có 2014/1+2013/2+2012/3+...+2/2013+1/2014=[1+(2013/2)]+[1+(2012/3)]+...+[1+(2/2013)]+[1+(1/2014)]+1

=2015/2+2015/3+...+2015/2014+2015/2015=2015.[1/2+1/3+..+1/2015)

vậy (1/2+1/3+...+1/2015).x=(1/2+1/3+...+1/2015).2015

x=2015

\(\frac{1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2012}}{\frac{2013}{1}+\frac{2014}{2}+\frac{2015}{3}+...+\frac{4024}{2012}-2012}\)

\(=\frac{1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2012}}{\left(\frac{2013}{1}-1\right)+\left(\frac{2014}{2}-1\right)+\left(\frac{2015}{3}-1\right)+...+\left(\frac{4024}{2012}-1\right)}\)

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

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

\(=\frac{1}{2012}\)

Ủng hộ mk nha ^_-

ĐẶT \(A=1+2+2^2+...+2^{2012}\)

\(\Rightarrow2A=2+2^2+2^3+...+2^{2013}\)

\(\Rightarrow2A-A=2^{2013}-1\Rightarrow A=2^{2013}-1\)

vẬY \(\frac{1+2+2^2+...+2^{2012}}{2^{2014}-2}=\frac{2^{2013}-1}{2\left(2^{2013}-1\right)}=\frac{1}{2}\)

Đặt \(S=1+2+2^2+2^3+...+2^{2012}\)

\(\Leftrightarrow2S=2+2^2+2^3+2^4+...+2^{2013}\)

\(\Leftrightarrow2S-S=2+2^2+2^3+2^4+...+2^{2013}-1-2-2^2-2^3-2^{2012}\)

\(\Leftrightarrow S=2^{2013}-1\)

Thay \(S=2^{2013}-1\) vào phân thức \(\frac{1+2+2^2+2^3+...+2^{2012}}{2^{2014}-2}\), ta được:

\(\frac{1+2+2^2+2^3+...+2^{2012}}{2^{2014}-2}=\frac{2^{2013}-1}{2^{2014}-2}=\frac{2^{2013}-1}{2\left(2^{2013}-1\right)}=\frac{1}{2}\)