Tính \(\frac{{\frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \ldots + \frac{1}{{2012}}}}{{\frac{{2011}}{1} + \frac{{2010}}{2} + \frac{{2009}}{3} + \ldots + \frac{1}{{2011}}}}\)
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Lời giải:
Báo saiTa có
\(\begin{array}{l} {{\frac{{2011}}{1} + \frac{{2010}}{2} + \frac{{2009}}{3} + \ldots + \frac{1}{{2011}}}}\\ =2011 + \frac{{2010}}{2} + \frac{{2009}}{3} + \ldots + \frac{1}{{2011}} = \left( {1 + \frac{{2010}}{2}} \right) + \left( {1 + \frac{{2009}}{3}} \right) + \ldots + \left( {1 + \frac{1}{{2011}}} \right) + 1 = \frac{{2012}}{2} + \frac{{2012}}{3} + \ldots + \frac{{2012}}{{2011}} + \frac{{2012}}{{2012}}\\ = 2012\left( {\frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \ldots + \frac{1}{{2012}}} \right)\\ \Rightarrow A = \frac{{\frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \ldots + \frac{1}{{2012}}}}{{\frac{{2011}}{1} + \frac{{2010}}{2} + \frac{{2009}}{3} + \ldots + \frac{1}{{2011}}}} = \frac{{\frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \ldots + \frac{1}{{2012}}}}{{2012\left( {\frac{1}{2} + \frac{1}{3} + \ldots + \frac{1}{{2012}}} \right)}} = \frac{1}{{2012}} \end{array}\)