Đáp án:
$\begin{array}{l}
\dfrac{{{4^6}{{.27}^2}.120}}{{{8^4}{{.3}^{12}}{{.6}^5}}}\\
= \dfrac{{{{\left( {{2^2}} \right)}^6}.{{\left( {{3^3}} \right)}^2}{{.2}^3}.3.5}}{{{{\left( {{2^3}} \right)}^4}{{.3}^{12}}{{.2}^5}{{.3}^5}}}\\
= \dfrac{{{2^{12}}{{.3}^6}{{.2}^3}.3.5}}{{{2^{12}}{{.3}^{12}}{{.2}^5}{{.3}^5}}}\\
= \dfrac{{{2^{15}}{{.3}^7}.5}}{{{2^{17}}{{.3}^{17}}}}\\
= \dfrac{5}{{{2^2}{{.3}^{10}}}}\\
= \dfrac{5}{{236196}}
\end{array}$
