Bài 2:

\(A=\dfrac{x\left(x^3+1\right)}{x^2-x+1}-\dfrac{x\left(x^3-1\right)}{x^2+x+1}\)

\(=x\left(x+1\right)-x\left(x-1\right)\)

=x^2+x-x^2+x

=2x

\(VP=1+\frac{2014}{2}+\frac{2015}{3}+...+\frac{4023}{2011}+\frac{4024}{2012}\)

\(=1-1+\left(\frac{2014}{2}-1\right)+\left(\frac{2015}{3}-1\right)+...+\left(\frac{4023}{2011}-1\right)+\left(\frac{40024}{2012}-1\right)+2012\)

\(=\frac{2012}{2}+\frac{2012}{3}+...+\frac{2012}{2011}+\frac{2012}{2012}+\frac{2012}{1}\)

\(=2012.\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2011}+\frac{1}{2012}\right)\)

\(\Rightarrow2012=503.x\Rightarrow x=\frac{2012}{503}=4\)

\(\frac{1+2+...+n}{n}=\frac{n\left(n+1\right)}{2n}=\frac{n+1}{2}\)

\(\Rightarrow A=1+\frac{1}{2}\left(3+4+...+2012\right)\)

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

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

\(=\frac{1}{2}.\frac{2012.2013}{2}-\frac{1}{2}=503.2013-\frac{1}{2}=...\)

Tổng các số tự nhiên từ 1 đến n là \(\frac{n\left(n+1\right)}{2}\)

Do đó \(A=1+\frac{1}{2}.\frac{2.3}{2}+\frac{1}{3}.\frac{3.4}{2}+....+\frac{1}{2011}.\frac{2011.2012}{2}\)

\(=1+\frac{3}{2}+\frac{4}{2}+\frac{5}{2}+...+\frac{2012}{2}\)

\(=\left(\frac{1}{2}+\frac{2}{2}+\frac{3}{2}+\frac{4}{2}+...+\frac{2012}{2}\right)-\frac{1}{2}\)

\(=\frac{1+2+3+...+2012}{2}-\frac{1}{2}\)

\(=\frac{\frac{2012.2013}{2}}{2}-\frac{1}{2}\)

\(=1012538,5\)

Vậy ....

A=(n+1)(n+2)/4=2012.2013/4=503.2013

A = 0 

B= 3/11

C= -1 

D= -9/10