Đáp án:
\[\frac{{83}}{{18}}\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
A = \frac{{{2^{12}}{{.3}^5} - {4^6}{{.9}^2}}}{{{{\left( {{2^2}.3} \right)}^6} + {8^4}{{.3}^5}}} + \frac{{{5^{10}}{{.7}^3} + {{25}^5}{{.49}^2}}}{{{{\left( {125.7} \right)}^3} + {5^9}{{.14}^3}}}\\
= \frac{{{2^{12}}{{.3}^5} - {{\left( {{2^2}} \right)}^6}.{{\left( {{3^2}} \right)}^2}}}{{{{\left( {{2^2}.3} \right)}^6} + {{\left( {{2^3}} \right)}^4}{{.3}^5}}} + \frac{{{5^{10}}{{.7}^3} + {{\left( {{5^2}} \right)}^5}.{{\left( {{7^2}} \right)}^2}}}{{{{\left( {{5^3}.7} \right)}^3} + {5^9}.{{\left( {2.7} \right)}^3}}}\\
= \frac{{{2^{12}}{{.3}^5} - {2^{12}}{{.3}^4}}}{{{2^{12}}{{.3}^6} + {2^{12}}{{.3}^5}}} + \frac{{{5^{10}}{{.7}^3} + {5^{10}}{{.7}^4}}}{{{5^9}{{.7}^3} + {5^9}{{.2}^3}{{.7}^3}}}\\
= \frac{{{2^{12}}{{.3}^4}\left( {3 - 1} \right)}}{{{2^{12}}{{.3}^5}\left( {3 + 1} \right)}} + \frac{{{5^{10}}{{.7}^3}\left( {1 + 7} \right)}}{{{5^9}{{.7}^3}\left( {1 + {2^3}} \right)}}\\
= \frac{2}{{3.4}} + \frac{{5.8}}{9}\\
= \frac{{83}}{{18}}
\end{array}\)
