Đáp án:
$\begin{array}{l}
Do:\frac{{{2^{19}} - 3}}{{{2^{21}} - 3}} < 1 \Rightarrow \frac{{{2^{19}} - 3}}{{{2^{21}} - 3}} < \frac{{{2^{19}} - 3 - 1}}{{{2^{21}} - 3 - 1}}\\
Ta\,có:\frac{{{2^{19}} - 3 - 1}}{{{2^{21}} - 3 - 1}} = \frac{{{2^{19}} - 4}}{{{2^{21}} - 4}}\\
= \frac{{{2^{19}} - {2^2}}}{{{2^{21}} - {2^2}}} = \frac{{{2^2}\left( {{2^{17}} - 1} \right)}}{{{2^2}\left( {{2^{19}} - 1} \right)}} = \frac{{{2^{17}} - 1}}{{{2^{19}} - 1}}\\
Do\,{2^{18}} - 1 < {2^{19}} - 1\\
\Rightarrow \frac{{{2^{17}} - 1}}{{{2^{19}} - 1}} < \frac{{{2^{17}} - 1}}{{{2^{18}} - 1}}\\
\Rightarrow \frac{{{2^{19}} - 3}}{{{2^{21}} - 3}} < \frac{{{2^{17}} - 1}}{{{2^{19}} - 1}} < \frac{{{2^{17}} - 1}}{{{2^{18}} - 1}}\\
Vậy\,\frac{{{2^{19}} - 3}}{{{2^{21}} - 3}} < \frac{{{2^{17}} - 1}}{{{2^{18}} - 1}}
\end{array}$
