Bài giải :

$A=10.\frac{{{4}^{6}}{{.9}^{5}}+{{6}^{9}}.120}{{{8}^{4}}{{.3}^{12}}-{{6}^{11}}}=\left( 2.5 \right).\frac{{{\left( {{2}^{2}} \right)}^{6}}.{{\left( {{3}^{2}} \right)}^{5}}+{{\left( 2.3 \right)}^{9}}.\left( {{2}^{3}}.3.5 \right)}{{{\left( {{2}^{3}} \right)}^{4}}{{.3}^{12}}-{{\left( 2.3 \right)}^{11}}}=\frac{{{2.5.2}^{12}}{{.3}^{10}}+{{2.5.2}^{9}}{{.3}^{9}}{{.2}^{3}}.3.5}{{{2}^{12}}{{.3}^{12}}-{{2}^{11}}{{.3}^{11}}}$

$A=\frac{{{2}^{13}}{{.3}^{10}}.5+{{2}^{13}}{{.3}^{10}}{{.5}^{2}}}{{{2}^{12}}{{.3}^{12}}-{{2}^{11}}{{.3}^{11}}}=\frac{{{2}^{13}}{{.3}^{10}}.5.\left( 1+5 \right)}{{{2}^{11}}{{.3}^{11}}.\left( 2.3-1 \right)}=\frac{{{2}^{2}}.5.6}{3.5}=8$