Đáp án:
$\begin{array}{l}
1)Dkxd:x > 0\\
P = \left( {\dfrac{1}{{\sqrt x }} + \dfrac{{\sqrt x }}{{\sqrt x + 1}}} \right):\dfrac{{\sqrt x }}{{x + \sqrt x }}\\
= \dfrac{{\sqrt x + 1 + x}}{{\sqrt x \left( {\sqrt x + 1} \right)}}:\dfrac{{\sqrt x }}{{\sqrt x \left( {\sqrt x + 1} \right)}}\\
= \dfrac{{x + \sqrt x + 1}}{{\sqrt x .\left( {\sqrt x + 1} \right)}}.\left( {\sqrt x + 1} \right)\\
= \dfrac{{x + \sqrt x + 1}}{{\sqrt x }}\\
2)x = 9\left( {tmdk} \right)\\
\Rightarrow \sqrt x = 3\\
\Rightarrow P = \dfrac{{9 + 3 + 1}}{3} = \dfrac{{13}}{3}\\
3)P = \dfrac{7}{2}\\
\Rightarrow \dfrac{{x + \sqrt x + 1}}{{\sqrt x }} = \dfrac{7}{2}\\
\Rightarrow 2x + 2\sqrt x + 2 = 7\sqrt x \\
\Rightarrow 2x - 5\sqrt x + 2 = 0\\
\Rightarrow \left( {2\sqrt x - 1} \right)\left( {\sqrt x - 2} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
\sqrt x = \dfrac{1}{2}\\
\sqrt x = 2
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = \dfrac{1}{4}\left( {tmdk} \right)\\
x = 4\left( {tmdk} \right)
\end{array} \right.\\
4)P = \dfrac{{x + \sqrt x + 1}}{{\sqrt x }}\\
= \sqrt x + 1 + \dfrac{1}{{\sqrt x }}\\
THeo\,Co - si:\sqrt x + \dfrac{1}{{\sqrt x }} \ge 2\sqrt {\sqrt x .\dfrac{1}{{\sqrt x }}} = 2\\
\Rightarrow P \ge 2 + 1 = 3\\
\Rightarrow GTNN:P = 3 \Leftrightarrow x = 1
\end{array}$
