Đáp án:
$\begin{array}{l}
a)Dkxd:x > 0\\
A = \dfrac{{2011x - 2\sqrt x + 1}}{{\sqrt x }}\\
= 2011\sqrt x - 2 + \dfrac{1}{{\sqrt x }}\\
Theo\,Co - si:\\
2011\sqrt x + \dfrac{1}{{\sqrt x }} \ge 2\sqrt {2011\sqrt x .\dfrac{1}{{\sqrt x }}} \\
\Leftrightarrow 2011\sqrt x + \dfrac{1}{{\sqrt x }} \ge 2\sqrt {2011} \\
\Leftrightarrow 2011\sqrt x + \dfrac{1}{{\sqrt x }} - 2 \ge 2\sqrt {2011} - 2\\
\Leftrightarrow A \ge 2\sqrt {2011} - 2\\
\Leftrightarrow GTNN:A = 2\sqrt {2011} - 2\\
Khi:2011\sqrt x = \dfrac{1}{{\sqrt x }}\\
\Leftrightarrow x = \dfrac{1}{{2011}}\left( {tmdk} \right)\\
f)Dkxd:x \ge 0\\
A = \dfrac{{x + 2\sqrt x + 5}}{{\sqrt x + 1}}\\
= \dfrac{{x + 2\sqrt x + 1 + 4}}{{\sqrt x + 1}}\\
= \dfrac{{{{\left( {\sqrt x + 1} \right)}^2} + 4}}{{\sqrt x + 1}}\\
= \sqrt x + 1 + \dfrac{4}{{\sqrt x + 1}}\\
\ge 2\sqrt {\left( {\sqrt x + 1} \right).\dfrac{4}{{\sqrt x + 1}}} = 4\\
\Leftrightarrow A \ge 4\\
\Leftrightarrow GTNN:A = 4\\
Khi:\sqrt x + 1 = \dfrac{4}{{\sqrt x + 1}} \Leftrightarrow x = 1
\end{array}$
