tính hợp lý a,A=6/2.5+6/5.8+6/8.11+….+6/29.32 b,B=15.9^5-4.9^6/9.9^4.2^5-3^9.8

0 bình luận về “tính hợp lý a,A=6/2.5+6/5.8+6/8.11+….+6/29.32 b,B=15.9^5-4.9^6/9.9^4.2^5-3^9.8”

1. Đáp án:

   $\begin{array}{l}
   a)A = \dfrac{6}{{2.5}} + \dfrac{6}{{5.8}} + \dfrac{6}{{8.11}} + … + \dfrac{6}{{29.32}}\\
    = 2.\left( {\dfrac{3}{{2.5}} + \dfrac{3}{{5.8}} + \dfrac{3}{{8.11}} + … + \dfrac{3}{{29.32}}} \right)\\
    = 2.\left( {\dfrac{{5 – 2}}{{2.5}} + \dfrac{{8 – 5}}{{5.8}} + \dfrac{{11 – 8}}{{8.11}} + … + \dfrac{{32 – 29}}{{29.32}}} \right)\\
    = 2.\left( {\dfrac{1}{2} – \dfrac{1}{5} + \dfrac{1}{5} – \dfrac{1}{8} + … + \dfrac{1}{{29}} – \dfrac{1}{{32}}} \right)\\
    = 2.\left( {\dfrac{1}{2} – \dfrac{1}{{32}}} \right)\\
    = 2.\dfrac{{16 – 1}}{{32}}\\
    = \dfrac{{15}}{{16}}\\
   b)B = \dfrac{{{{15.9}^5} – {{4.9}^6}}}{{{{9.9}^4}{{.2}^5} – {3^9}.8}}\\
    = \dfrac{{{{5.3.3}^{2.5}} – {{4.3}^{2.6}}}}{{{3^2}{{.3}^{2.4}}{{.2}^5} – {3^9}.8}}\\
    = \dfrac{{{3^{11}}.\left( {5 – 4.3} \right)}}{{{3^{10}}.32 – {3^9}.8}}\\
    = \dfrac{{{3^{11}}.\left( {5 – 12} \right)}}{{{3^9}.\left( {3.32 – 8} \right)}}\\
    = \dfrac{{{3^2}.\left( { – 7} \right)}}{{88}}\\
    = \dfrac{{ – 63}}{{88}}
   \end{array}$

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