Đáp án:
$\begin{array}{l}
a)Dkxd:\left\{ \begin{array}{l}
x \ge 0\\
x \ne 4
\end{array} \right.\\
Vay\,TXD:D = \left[ {0; + \infty } \right)\backslash \left\{ 4 \right\}\\
b)P = \dfrac{{\sqrt x + 1}}{{\sqrt x - 2}} + \dfrac{{2\sqrt x }}{{\sqrt x + 2}} + \dfrac{{2 + 5\sqrt x }}{{4 - x}}\\
= \dfrac{{\left( {\sqrt x + 1} \right)\left( {\sqrt x + 2} \right) + 2\sqrt x \left( {\sqrt x - 2} \right) - 2 - 5\sqrt x }}{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 2} \right)}}\\
= \dfrac{{x + 3\sqrt x + 2 + 2x - 4\sqrt x - 2 - 5\sqrt x }}{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 2} \right)}}\\
= \dfrac{{3x - 6\sqrt x }}{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 2} \right)}}\\
= \dfrac{{3\sqrt x \left( {\sqrt x - 2} \right)}}{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 2} \right)}}\\
= \dfrac{{3\sqrt x }}{{\sqrt x + 2}}\\
c)P = 2\\
\Leftrightarrow \dfrac{{3\sqrt x }}{{\sqrt x + 2}} = 2\\
\Leftrightarrow 3\sqrt x = 2\sqrt x + 4\\
\Leftrightarrow \sqrt x = 4\\
\Leftrightarrow x = 16\left( {tmdk} \right)\\
Vậy\,x = 16
\end{array}$
