\(=\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{9.11}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{11}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{11}\right)\)

\(=\frac{1}{2}.\frac{10}{11}\)

\(=\frac{5}{11}\)

\(=\frac{1}{2}\times\left(\frac{2}{1\times3}+\frac{2}{3\times5}+....+\frac{2}{9\times11}\right)\)

\(=\frac{1}{2}\times\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{9}-\frac{1}{11}\right)\)

\(=\frac{1}{2}\times\left(1-\frac{1}{11}\right)\)

\(=\frac{1}{2}\times\frac{10}{11}\)

\(=\frac{5}{11}\)

           p=1/(3*5)+1/(5*7)+.....+1/(2015*2017)+1/(2017*2019)

<=> p = 1/3-1/5+1/5-1/7+1/7-......+1/2017-1/2019

<=> p = 1/3 - 1/2019

<=> p = 224/673

\(P=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{2015.2017}+\frac{1}{2017.2019}\)

\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{2017}-\frac{1}{2019}\right)\)

\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{2019}\right)\)

\(=\frac{112}{673}\)

\(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{2009.2011}\)

\(=\frac{1}{2}.\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{2009.2011}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+....+\frac{1}{2009}-\frac{1}{2011}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{2011}\right)=\frac{1}{2}.\frac{2008}{6033}=\frac{1004}{6033}\)

\(\frac{1}{3x5}+\frac{1}{5x7}+\frac{1}{7x9}+.....+\frac{1}{2009x2011}\)

\(=\frac{1.2}{3.5.2}+\frac{1.2}{5.7.2}+\frac{1.2}{7.9.2}+....+\frac{1.2}{2009.2011.2}\)

\(=\frac{1}{2}.\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+.....+\frac{2}{2009.2011}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{2011}\right)\)

\(=\frac{1}{2}.\frac{2008}{6033}=\frac{2008}{12066}\)

\(S.2=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\)

\(S.2=\frac{1}{1}-\frac{1}{11}\)

\(S.2=\frac{10}{11}\)

\(S=\frac{10}{11}:2\)

\(S=\frac{5}{11}\)

S = 5/11

\(\left(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{19.21}\right)x=\frac{9}{7}\)

\(\left[\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{19}-\frac{1}{21}\right)\right]x=\frac{9}{7}\)

\(\left[\frac{1}{2}\left(\frac{1}{3}-\frac{1}{21}\right)\right]x=\frac{9}{7}\)

\(\left(\frac{1}{2}.\frac{2}{7}\right)x=\frac{9}{7}\)

\(\frac{1}{7}.x=\frac{9}{7}\)

\(x=\frac{9}{7}\div\frac{1}{7}\)

\(x=9\)

Vậy ...

\(S=\frac{1}{1.3}-\frac{1}{2.4}+\frac{1}{3.5}-\frac{1}{4.6}+\frac{1}{5.7}-\frac{1}{6.8}+\frac{1}{7.9}-\frac{1}{8.10}\)

\(\Rightarrow S=\frac{1}{2}\left(1-\frac{1}{3}-\frac{1}{2}+\frac{1}{4}+\frac{1}{3}-\frac{1}{5}-\frac{1}{4}+\frac{1}{6}+\frac{1}{5}-\frac{1}{7}-\frac{1}{6}+\frac{1}{8}+\frac{1}{7}-\frac{1}{9}-\frac{1}{8}+\frac{1}{10}\right)\)

\(\Rightarrow S=\frac{1}{2}\left(1+\frac{1}{10}\right)\)

\(\Rightarrow S=\frac{1}{2}.\frac{11}{10}\)

\(\Rightarrow S=\frac{11}{20}\)

ko bao giờ 323445465

2A = 2/3x5 + 2/5x7 + ... + 2/47x49 + 2/49x51

2A = 1/3 - 1/5 + 1/5 - 1/7 + ... + 1/47 - 1/49 + 1/49 - 1/51

2A = 1/3 - 1/51

2A = 16/51

  A = 16/51 : 2 =8/51

A = 1/2 . ( 1/3 -1/5 + 1/5-1/7 + ...+1/47 - 1/49 + 1/49 - 1/51)

A = 1/2 .(1/3 -1/51)

A=1/2 . 16/51

A= 8/51

\(=\frac{1}{1x3}+\frac{1}{3x5}+\frac{1}{5x7}+...+\frac{1}{2013x2015}\)

\(=\frac{1}{2}x\left(\frac{2}{1x3}+\frac{2}{3x5}+\frac{2}{5x7}+...+\frac{2}{2013x2015}\right)\)

\(=\frac{1}{2}x\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2013}-\frac{1}{2015}\right)\)

\(=\frac{1}{2}x\left(1-\frac{1}{2015}\right)\)

\(=\frac{1}{2}x\frac{2014}{2015}\)

\(=\frac{1007}{2015}\)

1/1-1/3+1/3-1/5+1/5-1/7+....+1/2013-1/2015

=1/1-1/2015

=2014/2015

\(2A=\frac{2}{1x3}+\frac{2}{3x5}+\frac{2}{5x7}+...+\frac{2}{999x1001}\)

\(2A=\frac{3-1}{1x3}+\frac{5-3}{3x5}+\frac{7-5}{5x7}+...+\frac{1001-999}{999x1001}\)

\(2A=\frac{3}{1x3}-\frac{1}{1x3}+\frac{5}{3x5}-\frac{3}{3x5}+\frac{7}{5x7}-\frac{5}{5x7}+...+\frac{1001}{999x1001}-\frac{999}{999x1001}\)

\(2A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{999}-\frac{1}{1001}\)

\(2A=1-\frac{1}{1001}=\frac{1000}{1001}\)=> A = 500/1001

\(\frac{500}{1001}\)!