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\(=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2007}-\dfrac{1}{2009}\right)=\dfrac{1}{2}\cdot\dfrac{2008}{2009}=\dfrac{1004}{2009}\)
\(\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+\dfrac{1}{5\cdot7}+...+\dfrac{1}{2009\cdot2011}\)
\(=\dfrac{1}{2}\cdot\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{2009\cdot2011}\right)\)
\(=\dfrac{1}{2}\cdot\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2009}-\dfrac{1}{2011}\right)\)
\(=\dfrac{1}{2}\cdot\left(1-\dfrac{1}{2011}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{2010}{2011}=\dfrac{1005}{2011}\)
= \(\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+....+\frac{2}{2009.2011}\right)\)
= \(\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{2009}-\frac{1}{2011}\right)\)
= \(\frac{1}{2}\left(1-\frac{1}{2011}\right)=\frac{1}{2}.\frac{2010}{2011}=\frac{1005}{2011}\)
\(\frac{1}{1\times3}+\frac{1}{3\times5}+\frac{1}{5\times7}+...+\frac{1}{2009\times2011}\)
= \(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2009}-\frac{1}{2011}\)
= \(1-\frac{1}{2011}\)
= \(\frac{2010}{2011}\)
Đặt A=1/1.3+1/3.5+1/5.7+...+1/2009.2011
2A=2/1.3+2/3.5+2/5.7+...+2/2009.2011
2A=1/1-1/3+1/3-1/5+1/5-1/7+...+1/2009-1/2011
2A=1-1/2011=2011/2011-1/2011=2010/2011
A=2010/2011.1/2=1005/2011
\(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{97.99}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{97}-\frac{1}{99}\)
\(=\frac{1}{3}+\left(\frac{1}{5}-\frac{1}{5}\right)+\left(\frac{1}{7}-\frac{1}{7}\right)+...+\left(\frac{1}{97}-\frac{1}{97}\right)-\frac{1}{99}\)
\(=\frac{1}{3}-\frac{1}{99}=\frac{32}{99}\)
~ Hok tốt ~
\(\)
1/1*3 + 1/3*5 + 1/5*7 + ... + 1/2007*2009
= 1/2(2/1*3 + 2/3*5 + 2/5*7 + ... + 2/2007*2009)
= 1/2(1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 + ... + 1/2007 - 1/2009)
= 1/2( 1- 1/2009)
= 1/2 * 2008/2009
= 1009/2009
\(p=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{2009.2011}\)
\(p=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{2009}-\frac{1}{2011}\)
\(p=\frac{1}{3}-\frac{1}{2011}\)
\(p=\frac{2011}{6033}-\frac{3}{6033}\)
\(p=\frac{2008}{6033}\)
1/1.3+1/3.5+1/5.7+...=1/2009.2011
=1/2.(1-1/3+1/3-1/5+1/5-1/7+...+1/2009-1/2011)
=1/2.(1-1/2011)
=1/2.2010/2011
=1005/2011
Gọi tổng trên là A
2A = 2/1.3 + 2/3.5 + 2/5.7 +......+ 2/2009.2011
2A = 1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 +..........+ 1/2009 - 1/2011
2A = 1 - 1/2011
2A = 2010/2011
A = 1005/2011
Vậy................
= \(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{97}-\dfrac{1}{99}\)
\(=\dfrac{1}{3}-\dfrac{1}{99}=\dfrac{32}{99}\)