Đáp án:
$\begin{array}{l}
a)Dkxd:\left\{ \begin{array}{l}
x > 0\\
\sqrt x \ne 2\\
x \ne 4
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
x > 0\\
x \ne 4
\end{array} \right.\\
b)P = \left( {\frac{1}{{\sqrt x - 2}} + \frac{1}{{\sqrt x + 2}}} \right):\frac{{2x}}{{x - 4}}\\
= \frac{{\sqrt x + 2 + \sqrt x - 2}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}.\frac{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}{{2x}}\\
= \frac{{2\sqrt x }}{{2x}}\\
= \frac{1}{{\sqrt x }}\\
d)x > 0;x \ne 4\\
P < 1\\
\Rightarrow \frac{1}{{\sqrt x }} < 1\\
\Rightarrow \frac{{1 - \sqrt x }}{{\sqrt x }} < 0\\
\Rightarrow 1 - \sqrt x < 0\\
\Rightarrow \sqrt x > 1\\
\Rightarrow x > 1\\
Vậy\,x > 1;x \ne 4
\end{array}$
