Đáp án:
b. B=70
Giải thích các bước giải:
\(\begin{array}{l}
a.A = \dfrac{{{{\left( {{2^2}} \right)}^6}.{{\left( {{3^2}} \right)}^5} + {2^9}{{.3}^9}{{.2}^3}.3.5}}{{{{\left( {{2^2}} \right)}^4}{{.3}^{12}} - {2^{11}}{{.3}^{11}}}}\\
= \dfrac{{{2^{12}}{{.3}^{10}} + {2^{12}}{{.3}^{10}}.5}}{{{2^8}{{.3}^{12}} - {2^{11}}{{.3}^{11}}}}\\
= \dfrac{{{2^{12}}{{.3}^{10}}\left( {1 + 5} \right)}}{{{2^8}{{.3}^{11}}\left( {3 - {2^3}} \right)}}\\
= \dfrac{{{2^4}.6}}{{3.\left( { - 5} \right)}} = \dfrac{{{2^5}}}{{ - 5}} = - \dfrac{{32}}{5}\\
b.B = \dfrac{{{2^{2.2}}{{.5}^{2.2}} + {2^5}{{.5}^3}}}{{{2^3}{{.5}^2}}}\\
= \dfrac{{{2^4}{{.5}^3}\left( {5 + 2} \right)}}{{{2^3}{{.5}^2}}}\\
= \dfrac{{2.5.7}}{1} = 70
\end{array}\)
