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\(A=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2014}}{\dfrac{2013}{1}+\dfrac{2012}{2}+...+\dfrac{1}{2013}}\)
\(=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2014}}{\left(\dfrac{2012}{2}+1\right)+\left(\dfrac{2011}{3}+1\right)+...+\left(\dfrac{1}{2013}+1\right)+\dfrac{2014}{2014}}\)
\(=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2014}}{2014\left(\dfrac{1}{2}+\dfrac{1}{.3}+...+\dfrac{1}{2014}\right)}\)
\(=\dfrac{1}{2014}\)
\(\left(\frac{2013}{2011}+\frac{2014}{2012}+\frac{2015}{2013}\right).\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{6}\right)=0\)
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
Xét mẫu số của A :
\(\frac{2013}{1}+\frac{2012}{2}+\frac{2011}{3}+...+\frac{1}{2013}\) \(=\left(1+1+1+...+1\right)+\frac{2012}{2}+\frac{2011}{3}+...+\frac{1}{2013}\)
\(=\left(\frac{2012}{2}+1\right)+\left(\frac{2011}{3}+1\right)+...+\left(\frac{1}{2013}+1\right)+1\)\(=\frac{2014}{2}+\frac{2014}{3}+...+\frac{2014}{2013}+\frac{2014}{2014}\)\(=2014.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2013}+\frac{1}{2014}\right)\)
Mẫu số gấp 2014 lần tử số nên A = \(\frac{1}{2014}\)
\(A=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2014}}{\frac{2013}{1}+\frac{2012}{2}+\frac{2011}{3}+...+\frac{1}{2013}}\)
\(A=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2014}}{\left(\frac{2012}{2}+1\right)+\left(\frac{2011}{3}+1\right)+...+\left(\frac{1}{2013}+1\right)+1}\)
\(A=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2014}}{\frac{2014}{2}+\frac{2014}{3}+...+\frac{2014}{2013}+\frac{2014}{2014}}\)
\(A=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2014}}{2014.\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2014}\right)}\)\
\(A=\frac{1}{2014}\)