a) Điều kiện xác định:
$\begin{array}{l} \left\{ \begin{array}{l} x \ge 0\\ \sqrt x - 3 \ne 0\\ \dfrac{{2\sqrt x - 2}}{{\sqrt x - 3}} - 1 \ne 0 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} x \ge 0\\ x \ne 9\\ \dfrac{{2\sqrt x - 2 - \sqrt x + 3}}{{\sqrt x - 3}} \ne 0 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} x \ge 0\\ x \ne 9\\ \dfrac{{\sqrt x + 1}}{{\sqrt x - 3}} \ne 0 \end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l} x \ge 0\\ x \ne 9 \end{array} \right.\left( {do\,\sqrt x + 1 > 0} \right) \end{array}$
b)
$\begin{array}{l} A = \left( {\dfrac{{2\sqrt x }}{{\sqrt x + 3}} + \dfrac{{\sqrt x }}{{\sqrt x - 3}} - \dfrac{{3x + 3}}{{x - 9}}} \right):\left( {\dfrac{{2\sqrt x - 2}}{{\sqrt x - 3}} - 1} \right)\\ A = \dfrac{{2\sqrt x \left( {\sqrt x - 3} \right) + \sqrt x \left( {\sqrt x + 3} \right) - 3x - 3}}{{x - 9}}:\dfrac{{\sqrt x + 1}}{{\sqrt x - 3}}\\ A = \dfrac{{2x - 6\sqrt x + x + 3\sqrt x - 3x - 3}}{{x - 9}}.\dfrac{{\sqrt x - 3}}{{\sqrt x + 1}}\\ A = \dfrac{{ - 3\sqrt x - 3}}{{x - 9}}.\dfrac{{\sqrt x - 3}}{{\sqrt x + 1}} = \dfrac{{ - 3\left( {\sqrt x + 1} \right)\left( {\sqrt x - 3} \right)}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)\left( {\sqrt x + 1} \right)}}\\ A = \dfrac{{ - 3}}{{\sqrt x + 3}} < - \dfrac{1}{2}\\ \Rightarrow \dfrac{3}{{\sqrt x + 3}} > \dfrac{1}{2} \Leftrightarrow 6 > \sqrt x + 3\\ \Leftrightarrow \sqrt x < 3 \Leftrightarrow x < 9 \Rightarrow 0 \le x < 9\\ c)\\ A = \dfrac{{ - 3}}{{\sqrt x + 3}}\min \Rightarrow \sqrt x + 3\min \Rightarrow \min \sqrt x + 3 = 3 \Rightarrow x = 0\\ \Rightarrow \min A = - 1 \end{array}$
