Đáp án đúng: A
Giải chi tiết:Điều kiện: \(x \ge 0,\,\,\,x \ne 4,\,\,x \ne 9.\)
\(\begin{array}{l}P = \left( {\frac{{\sqrt x + 3}}{{\sqrt x - 2}} + \frac{{\sqrt x + 2}}{{3 - \sqrt x }} + \frac{{\sqrt x + 2}}{{x - 5\sqrt x + 6}}} \right):\left( {1 - \frac{{\sqrt x }}{{\sqrt x + 1}}} \right)\\\,\,\,\,\, = \left[ {\frac{{\sqrt x + 3}}{{\sqrt x - 2}} - \frac{{\sqrt x + 2}}{{\sqrt x - 3}} + \frac{{\sqrt x + 2}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 3} \right)}}} \right]:\frac{{\sqrt x + 1 - \sqrt x }}{{\sqrt x + 1}}\\\,\,\,\,\, = \frac{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right) - \left( {\sqrt x + 2} \right)\left( {\sqrt x - 2} \right) + \sqrt x + 2}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 3} \right)}}:\frac{1}{{\sqrt x + 1}}\\\,\,\,\,\, = \frac{{x - 9 - x + 4 + \sqrt x + 2}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 3} \right)}}.\left( {\sqrt x + 1} \right)\\\,\,\,\,\, = \frac{{\sqrt x - 3}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 3} \right)}}.\left( {\sqrt x + 1} \right) = \frac{{\sqrt x + 1}}{{\sqrt x - 2}}.\end{array}\)
Chọn A.
