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xét mẫu ta được
(2012/2+1)+(2011/3+1)+...+(1/2013+1)
=2014/2+2014/3+...+2014/2013
=2014(1/2+1/3+...+1/2013) (1)
mà tử bằng 1/2+1/3+1/4+..+1/2013 (2)
(1),(2)=> A=1/2014
xét mẫu
2012+2012/2+2011/3+...+1/2013
=(1+1+1+…+1) + 2012/2+2011/3+...+1/2013
2012 số hạng
=(1 + 2012/2) + (1 + 2011/3) + ….+ (1+1/2013)
=2014/2 + 2014/3 + …. + 2014/2013
=2014 x (1/2 + 1/3 + … + 1/2013)
=))
(1/2+1/3+1/4+...+1/2013)/(2012+2012/2+2011/3+...+1/2013) =
(1/2+1/3+1/4+...+1/2013)/ 2014 x (1/2+1/3+1/4+...+1/2013) = 1/2014
1/2+1/3+1/4+..+1/2013
2012+2012/2+2011/3+...+1/2013
=1/2+1/3+1/4+...+1/2013
(2012/2+1)+(2011/3+1)+...+(1/2013+1)
=1/2+1/3+1/4+...+1/2013
2014/2+2014/3+...+2014/2013
=1/2+1/3+1/4+...+1/2013
2014(1/2+1/3+...+1/2013)
=1/2014
=1/2014 còn đâu tự làm nhé!!!!!!!!!!!!!^^^^^^^^^^^@@@@@@@########$$$$$$*********
\(Đặt\)
\(A=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{2013}}{2012+\frac{2012}{2}+\frac{2011}{3}+......+\frac{1}{2013}}\)
\(A=\frac{\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+.....+\frac{1}{2013}\right)}{\left(\frac{2012}{2}+1\right)+\left(\frac{2011}{3}+1\right)+......+\left(\frac{1}{2013}+1\right)}\)
\(A=\frac{\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+.....+\frac{1}{2013}\right)}{\frac{2014}{2}+\frac{2014}{3}+\frac{2014}{4}+....+\frac{2014}{2013}}\)
\(A=\frac{\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+.....+\frac{1}{2013}\right)}{2014.\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{2013}\right)}=\frac{1}{2014}\)
Xét mẫu số ta có: \(2012+\frac{2012}{2}+\frac{2011}{3}+...+\frac{1}{2013}\)
=\(2012+\left(\frac{2014-2}{2}+\frac{2014-3}{3}+...+\frac{2014-2013}{2013}\right)\)
= \(2012+\left(\frac{2014}{2}+\frac{2014}{3}+\frac{2014}{4}+...+\frac{2014}{2013}\right)-\left(\frac{2}{2}+\frac{3}{3}+\frac{4}{4}+...+\frac{2013}{2013}\right)\)
= \(2012+2014\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2013}\right)-2012\)
= \(2014\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2013}\right)\)
\(\Rightarrow A=\frac{1}{2014}\)
\(\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2013}}{2012+\frac{2012}{2}+\frac{2011}{3}+...+\frac{1}{2013}}\)
\(=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2013}}{\left(\frac{2012}{2}+1\right)+\left(\frac{2011}{3}+1\right)+....+\left(\frac{1}{2013}+1\right)}\)
\(=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2003}}{\frac{2014}{2}+\frac{2014}{3}+...+\frac{2014}{2013}}\)
\(=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2003}}{2014\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2003}\right)}\)
\(=\frac{1}{2014}\)
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A=\(\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2013}}{2014\cdot\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2013}\right)}=\frac{1}{2014}\)
A=\(\frac{1}{2014}\)