\(A=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{2011.2013}\right)\)

\(A=\frac{1}{2}.\left(1-\frac{1}{2013}\right)\)

\(A=\frac{1}{2}.\frac{2012}{2013}\)

\(A=\frac{1006}{2013}\)

\(A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2011.2013}\)

\(A=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2011}-\frac{1}{2013}\right)\)

\(A=\frac{1}{2}.\left(1-\frac{1}{2013}\right)\)

\(A=\frac{1}{2}.\frac{2012}{2013}\)

\(A=\frac{1006}{2013}\)

A = 1/1. x 1/3 + 1/3 x 1/5 + 1/5 x 1/7 x.....+1/2011 x 1/2013

A=1/1 x 1/2013(bạn triệt tiêu 1/3 ,1/5 ,1/7,1/2011 nha )

A =1/2013 

\(S=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2011.2013}\)

\(\Rightarrow2S=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{2011.2013}\)

\(\Rightarrow2S=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2011}-\frac{1}{2013}\)

\(\Rightarrow2S=1-\frac{1}{2013}\)

\(\Rightarrow2S=\frac{2012}{2013}\)

\(\Rightarrow S=\frac{2012}{2013}\div2\)

\(\Rightarrow S=\frac{1006}{2013}\)

\(2S=\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+...+\frac{2}{2011\cdot2013}\)

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

\(2S=1-\frac{1}{2013}\)

\(2S=\frac{2012}{2013}\)

\(S=\frac{2012}{2013}\div2=\frac{1006}{2013}\)

                                #Louis

\(S=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2011.2013}\)

   \(=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{2011.2013}\right)\)

   \(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2011}-\frac{1}{2013}\right)\)

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

    \(=\frac{1}{2}.\frac{2012}{2013}\)

   \(=\frac{1006}{2013}\)

Study well ! >_<

Đặt \(A=\dfrac{1^2}{1.3}+\dfrac{2^2}{3.5}+\dfrac{3^3}{5.7}+...+\dfrac{1006^2}{2011.2013}\)

\(\Rightarrow4A=\dfrac{4.1^2}{1.3}+\dfrac{4.2^2}{3.5}+\dfrac{4.3^3}{5.7}+...+\dfrac{4.1006^2}{2011.2013}\)

\(\Rightarrow4A=1006+\dfrac{1}{2}.\left[1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-...+\dfrac{1}{2011}-\dfrac{1}{2013}\right]\)

\(\Rightarrow A=\dfrac{1006+\dfrac{1}{2}.\left(1-\dfrac{1}{2013}\right)}{4}\)

\(\Rightarrow A=251,6249\)

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

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

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

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

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

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

\(\frac{4}{1.3}\)+\(\frac{4}{3.5}\)+\(\frac{4}{5.7}\)+\(\frac{4}{7.9}\)+...+\(\frac{4}{2011.2013}\)

= 1+\(\frac{1}{3}\)-\(\frac{1}{3}\)+\(\frac{1}{5}\)-\(\frac{1}{5}\)+\(\frac{1}{7}\)-\(\frac{1}{7}\)+\(\frac{1}{9}\)+...+\(\frac{1}{2011}\)+\(\frac{1}{2013}\)

=1+       0          +        0        +        0         +...+        0          +         \(\frac{1}{2013}\)

=1+\(\frac{1}{2013}\)

=\(\frac{2014}{2013}\)

k dùm nha

\(\frac{4}{1\cdot3}+\frac{4}{3\cdot5}+\frac{4}{5\cdot7}+...+\frac{4}{2011\cdot2013}\)

\(=2\cdot\left(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{2011\cdot2013}\right)\)

\(=2\cdot\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2011}-\frac{1}{2013}\right)\)

\(=2\cdot\left(1-\frac{1}{2013}\right)\)

\(=2\cdot\frac{2012}{2013}\)

\(=\frac{4024}{2013}\)

Đặt A ta có : \(A=\frac{4}{1.3}+\frac{4}{3.5}+\frac{4}{5.7}+\frac{4}{7.9}+...+\frac{4}{2011.2013}\)

\(2A=4\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}...+\frac{1}{2011.2013}\right)\)

\(2A=4\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2011}-\frac{1}{2013}\right)\)

\(2A=4\left(1-\frac{1}{2013}\right)\)

\(2A=4.\frac{2012}{2013}\)

\(2A=\frac{8048}{2013}\)

\(\Rightarrow A=\frac{4024}{2013}\)

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

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

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

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

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

đề bài sải rồi

\(A=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{2011.2013}\)

\(=\dfrac{1}{2}\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{2011.2013}\right)\)

\(=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2011}-\dfrac{1}{2013}\right)\)

\(=\dfrac{1}{2}\left(1-\dfrac{1}{2013}\right)\)

\(=\dfrac{1}{2}.\dfrac{2012}{2013}\)

\(=\dfrac{1006}{2013}\)

Ý bạn là: \(A=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{2011.2013}\)

Lời giải:\(A=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{2011+2013}\)

\(\Rightarrow A=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2011}-\dfrac{1}{2013}\)

\(\Rightarrow A=1-\dfrac{1}{2013}\)

\(\Rightarrow A=\dfrac{2012}{2013}\)

Chà! Khó quá nhỉ!