\(\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+\dfrac{1}{5\cdot7}+...+\dfrac{1}{2009\cdot2011}\)

\(=\dfrac{1}{2}\cdot\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{2009\cdot2011}\right)\)

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

\(=\dfrac{1}{2}\cdot\left(1-\dfrac{1}{2011}\right)\)

\(=\dfrac{1}{2}\cdot\dfrac{2010}{2011}=\dfrac{1005}{2011}\)

= 1/2 . (1/1 - 1/3 + 1/3 - 1/5 +... + 1/2009 - 1/2011)

= 1/2 . (1/1 - 1/2011)

= 1/2 . 2010 / 2011

= 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
 

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................

đặt tổng trên là S nhân S với 2 rồi khử đi ta đc

=1-1/2011

=2010/2011

=(1-1/2011):2

\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2009.2011}\)

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

=\(1-\frac{1}{2011}\)

=\(\frac{2010}{2011}\)

\(\frac{1}{1\cdot3}+\frac{1}{3\cdot5}\frac{1}{5\cdot7}+...+\frac{1}{2009\cdot2011}\)

\(=\frac{1\cdot2}{2\cdot1\cdot3}+\frac{1\cdot2}{2\cdot3\cdot5}+\frac{1\cdot2}{2\cdot5\cdot7}+...+\frac{1\cdot2}{2\cdot2009\cdot2011}\)

\(=\frac{1}{2}\cdot\left(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{2009\cdot2011}\right)\)

\(=\frac{1}{2}\cdot\left(\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+...+\frac{1}{2009\cdot2011}\right)\)

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

Mình không chắc là đúng đâu nha

=1/2(2/1.3+2/3.5+2/5.7+....+2/2009.2011

=1/2(1/1-1/3+1/3-1/5+1/5-1/7+....+1/2009-1/2011

=1/2(1/1-1/2011)

=1/2.2010/2011

=1005/2011

=1/1-1/3+1/3-1/5+1/5-1/7+....+1/2009-2011

=1-1/2011

=2010/1011

\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{2009.2011}=(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{2009.2011}):2\)

\(=\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2009}-\frac{1}{2011}\right):2=\left(1-\frac{1}{2011}\right):2=\frac{1}{2}-\frac{1}{4022}=...\)

\(\frac{1}{2}\cdot\left(\frac{2}{1\cdot3}+\cdot\cdot\cdot+\frac{2}{2009\cdot2011}\right)\)

\(=\frac{1}{2}\cdot\left(1-\frac{1}{3}+\cdot\cdot\cdot+\frac{1}{2009}-\frac{1}{2011}\right)\)

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

\(=\frac{1}{2}\cdot\frac{2010}{2011}\)

\(=\frac{1005}{2011}\)

Giải:

\(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{2009.2011}.\)

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

\(=\dfrac{1}{2}\left[\left(\dfrac{1}{3}-\dfrac{1}{3}\right)+\left(\dfrac{1}{5}-\dfrac{1}{5}\right)+...+\left(\dfrac{1}{2009}-\dfrac{1}{2009}\right)+\left(1-\dfrac{1}{2011}\right)\right].\)

\(=\dfrac{1}{2}\left[0+0+0+...+\left(1-\dfrac{1}{2011}\right)\right].\)

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

\(=\dfrac{1}{2}.\dfrac{2010}{2011}=\dfrac{2010}{4022}=\dfrac{1005}{2011}.\)

~ Học tốt nha bn!!! ~

Bài mik đúng thì nhớ tick mik nha!!!

1\1-1\3+1\3-1\5+1\5-1\7+...+ 1\2009- 1\2011

=1- 1\2011

=2010\2011

dấu \ là 1 trên

Lời giải:

Ta có: \(A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+....+\frac{1}{2009.2011}\)

\(2A=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+....+\frac{2}{2009.2011}\)

\(2A=\frac{3-1}{1.3}+\frac{5-3}{3.5}+\frac{7-5}{5.7}+....+\frac{2011-2009}{2009.2011}\)

\(2A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-....+\frac{1}{2009}-\frac{1}{2011}\)

\(2A=1-\frac{1}{2011}=\frac{2010}{2011}\Rightarrow A=\frac{1005}{2011}\)