Đáp án:
$\begin{array}{l}
x \ge 0;x \ne 1\\
P = \frac{{ - 1 + \sqrt x }}{{\sqrt x + 2}}\\
= \frac{{\sqrt x + 2 - 3}}{{\sqrt x + 2}}\\
= 1 - \frac{3}{{\sqrt x + 2}}\\
Do:\sqrt x + 2 \ge 2\forall x\\
\Rightarrow \frac{3}{{\sqrt x + 2}} \le \frac{3}{2}\\
\Rightarrow - \frac{3}{{\sqrt x + 2}} \ge - \frac{3}{2}\\
\Rightarrow 1 - \frac{3}{{\sqrt x + 2}} \ge 1 - \frac{3}{2} = - \frac{1}{2}\\
\Rightarrow P \ge - \frac{1}{2}\\
\Rightarrow GTNN:P = - \frac{1}{2} \Leftrightarrow x = 0
\end{array}$
