Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\dfrac{{{5^2}{{.6}^{11}}{{.16}^2} + {6^2}{{.12}^6}{{.15}^2}}}{{{{2.6}^{12}}{{.10}^4} - {{81}^2}{{.960}^3}}}\\
= \dfrac{{{5^2}.{{\left( {2.3} \right)}^{11}}.{{\left( {{2^4}} \right)}^2} + {{\left( {2.3} \right)}^2}.{{\left( {{2^2}.3} \right)}^6}.{{\left( {3.5} \right)}^2}}}{{2.{{\left( {2.3} \right)}^{12}}.{{\left( {2.5} \right)}^4} - {{\left( {{3^4}} \right)}^2}.{{\left( {{2^6}.3.5} \right)}^3}}}\\
= \dfrac{{{5^2}{{.2}^{11}}{{.3}^{11}}{{.2}^{4.2}} + {2^2}{{.3}^2}{{.2}^{2.6}}{{.3}^6}{{.3}^2}{{.5}^2}}}{{{{2.2}^{12}}{{.3}^{12}}{{.2}^4}{{.5}^4} - {3^{4.2}}{{.2}^{6.3}}{{.3}^3}{{.5}^3}}}\\
= \dfrac{{{5^2}{{.2}^{11}}{{.3}^{11}}{{.2}^8} + {2^2}{{.3}^2}{{.2}^{12}}{{.3}^6}{{.3}^2}{{.5}^2}}}{{{{2.2}^{12}}{{.3}^{12}}{{.2}^4}{{.5}^4} - {3^8}{{.2}^{18}}{{.3}^3}{{.5}^3}}}\\
= \dfrac{{{5^2}{{.2}^{11 + 8}}{{.3}^{11}} + {2^{2 + 12}}{{.3}^{2 + 6 + 2}}{{.5}^2}}}{{{2^{1 + 12 + 4}}{{.3}^{12}}{{.5}^4} - {2^{18}}{{.3}^{8 + 3}}{{.5}^3}}}\\
= \dfrac{{{2^{19}}{{.3}^{11}}{{.5}^2} + {2^{14}}{{.3}^{10}}{{.5}^2}}}{{{2^{17}}{{.3}^{12}}{{.5}^4} - {2^{18}}{{.3}^{11}}{{.5}^3}}}\\
= \dfrac{{{2^{14}}{{.3}^{10}}{{.5}^2}.\left( {{2^5}.3 + 1} \right)}}{{{2^{17}}{{.3}^{11}}{{.5}^3}.\left( {3.5 - 2} \right)}}\\
= \dfrac{{{2^{14}}{{.3}^{10}}{{.5}^2}.97}}{{{2^{17}}{{.3}^{11}}{{.5}^3}.13}}\\
= \dfrac{{1.1.1.97}}{{{2^3}.3.5.13}}\\
= \dfrac{{97}}{{1560}}
\end{array}\)
