A= \dfrac{7^2017+1}{7^2018+1}$ và B=\dfrac{7^2018 +1}{7^2019+1}$ giải giúp

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1. Đáp án:

    $A>B$

   Giải thích các bước giải:

   $A=\dfrac{7^{2017}+1}{7^{2018}+1}\\
   \Rightarrow 7A=\dfrac{7.(7^{2017}+1)}{7^{2018}+1}
   =\dfrac{7^{2018}+7}{7^{2018}+1}
   =\dfrac{7^{2018}+1+6}{7^{2018}+1}
   =\dfrac{7^{2018}+1}{7^{2018}+1}+\dfrac{6}{7^{2018}+1}\\
   =1+\dfrac{6}{7^{2018}+1}\\$

   Tương tự ta có B

   $B=\dfrac{7^{2018}+1}{7^{2019}+1}\\
   \Rightarrow 7B=\dfrac{7.(7^{2018}+1)}{7^{2019}+1}
   =1+\dfrac{6}{7^{2019}+1}$
   Vì $7^{2019}+1>7^{2018}+1$
   $\Rightarrow \dfrac{6}{7^{2019}+1}<\dfrac{6}{7^{2018}+1}\\
   \Rightarrow 1+\dfrac{6}{7^{2019}+1}<1+\dfrac{6}{7^{2018}+1}\\
   \Rightarrow 7B<7A\\
   \Rightarrow B<A$

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