Đáp án:
$A>B$
Giải thích các bước giải:
$A=\dfrac{7^{2017}+1}{7^{2018}+1}\\
\Rightarrow 7.A=\dfrac{7.(7^{2017}+1)}{7^{2018}+1}\\
=\dfrac{7^{2018}+7}{7^{2018}+1}=\dfrac{7^{2018}+1+6}{7^{2018}+1}\\
=1+\dfrac{6}{7^{2018}+1}\\
B=\dfrac{7^{2018}+1}{7^{2019}+1}\\
\Rightarrow 7B=\dfrac{7.(7^{2018}+1)}{7^{2019}+1}\\
=\dfrac{7^{2019}+7}{7^{2019}+1}=\dfrac{7^{2019}+1+6}{7^{2019}+1}\\
=1+\dfrac{6}{7^{2019}+1}$
Vì $7^{2018}<7^{2019}\Rightarrow 1+7^{2018}<1+7^{2019}$
$\Rightarrow \dfrac{1}{1+7^{2018}}>\dfrac{1}{1+7^{2019}}\\
\Rightarrow \dfrac{6}{1+7^{2018}}>\dfrac{6}{1+7^{2019}}\\
\Rightarrow 1+\dfrac{6}{1+7^{2018}}>1+\dfrac{6}{1+7^{2019}}\\
\Rightarrow A>B$
