Đáp án:

$\begin{array}{l}
d)D = \dfrac{{x + 15}}{{\sqrt x  + 1}} = \dfrac{{x - 1 + 16}}{{\sqrt x  + 1}}\\
 = \sqrt x  - 1 + \dfrac{{16}}{{\sqrt x  + 1}}\\
 = \sqrt x  + 1 + \dfrac{{16}}{{\sqrt x  + 1}} - 2\\
Theo\,Co - si:\\
\sqrt x  + 1 + \dfrac{{16}}{{\sqrt x  + 1}} \ge 2\sqrt {\left( {\sqrt x  + 1} \right).\dfrac{{16}}{{\sqrt x  + 1}}}  = 8\\
 \Leftrightarrow \sqrt x  + 1 + \dfrac{{16}}{{\sqrt x  + 1}} - 2 \ge 6\\
 \Leftrightarrow D \ge 6\\
 \Leftrightarrow GTNN:D = 6\,\\
Khi:\sqrt x  + 1 = \dfrac{{16}}{{\sqrt x  + 1}}\\
 \Leftrightarrow \sqrt x  + 1 = 4\\
 \Leftrightarrow \sqrt x  = 3\\
 \Leftrightarrow x = 9\left( {tm} \right)\\
d)D = \dfrac{{x + 21}}{{\sqrt x  + 2}}\\
 = \dfrac{{x - 4 + 25}}{{\sqrt x  + 2}}\\
 = \sqrt x  - 2 + \dfrac{{25}}{{\sqrt x  + 2}}\\
 = \sqrt x  + 2 + \dfrac{{25}}{{\sqrt x  + 2}} - 4\\
 \ge 2\sqrt {\left( {\sqrt x  + 2} \right).\dfrac{{25}}{{\sqrt x  + 2}}}  - 4\\
 \Leftrightarrow D \ge 2\sqrt {25}  - 4 = 6\\
 \Leftrightarrow GTNN:D = 6\\
Khi:\left( {\sqrt x  + 2} \right) = \dfrac{{25}}{{\sqrt x  + 2}}\\
 \Leftrightarrow \sqrt x  + 2 = 5\\
 \Leftrightarrow \sqrt x  = 3 \Leftrightarrow x = 9\\
e)E = \dfrac{{x + 6\sqrt x  + 34}}{{\sqrt x  + 3}}\\
 = \dfrac{{x + 6\sqrt x  + 9 + 25}}{{\sqrt x  + 3}}\\
 = \dfrac{{{{\left( {\sqrt x  + 3} \right)}^2} + 25}}{{\sqrt x  + 3}}\\
 = \sqrt x  + 3 + \dfrac{{25}}{{\sqrt x  + 3}}\\
 \ge 2\sqrt {\left( {\sqrt x  + 3} \right).\dfrac{{25}}{{\sqrt x  + 3}}}  = 10\\
 \Leftrightarrow E \ge 10\\
 \Leftrightarrow GTNN:E = 10\,khi:\sqrt x  + 3 = 5 \Leftrightarrow x = 4
\end{array}$