Đáp án:
$\begin{array}{l}
A = \frac{{\sqrt x - 2}}{{x + 3}}\left( {dkxd:x \ge 0} \right)\\
a)x = 16\\
\Rightarrow \sqrt x = 4\\
\Rightarrow A = \frac{{4 - 2}}{{16 + 3}} = \frac{2}{{19}}\\
b)P = A.B\\
= \frac{{\sqrt x - 2}}{{x + 3}}.\left( {\frac{{\sqrt x - 1}}{{\sqrt x + 2}} - \frac{{5\sqrt x - 2}}{{4 - x}}} \right)\\
= \frac{{\sqrt x - 2}}{{x + 3}}.\frac{{\left( {\sqrt x - 1} \right)\left( {\sqrt x - 2} \right) + 5\sqrt x - 2}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}\\
= \frac{{x - 3\sqrt x + 2 + 5\sqrt x - 2}}{{\left( {x + 3} \right)\left( {\sqrt x + 2} \right)}}\\
= \frac{{x + 2\sqrt x }}{{\left( {x + 3} \right)\left( {\sqrt x + 2} \right)}}\\
= \frac{{\sqrt x }}{{x + 3}}\\
c)ĐKxđ:x \ge 0;x \ne 4\\
\left( {6x + 18} \right).P \ge x + 9\\
\Rightarrow 6\left( {x + 3} \right).\frac{{\sqrt x }}{{x + 3}} \ge x + 9\\
\Rightarrow 6\sqrt x \ge x + 9\\
\Rightarrow {x^2} - 6\sqrt x + 9 \le 0\\
\Rightarrow {\left( {\sqrt x - 3} \right)^2} \le 0\\
Do:{\left( {\sqrt x - 3} \right)^2} \ge 0\forall x \ge 0;x \ne 4\\
\Rightarrow \sqrt x - 3 = 0\\
\Rightarrow x = 9\left( {tmdk} \right)
\end{array}$
