Đáp án:

$\begin{array}{l}
a)Dkxd:x > 0\\
P = \left( {\sqrt x  - \dfrac{1}{{\sqrt x }}} \right):\left( {\dfrac{{\sqrt x  - 1}}{{\sqrt x }} + \dfrac{{1 - \sqrt x }}{{x + \sqrt x }}} \right)\\
 = \dfrac{{x - 1}}{{\sqrt x }}:\dfrac{{\left( {\sqrt x  - 1} \right)\left( {\sqrt x  + 1} \right) + 1 - \sqrt x }}{{\sqrt x \left( {\sqrt x  + 1} \right)}}\\
 = \dfrac{{x - 1}}{{\sqrt x }}.\dfrac{{\sqrt x \left( {\sqrt x  + 1} \right)}}{{x - 1 + 1 - \sqrt x }}\\
 = \dfrac{{\left( {\sqrt x  + 1} \right)\left( {\sqrt x  - 1} \right)}}{1}.\dfrac{{\sqrt x  + 1}}{{x - \sqrt x }}\\
 = \dfrac{{{{\left( {\sqrt x  + 1} \right)}^2}}}{{\sqrt x }}\\
b)x = \dfrac{2}{{2 + \sqrt 3 }}\left( {tmdk} \right)\\
 = \dfrac{{2.\left( {2 - \sqrt 3 } \right)}}{{4 - 3}}\\
 = 4 - 2\sqrt 3 \\
 = {\left( {\sqrt 3  - 1} \right)^2}\\
 \Leftrightarrow \sqrt x  = \sqrt 3  - 1\\
 \Leftrightarrow P = \dfrac{{{{\left( {\sqrt x  + 1} \right)}^2}}}{{\sqrt x }} = \dfrac{{{{\left( {\sqrt 3  - 1 + 1} \right)}^2}}}{{\sqrt 3  - 1}}\\
 = \dfrac{3}{{\sqrt 3  - 1}} = \dfrac{{3\left( {\sqrt 3  + 1} \right)}}{{3 - 1}} = \dfrac{{3\sqrt 3  + 3}}{2}\\
c)P - 2\\
 = \dfrac{{{{\left( {\sqrt x  + 1} \right)}^2}}}{{\sqrt x }} - 2\\
 = \dfrac{{x + 2\sqrt x  + 1 - 2\sqrt x }}{{\sqrt x }}\\
 = \dfrac{{x + 1}}{{\sqrt x }} > 0\\
 \Leftrightarrow P > 2\\
d)P.\sqrt x  = 6\sqrt x  - 3 - \sqrt {x - 4} \left( {dkxd:x \ge 4} \right)\\
 \Leftrightarrow \dfrac{{{{\left( {\sqrt x  + 1} \right)}^2}}}{{\sqrt x }}.\sqrt x  = 6\sqrt x  - 3 - \sqrt {x - 4} \\
 \Leftrightarrow {\left( {\sqrt x  + 1} \right)^2} = 6\sqrt x  - 3 - \sqrt {x - 4} \\
 \Leftrightarrow x + 2\sqrt x  + 1 = 6\sqrt x  - 3 - \sqrt {x - 4} \\
 \Leftrightarrow x - 4\sqrt x  + 4 + \sqrt {x - 4}  = 0\\
 \Leftrightarrow {\left( {\sqrt x  - 2} \right)^2} + \sqrt {x - 4}  = 0\\
 \Leftrightarrow \left\{ \begin{array}{l}
\sqrt x  - 2 = 0\\
x - 4 = 0
\end{array} \right.\\
 \Leftrightarrow x = 4\left( {tmdk} \right)\\
Vậy\,x = 4
\end{array}$