Đáp án:
$\begin{array}{l}
a)Dkxd:x > 0\\
P = \left( {\dfrac{1}{{\sqrt x }} + \dfrac{{\sqrt x }}{{\sqrt x + 1}}} \right):\dfrac{{\sqrt x }}{{x + \sqrt x }}\\
= \dfrac{{\sqrt x + 1 + x}}{{\sqrt x \left( {\sqrt x + 1} \right)}}.\dfrac{{\sqrt x \left( {\sqrt x + 1} \right)}}{{\sqrt x }}\\
= \dfrac{{x + \sqrt x + 1}}{{\sqrt x }}\\
b)x = 25\left( {tmdk} \right)\\
\Rightarrow \sqrt x = 5\\
\Rightarrow P = \dfrac{{25 + 5 + 1}}{5} = \dfrac{{31}}{5}\\
c)P = - 1\\
\Rightarrow \dfrac{{x + \sqrt x + 1}}{{\sqrt x }} = - 1\\
\Rightarrow x + \sqrt x + 1 = - \sqrt x \\
\Rightarrow x + 2\sqrt x + 1 = 0\\
\Rightarrow {\left( {\sqrt x + 1} \right)^2} = 0\\
\Rightarrow \sqrt x = - 1\left( {vo\,nghiem} \right)\\
d)P > \sqrt x + 2\\
\Rightarrow \dfrac{{x + \sqrt x + 1}}{{\sqrt x }} > \sqrt x + 2\\
\Rightarrow x + \sqrt x + 1 > \sqrt x \left( {\sqrt x + 2} \right)\\
\Rightarrow x + \sqrt x + 1 > x + 2\sqrt x \\
\Rightarrow \sqrt x < 1\\
\Rightarrow x < 1\\
Vay\,0 < x < 1\\
e)P = \dfrac{{x + \sqrt x + 1}}{{\sqrt x }} = \sqrt x + 1 + \dfrac{1}{{\sqrt x }}\\
Do:x > 0\\
\Rightarrow \sqrt x + \dfrac{1}{{\sqrt x }} > 0\\
\Rightarrow \sqrt x + \dfrac{1}{{\sqrt x }} + 1 > 1\\
\Rightarrow P > 1\\
\Rightarrow P > \sqrt P \\
f)P = \sqrt x + 1 + \dfrac{1}{{\sqrt x }}\\
Theo\,Co - si:\sqrt x + \dfrac{1}{{\sqrt x }} \ge 2\sqrt {\sqrt x .\dfrac{1}{{\sqrt x }}} = 2\\
\Rightarrow P \ge 2 + 1 = 3\\
\Rightarrow GTNN:P = 3\\
Khi:\sqrt x = \dfrac{1}{{\sqrt x }} \Rightarrow x = 1
\end{array}$
