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 $7^{2019}+1>7^{2018}+1$
$\Rightarrow \dfrac{6}{7^{2019}+1}<\dfrac{6}{7^{2018}+1}\\
\Rightarrow 1+\dfrac{6}{7^{2019}+1}<1+\dfrac{6}{7^{2018}+1}\\
\Rightarrow 7B<7A\\
\Rightarrow B<A$
