Xét \(\frac{n}{1+n^2+n^4}=\frac{n}{n^4+2n^2+1-n^2}=\frac{n}{\left(n^2+1\right)^2-n^2}=\frac{n}{\left(n^2-n+1\right)\left(n^2+n+1\right)}=\frac{1}{2}\left(\frac{1}{n^2-n+1}-\frac{1}{n^2+n+1}\right)\)

\(\Rightarrow A=\frac{1}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{13}+...+\frac{1}{2013^2-2013+1}-\frac{1}{2013^2+2013+1}\right)\)

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

@_@ đề bài yêu cầu gì? So sáng hay tính vậy 

tui chả hiểu đề bài như nào cả

à anh Thiện ơi , muốn làm được thì anh hãy tạo đối số , đó là cách cô em chỉ

Ta có : 1 + 2 + 3 + ... + n = \(\frac{\left(n+1\right)n}{2}\)

Vậy nên : \(A=2013+\frac{2013}{\frac{3.2}{2}}+\frac{2013}{\frac{4.3}{2}}+...+\frac{2013}{\frac{2013.2012}{2}}\)

\(A=2013+\frac{4026}{2.3}+\frac{4016}{3.4}+...+\frac{4026}{2012.2013}\)

\(A=4026\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2012.2013}\right)\)

\(A=4026\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2012}-\frac{1}{2013}\right)\)

\(A=4026\left(1-\frac{1}{2013}\right)=4026.\frac{2012}{2013}=4024.\)

Ai trả lời đi please

A= 1+(\(\dfrac{1}{2014}\)+1)+(\(\dfrac{2}{2013}\)+1)+...+(\(\dfrac{2013}{2}\)+1)

= \(\dfrac{2015}{2015}\)+(\(\dfrac{1}{2014}\)+1)+(\(\dfrac{2}{2013}\)+1)+...+(\(\dfrac{2013}{2}\)+1)

= 2015.(\(\dfrac{1}{2015}\)+\(\dfrac{1}{2014}\)+\(\dfrac{1}{2013}\)+...+\(\dfrac{1}{2}\))=2015.B

\(\Rightarrow\) \(\dfrac{A}{B}\)=2015