tính hợp lý a) [5/3(-3/5+8/13-4/10)]:(-15)/11 b) (1-1/2).(1-1/3)….(1-1/2005) c) (1/99+12/949+123/9499).(1/2-1/3-1/6)

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1. Đáp án:

    $a) 
   \dfrac{55}{117}\\
   b)
   \dfrac{1}{2005}\\
   c)
   0$

   Giải thích các bước giải:

    $a) 
   \left [ \dfrac{5}{3}\left ( -\dfrac{3}{5}+\dfrac{8}{13}-\dfrac{4}{10} \right ) \right ]:\dfrac{-15}{11}\\
   =\left [ \dfrac{5}{3}\left ( -\dfrac{3}{5}+\dfrac{8}{13}-\dfrac{2}{5} \right ) \right ].\dfrac{-11}{15}\\
   =\left [ \dfrac{5}{3}\left ( -\dfrac{5}{5}+\dfrac{8}{13} \right ) \right ].\dfrac{-11}{15}\\
   =\left [ \dfrac{5}{3}\left ( -\dfrac{13}{13}+\dfrac{8}{13} \right ) \right ].\dfrac{-11}{15}\\
   =\left [ \dfrac{5}{3}.\dfrac{-5}{13} \right ].\dfrac{-11}{15}\\
   =\dfrac{-25}{39}.\dfrac{-11}{15}\\
   =\dfrac{55}{117}\\
   b)
   \left ( 1-\dfrac{1}{2} \right ).\left ( 1-\dfrac{1}{3} \right )….\left ( 1-\dfrac{1}{2005} \right )\\
   =\dfrac{1}{2}.\dfrac{2}{3}….\dfrac{2004}{2005}\\
   =\dfrac{1}{2005}\\
   c)
   \left ( \dfrac{1}{99}+\dfrac{12}{949}+\dfrac{123}{9499} \right ).\left ( \dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{6} \right )\\
   =\left ( \dfrac{1}{99}+\dfrac{12}{949}+\dfrac{123}{9499} \right ).\left ( \dfrac{3}{6}-\dfrac{2}{6}-\dfrac{1}{6} \right )\\
   =\left ( \dfrac{1}{99}+\dfrac{12}{949}+\dfrac{123}{9499} \right ). \dfrac{0}{6}\\
   =0$

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