Đáp án:
\[\frac{{{{7.4}^5}{{.3}^{11}} + {2^{13}}{{.9}^5}}}{{{6^{10}} + {2^{12}}{{.3}^{10}}}} = \frac{{29}}{5}\]
Giải thích các bước giải:
\(\begin{array}{l}
\frac{{{{7.4}^5}{{.3}^{11}} + {2^{13}}{{.9}^5}}}{{{6^{10}} + {2^{12}}{{.3}^{10}}}}\\
= \frac{{7.{{\left( {{2^2}} \right)}^5}{{.3}^{11}} + {2^{13}}.{{\left( {{3^2}} \right)}^5}}}{{{{\left( {2.3} \right)}^{10}} + {2^{12}}{{.3}^{10}}}}\\
= \frac{{{{7.2}^{10}}{{.3}^{11}} + {2^{13}}{{.3}^{10}}}}{{{2^{10}}{{.3}^{10}} + {2^{12}}{{.3}^{10}}}}\\
= \frac{{{2^{10}}{{.3}^{10}}.\left( {7.3 + {2^3}} \right)}}{{{2^{10}}{{.3}^{10}}.\left( {1 + {2^2}} \right)}}\\
= \frac{{7.3 + {2^3}}}{{1 + {2^2}}}\\
= \frac{{29}}{5}
\end{array}\)
