Đáp án:
c)$C=8$
d)$D=6$
Giải thích các bước giải:
$\begin{array}{l}
c)C = \dfrac{{{{72}^3}{{.54}^2}}}{{{{108}^4}}}\\
= \dfrac{{{{\left( {{2^3}{{.3}^2}} \right)}^3}.{{\left( {{{2.3}^3}} \right)}^2}}}{{{{\left( {{2^2}{{.3}^3}} \right)}^4}}}\\
= \dfrac{{{2^9}{{.3}^6}{{.2}^2}{{.3}^6}}}{{{2^8}{{.3}^{12}}}}\\
= \dfrac{{{2^{11}}{{.3}^{12}}}}{{{2^8}{{.3}^{12}}}}\\
= {2^3}\\
= 8\\
d)D = \dfrac{{{{11.3}^{22}}{{.3}^7} - {9^{15}}}}{{{{\left( {{{2.3}^{14}}} \right)}^2}}}\\
= \dfrac{{{{11.3}^{29}} - {{\left( {{3^2}} \right)}^{15}}}}{{{2^2}{{.3}^{28}}}}\\
= \dfrac{{{{11.3}^{29}} - {3^{30}}}}{{{2^2}{{.3}^{28}}}}\\
= \dfrac{{{3^{28}}\left( {11.3 - {3^2}} \right)}}{{{2^2}{{.3}^{28}}}}\\
= \dfrac{{24}}{{{2^2}}}\\
= 6
\end{array}$
