1/15

giup minh voi, di ma

\(\dfrac{1}{2}\times\left(\dfrac{1}{3}-\dfrac{1}{5}\right)=\dfrac{1}{2}\times\left(\dfrac{5}{15}-\dfrac{3}{15}\right)=\dfrac{1}{2}\times\dfrac{2}{15}=\dfrac{1}{15}\)

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

2A = 2/1.3 +2/3.5 + 2/5.7 + ... + 2/2007.2009 + 2/2009. 2011

2A = 1/1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 + ... + 1/ 2007 - 1/2009 + 1/2009 - 1/2011

Gian uoc het ta co: 2A = 1/1 - 1/2011

2A = 2010/2011

A = 2010/2011 X 1/2

A = 1005/2011

**** mình nha 

S =\(\frac{1}{2}\left(1-\frac{1}{3}\right)+\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}\right)+....\frac{1}{2}\left(\frac{1}{8}-\frac{1}{10}\right)\)

S = 1/2 ( 1 -1/3 +1/2-1/4+......+ 1/8-1/10)

S = 1/2(1+1/2-1/9-1/10)

S= 29/45

Bạn nói cô giáo sửa đề thành: 

Tính tổng S=1/1.3+1/2.4+1/3.5+.....+1/\(7\).9+1/8.10 

chứ không tổng S lẻ lắm, chẳng ai muốn tính cả.