Đáp án:
$\begin{array}{l}
a)P = \dfrac{{\sqrt x }}{{\sqrt x - 2}} - \dfrac{{\sqrt x + 1}}{{\sqrt x + 2}} + \dfrac{{\sqrt x + 5}}{{4 - x}}\\
= \dfrac{{\sqrt x \left( {\sqrt x + 2} \right) - \left( {\sqrt x + 1} \right)\left( {\sqrt x - 2} \right) - \sqrt x - 5}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}\\
= \dfrac{{x + 2\sqrt x - \left( {x - \sqrt x - 2} \right) - \sqrt x - 5}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}\\
= \dfrac{{2\sqrt x - 3}}{{x - 4}}\\
Q = \dfrac{{3 - \sqrt x }}{{\sqrt x - 2}} + 1\\
= \dfrac{{3 - \sqrt x + \sqrt x - 2}}{{\sqrt x - 2}}\\
= \dfrac{1}{{\sqrt x - 2}}\\
b)\dfrac{P}{Q} - 2\\
= \dfrac{{2\sqrt x - 3}}{{x - 4}}:\dfrac{1}{{\sqrt x - 2}} - 2\\
= \dfrac{{2\sqrt x - 3}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}.\left( {\sqrt x - 2} \right) - 2\\
= \dfrac{{2\sqrt x - 3}}{{\sqrt x + 2}} - 2\\
= \dfrac{{2\sqrt x - 3 - 2\sqrt x - 4}}{{\sqrt x + 2}}\\
= \dfrac{{ - 7}}{{\sqrt x + 2}}\\
Do:\sqrt x + 2 > 0\\
\Rightarrow \dfrac{{ - 7}}{{\sqrt x + 2}} < 0\\
\Rightarrow \dfrac{P}{Q} - 2 < 0\\
\Rightarrow \dfrac{P}{Q} < 2
\end{array}$
