Đáp án:
$\begin{array}{l}
A = x + \sqrt x \\
Dkxd:x \ge 0\\
\Rightarrow x + \sqrt x \ge 0\\
\Rightarrow A \ge 0\\
\Rightarrow GTNN:A = 0 \Leftrightarrow x = 0\\
b)A = \dfrac{{ - 3}}{{\sqrt {3{x^2} - 6x + 11} }}\\
3{x^2} - 6x + 11\\
= 3.\left( {{x^2} - 2x + 1} \right) + 8\\
= 3.{\left( {x - 1} \right)^2} + 8 \ge 8\\
\Rightarrow \sqrt {3{x^2} - 6x + 11} \ge \sqrt 8 = 2\sqrt 2 \\
\Rightarrow \dfrac{1}{{\sqrt {3{x^2} - 6x + 11} }} \le \dfrac{1}{{2\sqrt 2 }} = \dfrac{{\sqrt 2 }}{4}\\
\Rightarrow \dfrac{{ - 3}}{{\sqrt {3{x^2} - 6x + 11} }} \ge \dfrac{{ - 3\sqrt 2 }}{4}\\
\Rightarrow A \ge \dfrac{{ - 3\sqrt 2 }}{4}\\
\Rightarrow GTNN:A = - \dfrac{{3\sqrt 2 }}{4} \Leftrightarrow x = 1
\end{array}$
