Đáp án:
\[\frac{1}{3}\]
Giải thích các bước giải:
\(\begin{array}{l}
\frac{{{3^{2019}}{{.4}^{20}}}}{{{6^{40}}{{.3}^{1980}}}} = \frac{{{3^{2019}}.{{\left( {{2^2}} \right)}^{20}}}}{{{{\left( {2.3} \right)}^{40}}{{.3}^{1980}}}} = \frac{{{3^{2019}}{{.2}^{40}}}}{{{2^{40}}{{.3}^{40}}{{.3}^{1980}}}}\\
= \frac{{{3^{2019}}{{.2}^{40}}}}{{{2^{40}}{{.3}^{2020}}}} = \frac{1}{3}
\end{array}\)
