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

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

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

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

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

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

= 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

= 1 - 1/2011

= 2010 / 2011

   1/2 ( 2/1.3 + 2/3.5 +...+ 2 /2009.2011)

= 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

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

2S=\(\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{2007.2009}\)

=\(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-...+\dfrac{1}{2007}-\dfrac{1}{2009}\)

= 1- \(\dfrac{1}{2009}\)

= \(\dfrac{2008}{2009}\)

=> S=\(\dfrac{1004}{2009}\)

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

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

I: Để 3n+4/n+2 là số nguyên thì \(3n+4⋮n+2\)

\(\Leftrightarrow3n+6-2⋮n+2\)

\(\Leftrightarrow n+2\in\left\{1;-1;2;-2\right\}\)

hay \(n\in\left\{-1;-3;0;-4\right\}\)

II: \(D=2\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2007}-\dfrac{1}{2009}\right)\)

\(D=2\cdot\left(1-\dfrac{1}{2009}\right)=2\cdot\dfrac{2008}{2009}=\dfrac{4016}{2009}\)