Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
1,\\
\dfrac{{{{11.3}^{22}}{{.3}^7} - {9^{15}}}}{{{{\left( {{{2.3}^{14}}} \right)}^2}}}\\
= \dfrac{{{{11.3}^{22 + 7}} - {{\left( {{3^2}} \right)}^{15}}}}{{{2^2}.{{\left( {{3^{14}}} \right)}^2}}}\\
= \dfrac{{{{11.3}^{29}} - {3^{2.15}}}}{{{2^2}{{.3}^{14.2}}}}\\
= \dfrac{{{{11.3}^{29}} - {3^{30}}}}{{{2^2}{{.3}^{28}}}}\\
= \dfrac{{{{11.3}^{29}} - {{3.3}^{29}}}}{{{{4.3}^{28}}}}\\
= \dfrac{{{3^{29}}.\left( {11 - 3} \right)}}{{{{4.3}^{28}}}}\\
= \dfrac{{{3^{29}}.8}}{{{{4.3}^{28}}}} = \dfrac{{{{3.3}^{28}}.2.4}}{{{{4.3}^{28}}}} = 3.2 = 6\\
2,\\
{5^{299}} < {5^{300}} = {5^{2.150}} = {\left( {{5^2}} \right)^{150}} = {25^{150}} < {27^{150}} = {\left( {{3^3}} \right)^{150}} = {3^{3.150}} = {3^{450}} < {3^{453}}
\end{array}\)
