Vì \(A = \frac{{{{17}^{18}} + 1}}{{{{17}^{19}} + 1}} < 1 \)
\(\Rightarrow A= \frac{{{{17}^{18}} + 1}}{{{{17}^{19}} + 1}} < \frac{{{{17}^{18}} + 1 + 16}}{{{{17}^{19}} + 1 + 16}}\)
\(= \frac{{17\left( {{{17}^{17}} + 1\left. {} \right)} \right.}}{{17\left( {{{17}^{18}} + 1\left. {} \right)} \right.}}\)
\(= \frac{{{{17}^{17}} + 1}}{{{{17}^{18}} + 1}} = B\)
Vậy A < B
