Đáp án:
$\begin{array}{l}
a)Dkxd:x \ge 0;x \ne 1\\
P = \left( {\dfrac{{x - 1}}{{x + 3\sqrt x - 4}} + \dfrac{{\sqrt x + 1}}{{1 - \sqrt x }}} \right):\dfrac{{x + 2\sqrt x + 1}}{{x - 1}}\\
= \left[ {\dfrac{{x - 1}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 4} \right)}} - \dfrac{{\sqrt x + 1}}{{\sqrt x - 1}}} \right].\dfrac{{x - 1}}{{{{\left( {\sqrt x + 1} \right)}^2}}}\\
= \dfrac{{x - 1 - \left( {\sqrt x + 1} \right)\left( {\sqrt x + 4} \right)}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 4} \right)}}.\dfrac{{\sqrt x - 1}}{{\sqrt x + 1}}\\
= \dfrac{{x - 1 - x - 5\sqrt x - 4}}{{\sqrt x + 4}}.\dfrac{1}{{\sqrt x + 1}}\\
= \dfrac{{ - 5\sqrt x - 5}}{{\left( {\sqrt x + 4} \right)\left( {\sqrt x + 1} \right)}}\\
= - \dfrac{5}{{\sqrt x + 4}}\\
b)x = 13 + 4\sqrt 3 \left( {tmdk} \right)\\
= 12 + 2.2\sqrt 3 + 1\\
= {\left( {2\sqrt 3 + 1} \right)^2}\\
\Rightarrow \sqrt x = 2\sqrt 3 + 1\\
P = - \dfrac{5}{{\sqrt x + 4}} = \dfrac{{ - 5}}{{2\sqrt 3 + 1 + 4}}\\
= \dfrac{{ - 5}}{{2\sqrt 3 + 5}} = \dfrac{{ - 25 + 10\sqrt 3 }}{{13}}\\
c)P = - \dfrac{5}{{\sqrt x + 4}}\\
Do:\sqrt x + 4 \ge 4\\
\Rightarrow \dfrac{1}{{\sqrt x + 4}} \le \dfrac{1}{4}\\
\Rightarrow \dfrac{{ - 5}}{{\sqrt x + 4}} \ge \dfrac{{ - 5}}{4}\\
\Rightarrow P \ge - \dfrac{5}{4}\\
\Rightarrow GTNN:P = - \dfrac{5}{4} \Leftrightarrow x = 0
\end{array}$
