So sánh: A=2/60.63+2/63.66+…+2/117.120+2/2011 và B=5/40.44+5/44.48+5/48.52+…+5/16.80+5/20.10

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1. Đáp án: B>A

    

   Giải thích các bước giải:

   $\begin{array}{l}
   A = \frac{2}{{60.63}} + \frac{2}{{63.66}} + … + \frac{2}{{117.120}} + \frac{2}{{2011}}\\
    = \frac{2}{3}\left( {\frac{3}{{60.63}} + \frac{3}{{63.66}} + … + \frac{3}{{117.120}}} \right) + \frac{2}{{2011}}\\
    = \frac{2}{3}.\left( {\frac{1}{{60}} – \frac{1}{{63}} + \frac{1}{{63}} – \frac{1}{{66}} + … + \frac{1}{{117}} – \frac{1}{{120}}} \right) + \frac{2}{{2011}}\\
    = \frac{2}{3}\left( {\frac{1}{{60}} – \frac{1}{{120}}} \right) + \frac{2}{{2011}}\\
    = \frac{2}{3}.\frac{1}{{120}} + \frac{2}{{2011}}\\
    = \frac{1}{{180}} + \frac{2}{{2011}}\\
   B = \frac{5}{{40.44}} + \frac{5}{{44.48}} + … + \frac{5}{{76.80}} + \frac{5}{{20.10}}\\
    = \frac{5}{4}\left( {\frac{1}{{40}} – \frac{1}{{44}} + \frac{1}{{44}} – \frac{1}{{48}} + … + \frac{1}{{76}} – \frac{1}{{80}}} \right) + \frac{5}{{2010}}\\
    = \frac{5}{4}\left( {\frac{1}{{40}} – \frac{1}{{80}}} \right) + \frac{5}{{2010}}\\
    = \frac{5}{4}.\frac{1}{{80}} + \frac{5}{{2010}}\\
    = \frac{5}{{320}} + \frac{5}{{2010}} > \frac{1}{{180}} + \frac{2}{{2011}}\\
    \Rightarrow B > A
   \end{array}$

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