Đáp án:

$\begin{array}{l}
Dkxd:x \ge 0;x \ne 9\\
a)x = 2\left( {tmdk} \right)\\
 \Rightarrow A = \dfrac{{3\sqrt x  - 9}}{{\left( {x + 7} \right)\left( {\sqrt x  - 3} \right)}}\\
 = \dfrac{{3\left( {\sqrt x  - 3} \right)}}{{\left( {x + 7} \right)\left( {\sqrt x  - 3} \right)}}\\
 = \dfrac{3}{{x + 7}}\\
 = \dfrac{3}{{2 + 7}} = \dfrac{3}{9} = \dfrac{1}{3}\\
Vay\,A = \dfrac{1}{3}\\
b)B = \dfrac{{\sqrt x }}{{\sqrt x  + 3}} + \dfrac{{2\sqrt x }}{{\sqrt x  - 3}} - \dfrac{{3x + 9}}{{x - 9}}\\
 = \dfrac{{\sqrt x \left( {\sqrt x  - 3} \right) + 2\sqrt x \left( {\sqrt x  + 3} \right) - 3x - 9}}{{\left( {\sqrt x  + 3} \right)\left( {\sqrt x  - 3} \right)}}\\
 = \dfrac{{x - 3\sqrt x  + 2x + 6\sqrt x  - 3x - 9}}{{\left( {\sqrt x  + 3} \right)\left( {\sqrt x  - 3} \right)}}\\
 = \dfrac{{3\sqrt x  - 9}}{{\left( {\sqrt x  + 3} \right)\left( {\sqrt x  - 3} \right)}}\\
 = \dfrac{{3\left( {\sqrt x - 3} \right)}}{{\left( {\sqrt x  + 3} \right)\left( {\sqrt x  - 3} \right)}}\\
 = \dfrac{3}{{\sqrt x  + 3}}\\
Do:\sqrt x  + 3 > 0\\
 \Rightarrow B > 0\
\end{array}$

=> B luôn dương với mọi x thỏa mãn đk

$\begin{array}{l}
c)P = \dfrac{B}{A}\\
 = \dfrac{3}{{\sqrt x  + 3}}:\left( {\dfrac{3}{{x + 7}}} \right)\\
 = \dfrac{{x + 7}}{{\sqrt x  + 3}}\\
 \Rightarrow P.\sqrt x  + 3P = x + 7\\
 \Rightarrow x - P.\sqrt x  - 3P + 7 = 0\\
DK:\left\{ \begin{array}{l}
\Delta  \ge 0\\
P \ge 0
\end{array} \right.\\
 \Rightarrow \left\{ \begin{array}{l}
{P^2} - 4.\left( { - 3P + 7} \right) \ge 0\\
P \ge 0
\end{array} \right.\\
 \Rightarrow \left\{ \begin{array}{l}
{P^2} + 12P - 28 \ge 0\\
P \ge 0
\end{array} \right.\\
 \Rightarrow \left\{ \begin{array}{l}
\left( {P + 14} \right)\left( {P - 2} \right) \ge 0\\
P \ge 0
\end{array} \right.\\
 \Rightarrow P \ge 2\\
 \Rightarrow GTNN:P = 2\\
Khi:x - 2\sqrt x  - 3.2 + 7 = 0\\
 \Rightarrow x - 2\sqrt x  + 1 = 0\\
 \Rightarrow \sqrt x  = 1\\
 \Rightarrow x = 1\left( {tmdk} \right)
\end{array}$