Đáp án:
\[A > B\]
Giải thích các bước giải:
Ta có:
\[\begin{array}{l}
A = \frac{{{{10}^{2016}} + 1}}{{{{10}^{2017}} + 1}}\\
\Rightarrow 10A = \frac{{{{10}^{2017}} + 10}}{{{{10}^{2017}} + 1}} = \frac{{\left( {{{10}^{2017}} + 1} \right) + 9}}{{{{10}^{2017}} + 1}} = 1 + \frac{9}{{{{10}^{2017}} + 1}}\\
B = \frac{{{{10}^{2017}} + 1}}{{{{10}^{2018}} + 1}}\\
\Rightarrow 10B = \frac{{{{10}^{2018}} + 10}}{{{{10}^{2018}} + 1}} = \frac{{\left( {{{10}^{2018}} + 1} \right) + 9}}{{{{10}^{2018}} + 1}} = 1 + \frac{9}{{{{10}^{2018}} + 1}}\\
{10^{2017}} + 1 < {10^{2018}} + 1\\
\Rightarrow \frac{9}{{{{10}^{2017}} + 1}} > \frac{9}{{{{10}^{2018}} + 1}}\\
\Rightarrow 10A > 10B\\
\Leftrightarrow A > B
\end{array}\]
