Đáp án:
$\begin{array}{l}
c)A = \dfrac{{x - \sqrt x + 1}}{{\sqrt x }}\\
= \sqrt x - 1 + \dfrac{1}{{\sqrt x }}\\
= \sqrt x + \dfrac{1}{{\sqrt x }} - 1\\
Khi:\sqrt x > 0\\
Theo\,Co - si:\\
\sqrt x + \dfrac{1}{{\sqrt x }} \ge 2\sqrt {\sqrt x .\dfrac{1}{{\sqrt x }}} = 2\\
\Leftrightarrow \sqrt x + \dfrac{1}{{\sqrt x }} - 1 \ge 2 - 1\\
\Leftrightarrow A \ge 1 > 0\\
Vay\,A > 0\\
d)A \ge 1\\
\Leftrightarrow GTNN:A = 1\,\\
Khi:\sqrt x = \dfrac{1}{{\sqrt x }} \Leftrightarrow x = 1\left( {tmdk} \right)
\end{array}$
