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