Đáp án:

$\begin{array}{l}
1)\dfrac{{{3^{10}}.11 + {3^{10}}.5}}{{{{16}^4}.100}}\\
 = \dfrac{{{3^{10}}.\left( {11 + 5} \right)}}{{{{16}^4}.100}}\\
 = \dfrac{{{3^{10}}}}{{{{16}^3}.100}}\\
 = \dfrac{{{3^{10}}}}{{{2^{12}}{{.2}^2}{{.5}^2}}}\\
 = \dfrac{{{3^{10}}}}{{{2^{14}}{{.5}^2}}}\\
2)\dfrac{{{2^{10}}.13 + {2^{10}}.65}}{{{2^8}.104}}\\
 = \dfrac{{{2^{10}}.\left( {13 + 65} \right)}}{{{2^8}.8.13}}\\
 = \dfrac{{{2^{10}}.78}}{{{2^8}{{.2}^3}.13}}\\
 = \dfrac{{{2^{10}}.6}}{{{2^{11}}}}\\
 = 3\\
3)\dfrac{{{{72}^3}{{.54}^2}}}{{{{108}^4}}} = \dfrac{{{{\left( {{2^3}{{.3}^2}} \right)}^3}.{{\left( {{{2.3}^3}} \right)}^2}}}{{{{\left( {{2^2}{{.3}^3}} \right)}^4}}}\\
 = \dfrac{{{2^9}{{.3}^6}{{.2}^2}{{.3}^6}}}{{{2^8}{{.3}^{12}}}} = \dfrac{{{2^{11}}{{.3}^{12}}}}{{{2^8}{{.3}^{12}}}} = {2^3} = 8\\
4)\dfrac{{{{21}^2}.14.125}}{{{{35}^5}.6}} = \dfrac{{{3^2}{{.7}^2}{{.2.7.5}^3}}}{{{5^5}{{.7}^5}.2.3}} = \dfrac{3}{{{5^2}{{.7}^2}}} = \dfrac{3}{{1225}}\\
5)\dfrac{{{{5.4}^{15}}{{.9}^9} - {{4.3}^{20}}{{.8}^9}}}{{{{5.2}^9}{{.6}^{19}} - {{7.2}^{29}}{{.27}^6}}}\\
 = \dfrac{{{{5.2}^{30}}{{.3}^{18}} - {2^2}{{.3}^{20}}{{.2}^{27}}}}{{{{5.2}^9}{{.2}^{19}}{{.3}^{19}} - {{7.2}^{29}}{{.3}^{18}}}}\\
 = \dfrac{{{2^{29}}{{.3}^{18}}.\left( {5.2 - {3^2}} \right)}}{{{2^{28}}{{.3}^{18}}.\left( {5.3 - 7.2} \right)}}\\
 = \dfrac{{2.1}}{1} = 2\\
6)\dfrac{{{4^9}.36 + {{64}^4}}}{{{{16}^4}.100}} = \dfrac{{{2^{18}}{{.2}^2}{{.3}^2} + {2^{24}}}}{{{2^{16}}{{.2}^2}{{.5}^2}}}\\
 = \dfrac{{{2^{20}}.\left( {{3^2} + {2^4}} \right)}}{{{2^{18}}.25}}\\
 = \dfrac{{2.25}}{{25}} = 2\\
7)\dfrac{{{{11.3}^{22}}{{.3}^7} - {9^{15}}}}{{{{\left( {{{2.3}^{14}}} \right)}^2}}} = \dfrac{{{{11.3}^{29}} - {3^{30}}}}{{{2^2}{{.3}^{28}}}}\\
 = \dfrac{{{3^{29}}.\left( {11 - 3} \right)}}{{{{4.3}^{28}}}} = \dfrac{{3.8}}{4} = 6
\end{array}$