Đáp án:
$\begin{array}{l}
a)Dkxd:x > 0;x \ne 4\\
P = \left( {\dfrac{{\sqrt x - 4}}{{\sqrt x \left( {\sqrt x - 2} \right)}} + \dfrac{3}{{\sqrt x - 2}}} \right):\left( {\dfrac{{\sqrt x + 2}}{{\sqrt x }} - \dfrac{{\sqrt x }}{{\sqrt x - 2}}} \right)\\
= \dfrac{{\sqrt x - 4 + 3\sqrt x }}{{\sqrt x \left( {\sqrt x - 2} \right)}}:\dfrac{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 2} \right) - \sqrt x .\sqrt x }}{{\sqrt x \left( {\sqrt x - 2} \right)}}\\
= \dfrac{{4\sqrt x - 4}}{{\sqrt x \left( {\sqrt x - 2} \right)}}.\dfrac{{\sqrt x \left( {\sqrt x - 2} \right)}}{{x - 4 - x}}\\
= \dfrac{{4\left( {\sqrt x - 1} \right)}}{{ - 4}}\\
= 1 - \sqrt x \\
b)x = 6 - 2\sqrt 5 \\
= 5 - 2\sqrt 5 + 1\\
= {\left( {\sqrt 5 - 1} \right)^2}\\
\Leftrightarrow \sqrt x = \sqrt 5 - 1\\
\Leftrightarrow P = 1 - \left( {\sqrt 5 - 1} \right) = 1 - \sqrt 5 + 1 = 2 - \sqrt 5 \\
c)\\
P.\left( {\sqrt x + 1} \right) > \sqrt x + n\\
\Leftrightarrow \left( {1 - \sqrt x } \right).\left( {\sqrt x + 1} \right) - \sqrt x > n\\
\Leftrightarrow 1 - x - \sqrt x > n\\
Do: - x - \sqrt x < 0\\
\Leftrightarrow - x - \sqrt x + 1 < 1\\
Khi:1 - x - \sqrt x > n\\
\Leftrightarrow n < 1\\
Vậy\,n < 1
\end{array}$
