Đáp án đúng: A Giải chi tiết:Với \(x \ge 0,\,\,x \ne 9\) ta có: \(\begin{array}{l}P = \frac{{\sqrt x }}{{\sqrt x + 3}} + \frac{{2\sqrt x }}{{\sqrt x - 3}} - \frac{{3x + 9}}{{x - 9}}\\\;\;\; = \frac{{\sqrt x }}{{\sqrt x + 3}} + \frac{{2\sqrt x }}{{\sqrt x - 3}} - \frac{{3x + 9}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right)}}\\\;\;\; = \frac{{\sqrt x \left( {\sqrt x - 3} \right) + 2\sqrt x \left( {\sqrt x + 3} \right) - (3x + 9)}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right)}}\\\;\;\; = \frac{{x - 3\sqrt x + 2x + 6\sqrt x - 3x - 9}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right)}}\\\;\;\; = \frac{{3\sqrt x - 9}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right)}}\\\;\;\; = \frac{{3\left( {\sqrt x - 3} \right)}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right)}}\\\;\;\; = \frac{3}{{\sqrt x + 3}}.\end{array}\) Vậy \(P = \frac{3}{{\sqrt x + 3}}\) với \(x \ge 0,\,\,x \ne 9\) Chọn A.