4}{21.25}_{ }\)

=\(\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+....+\frac{1}{21}-\frac{1}{23}\)

=\(\frac{1}{5}-\frac{1}{25}=\frac{5}{25}-\frac{1}{25}=\frac{4}{25}\)

=> \(S=\frac{4}{25}:4=\frac{4}{25}.\frac{1}{4}=\frac{1}{25}\)

\(S=\frac{1}{5\times9}+\frac{1}{9\times13}+...+\frac{1}{21\times25}\)

\(S\times4=\frac{4}{5\times9}=\frac{4}{9\times13}+...+\frac{4}{21\times25}\)

\(S\times4=\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{21}-\frac{1}{25}\)

\(S\times4=\frac{1}{5}-\frac{1}{25}\)

\(S\times4=\frac{4}{25}\)

\(S=\frac{1}{25}\)

ta có S=  1/5.9+1/9.13+1/13.17+1/17.21+1/21.25

<=>4S=4.(1/5.9+1/9.13+1/13.17+1/17.21+1/21.25)

<=>4S=4/5.9+4/9.13+4/13.17+4/17.21+4/21.25

<=>4S=1/5-1/9+1/9-1/13+1/13-1/17+1/21-1/25

<=>4S=1/5-1/25

<=>4S=4/25

<=>S=4/25:4

<=>S=1/25

vậy S=1/25

a,\(\frac{7}{x}+\frac{4}{5.9}+\frac{4}{9.13}+\frac{4}{13.17}+...+\frac{4}{41.45}=\frac{29}{45}\)

\(\frac{7}{x}+\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}+...+\frac{1}{41}-\frac{1}{45}\right)=\frac{29}{45}\)

\(\frac{7}{x}+\left(\frac{1}{5}-\frac{1}{45}\right)=\frac{29}{45}\)

\(\frac{7}{x}+\frac{8}{45}=\frac{29}{45}\)

\(\frac{7}{x}=\frac{29}{45}-\frac{8}{45}=\frac{21}{45}\)

\(\frac{7}{x}=\frac{7}{15}\)

=> x = 15

b,\(\frac{x}{2008}-\left(\frac{2}{20}+\frac{2}{30}+\frac{2}{42}+....+\frac{2}{240}\right)=\frac{5}{8}\)

\(\frac{x}{2008}-2\left(\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+....+\frac{1}{15.16}\right)=\frac{5}{8}\)

\(\frac{x}{2008}-2\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+....+\frac{1}{15}-\frac{1}{16}\right)=\frac{5}{8}\)

\(\frac{x}{2008}-2\left(\frac{1}{4}-\frac{1}{16}\right)=\frac{5}{8}\)

\(\frac{x}{2008}-2.\frac{3}{16}=\frac{5}{8}\)

\(\frac{x}{2008}-\frac{3}{8}=\frac{5}{8}\)

\(\frac{x}{2008}=\frac{5}{8}+\frac{3}{8}=1=\frac{2008}{2008}\)

=> x = 2008

a) A = 4/5.9 + 4/9.13 + 4/13.17 + ... + 4/41/45

A = 1/5 - 1/9 + 1/9 - 1/13 + 1/13 - 1/17 + ... + 1/41 - 1/45

A = 1/5 - 1/45

A = 8/45

b) B = ( 1 - 1/2 ) . ( 1 - 1/3 ) . ( 1 - 1/4 ) . ..... . ( 1 - 1/100 )

B = 1/2 . 2/3 . 3/4 . .... . 99/100

B = \(\frac{1.2.3.......99}{2.3.4......100}\)

B = 1/100

B = \(\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{3}\right)+...+\left(1-\frac{1}{100}\right)\)

B = \(\frac{1}{2}.\frac{2}{3}.....\frac{99}{100}\)

B = \(\frac{1}{100}\)

\(\dfrac{1}{5.9}+\dfrac{1}{9.13}+...+\dfrac{1}{21.25}\\ =\dfrac{4\cdot\dfrac{1}{4}}{5.9}+\dfrac{4\cdot\dfrac{1}{4}}{9.13}+...+\dfrac{4\cdot\dfrac{1}{4}}{21.25}\\ =\dfrac{1}{4}\left(\dfrac{4}{5.9}+\dfrac{4}{9.13}+...+\dfrac{4}{21.25}\right)\\ =\dfrac{1}{4}\cdot\left(\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{21}-\dfrac{1}{25}\right)\\ =\dfrac{1}{4}\left(\dfrac{1}{5}-\dfrac{1}{25}\right)=\dfrac{1}{4}\left(\dfrac{5}{25}-\dfrac{1}{25}\right)\\ =\dfrac{1}{4}\cdot\dfrac{4}{25}=\dfrac{1}{25}\)

`1/(5.9) + 1/(9.13) + ...+ 1/(21.25)`

`= 1/5 - 1/9 + 1/9 - 1/13 + ... + 1/21 - 1/25`

`= 1/5 - 1/25`

`= 4/25`

Ta có : \(\left(x-1\right)^2+\dfrac{1}{5.9}+\dfrac{1}{9.13}+...+\dfrac{1}{41.45}=\dfrac{49}{900}\)

\(\Leftrightarrow\left(x-1\right)^2+\dfrac{1}{4}.\left(\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{41}-\dfrac{1}{45}\right)=\dfrac{49}{900}\)

\(\Leftrightarrow\left(x-1\right)^2+\dfrac{1}{4}\left(\dfrac{1}{5}-\dfrac{1}{45}\right)=\dfrac{49}{900}\)

\(\Leftrightarrow\left(x-1\right)^2=\dfrac{1}{100}\)  \(\Leftrightarrow\left[{}\begin{matrix}x-1=\dfrac{1}{10}\\x-1=-\dfrac{1}{10}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{11}{10}\\x=\dfrac{9}{10}\end{matrix}\right.\)

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