Đáp án:
$a){15^{21}} \ge {27^7}{.25^{10}}$
$b){600^{13}} < {127^{23}}$
$c){31^{11}} < {17^{15}}$
$d){3^{39}} < {11^{21}}$
Giải thích các bước giải:
Ta có:
$\begin{array}{l}
5){15^{21}} = {\left( {3.5} \right)^{21}} = {3^{21}}{.5^{21}} \ge {3^{21}}{.5^{20}} = {\left( {{3^3}} \right)^7}.{\left( {{5^2}} \right)^{10}} = {27^7}{.25^{10}}\\
\Rightarrow {15^{21}} \ge {27^7}{.25^{10}}\\
6)\dfrac{{{{600}^{13}}}}{{{{127}^{23}}}} = \dfrac{{{{600}^{13}}}}{{{{127}^{13}}{{.127}^{10}}}} = {\left( {\dfrac{{600}}{{127}}} \right)^{13}}.{\left( {\dfrac{1}{{127}}} \right)^{10}} < {\left( {\dfrac{{635}}{{127}}} \right)^{13}}.{\left( {\dfrac{1}{{127}}} \right)^{10}}\\
\Rightarrow \dfrac{{{{600}^{13}}}}{{{{127}^{23}}}} < {5^{13}}.{\left( {\dfrac{1}{{125}}} \right)^{10}} = {5^{13}}.\dfrac{1}{{{5^{30}}}} = \dfrac{1}{{{5^{17}}}} < 1\\
\Rightarrow {600^{13}} < {127^{23}}\\
7)\dfrac{{{{31}^{11}}}}{{{{17}^{15}}}} < \dfrac{{{{34}^{11}}}}{{{{17}^{15}}}} = \dfrac{1}{{{{17}^4}}}.{\left( {\dfrac{{34}}{{17}}} \right)^{11}} = \dfrac{{{2^{11}}}}{{{{17}^4}}} < \dfrac{{{2^{11}}}}{{{{16}^4}}} = \dfrac{{{2^{11}}}}{{{{\left( {{2^4}} \right)}^4}}} = \dfrac{{{2^{11}}}}{{{2^{16}}}} = \dfrac{1}{{{2^5}}} < 1\\
\Rightarrow \dfrac{{{{31}^{11}}}}{{{{17}^{15}}}} < 1\\
\Rightarrow {31^{11}} < {17^{15}}\\
8)\dfrac{{{3^{39}}}}{{{{11}^{21}}}} < \dfrac{{{3^{39}}}}{{{9^{21}}}} = \dfrac{{{3^{39}}}}{{{{\left( {{3^2}} \right)}^{21}}}} < \dfrac{{{3^{39}}}}{{{3^{42}}}} = \dfrac{1}{{{3^3}}} < 1\\
\Rightarrow \dfrac{{{3^{39}}}}{{{{11}^{21}}}} < 1\\
\Rightarrow {3^{39}} < {11^{21}}
\end{array}$
