\(=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+............+\frac{1}{2009}-\frac{1}{2011}=\frac{1}{1}-\frac{1}{2011}=\frac{2010}{2011}\)

sai rồi top scorer ạ tử trừ mẫu là 2 mà tử là 1 phải nhân 2 lên tử

Ta có:

\(2\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2007.2009}+\frac{1}{2009.2011}\right)\)

\(=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+....+\frac{2}{2007.2009}+\frac{2}{2009.2011}\)

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

\(=1-\frac{1}{2011}=\frac{2010}{2011}\)

\(\Rightarrow\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{2009.2011}=\frac{2010}{2011}\div2=\frac{1005}{2011}\)

Vậy giá trị của biểu thức là \(\frac{1005}{2011}\)

\(=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2007}-\dfrac{1}{2009}\right)=\dfrac{1}{2}\cdot\dfrac{2008}{2009}=\dfrac{1004}{2009}\)

a)\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}\)

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

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

\(=\frac{1}{2}.\frac{6}{7}\)

\(=\frac{3}{7}\)

b)\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2007.2009}+\frac{1}{2009.2011}\)

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

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

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

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

\(A=\frac{1^2}{1.3}+\frac{2^2}{3.5}+...+\frac{1006^2}{2011.2013}\)

\(\Leftrightarrow4A=\frac{2^2.1^2}{2^2-1}+\frac{2^2.2^2}{4^2-1}+...+\frac{2^2.1006^2}{2012^2-1}\)

\(=1006+\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{2011.2013}\right)\)

\(=1006+\frac{1}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2011}-\frac{1}{2013}\right)\)

\(=1006+\frac{1}{2}\left(1-\frac{1}{2013}\right)=\frac{2026084}{2013}\)

\(\Rightarrow A=\frac{506521}{2013}\)

=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

cậu có kết bạn với tớ không

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

#)Giải :

Gọi A = 1/1.3 + 1/3.5 + 1/5.7 + ... + 1/2007.2009

      A = 1/2 . ( 1/1 - 1/3 + 1/3 - 1/5 + ... + 1/2007 - 1/2009

      A = 1/2 . ( 1/1 - 1/2009 )

      A = 1/2 . 2008/2009

      A = 1004/2009

#)Chúc bn học tốt :D

\(\frac{1}{1.3}\)+\(\frac{1}{3.5}\)+...+\(\frac{1}{2007.2009}\)=(\(\frac{2}{1.3}\)+\(\frac{2}{3.5}\)+\(\frac{2}{2007.2009}\)) (*)

Ta có:\(\frac{2}{n.\left(n+2\right)}\)=\(\frac{\left(n+2\right)-n}{n.\left(n+2\right)}\)=\(\frac{n+2}{n.\left(n+2\right)}\)-\(\frac{n}{n.\left(n+2\right)}\)=\(\frac{1}{n}\)-\(\frac{1}{n+2}\)

Áp dụng vào (*),ta được:\(\frac{2}{1.3}\)+\(\frac{2}{3.5}\)+\(\frac{2}{2007.2009}\)=\(\frac{1}{1}\)-\(\frac{1}{3}\)+\(\frac{1}{3}\)-\(\frac{1}{5}\)+...+\(\frac{1}{2007}\)-\(\frac{1}{2009}\)=1-\(\frac{1}{2009}\)=\(\frac{2008}{2009}\)

Vậy:\(\frac{1}{1.3}\)+\(\frac{1}{3.5}\)+...+\(\frac{1}{2007.2009}\)=\(\frac{2008}{2009}\)

A nhé bn 

vậy kết quả là gì vậy giúp mình nha

\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2007.2009}+\frac{1}{2009.2011}\)

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

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

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

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

Vậy đáp án B là đúng