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