Đáp án:
$\begin{array}{l}
a){99^{20}} = {\left( {{{99}^2}} \right)^{10}} = {\left( {99.99} \right)^{10}}\\
{9999^{10}} = {\left( {99.101} \right)^{10}} > {\left( {99.99} \right)^{10}}\\
\Leftrightarrow {99^{20}} < {9999^{10}}\\
b){11^{1979}} < {11^{1980}} = {11^{3.660}} = {1331^{660}}\\
{37^{1320}} = {37^{2.660}} = {1369^{660}} > {1331^{660}}\\
\Leftrightarrow {11^{1979}} < {37^{1320}}\\
c){10^{10}}\\
{48.50^5} = {2^4}.3.{\left( {{{2.5}^2}} \right)^5} = {3.2^9}{.5^{10}} > {2.2^9}{.5^{10}}\\
\Leftrightarrow {10^{10}} < {48.50^5}\\
d){1990^{10}} + {1990^9}\\
= {1990^9}\left( {1990 + 1} \right)\\
= {1990^9}.1991 < {1991^9}.1991\\
\Leftrightarrow {1990^{10}} + {1990^9} < {1991^{10}}
\end{array}$
