Đáp án:
$\begin{array}{l}
Dkxd:x \ge 0;x \ne 9\\
a)P = \left( {\dfrac{{2\sqrt x }}{{\sqrt x + 3}} + \dfrac{{\sqrt x }}{{\sqrt x - 3}} - \dfrac{{3x + 3}}{{x - 9}}} \right):\left( {\dfrac{{2\sqrt x - 2}}{{\sqrt x - 3}} - 1} \right)\\
= \dfrac{{2\sqrt x \left( {\sqrt x - 3} \right) + \sqrt x \left( {\sqrt x + 3} \right) - 3x - 3}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}}\\
:\dfrac{{2\sqrt x - 2 - \sqrt x + 3}}{{\sqrt x - 3}}\\
= \dfrac{{2x - 6\sqrt x + x + 3\sqrt x - 3x - 3}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}}.\dfrac{{\sqrt x - 3}}{{\sqrt x + 1}}\\
= \dfrac{{ - 3\sqrt x - 3}}{{\sqrt x + 3}}.\dfrac{1}{{\sqrt x + 1}}\\
= \dfrac{{ - 3}}{{\sqrt x + 3}}\\
b)x = 4 - 2\sqrt 3 \left( {tmdk} \right) = {\left( {\sqrt 3 - 1} \right)^2}\\
\Leftrightarrow \sqrt x = \sqrt 3 - 1\\
P = \dfrac{{ - 3}}{{\sqrt x + 3}} = \dfrac{{ - 3}}{{\sqrt 3 - 1 + 3}} = \dfrac{{ - 3}}{{2 + \sqrt 3 }}\\
= \dfrac{{ - 3\left( {2 - \sqrt 3 } \right)}}{{{2^2} - 3}}\\
= - 6 + 3\sqrt 3 \\
c)P < - \dfrac{1}{2}\\
\Leftrightarrow \dfrac{{ - 3}}{{\sqrt x + 3}} + \dfrac{1}{2} < 0\\
\Leftrightarrow \dfrac{{ - 6 + \sqrt x + 3}}{{2\left( {\sqrt x + 3} \right)}} < 0\\
\Leftrightarrow \sqrt x - 3 < 0\\
\Leftrightarrow \sqrt x < 3\\
\Leftrightarrow x < 9\\
Vậy\,0 \le x < 9\\
d)P = \dfrac{{ - 3}}{{\sqrt x + 3}}\\
Do:\sqrt x + 3 \ge 3\\
\Leftrightarrow \dfrac{1}{{\sqrt x + 3}} \le \dfrac{1}{3}\\
\Leftrightarrow \dfrac{3}{{\sqrt x + 3}} \le 1\\
\Leftrightarrow - \dfrac{3}{{\sqrt x + 3}} \ge - 1\\
\Leftrightarrow P \ge - 1\\
\Leftrightarrow GTNN:P = - 1\,khi:x = 0\\
e)P \in Z\\
\Leftrightarrow \dfrac{{ - 3}}{{\sqrt x + 3}} \in Z\\
\Leftrightarrow \left( {\sqrt x + 3} \right) = 3\\
\Leftrightarrow \sqrt x = 0\\
\Leftrightarrow x = 0\left( {tmdk} \right)\\
Vậy\,x = 0
\end{array}$
