Đáp án đúng: A Giải chi tiết:Điều kiện: \(x \ge 0,\,\,x \ne 25.\) \(\begin{array}{l}B = \left( {\frac{{15 - \sqrt x }}{{x - 25}} + \frac{2}{{\sqrt x + 5}}} \right):\frac{{\sqrt x + 1}}{{\sqrt x - 5}} = \left[ {\frac{{15 - \sqrt x }}{{\left( {\sqrt x - 5} \right)\left( {\sqrt x + 5} \right)}} + \frac{2}{{\sqrt x + 5}}} \right].\frac{{\sqrt x - 5}}{{\sqrt x + 1}}\\ = \frac{{15 - \sqrt x + 2\left( {\sqrt x - 5} \right)}}{{\left( {\sqrt x - 5} \right)\left( {\sqrt x + 5} \right)}}.\frac{{\sqrt x - 5}}{{\sqrt x + 1}} = \frac{{15 - \sqrt x + 2\sqrt x - 10}}{{\sqrt x + 5}}.\frac{1}{{\sqrt x + 1}}\\ = \frac{{\sqrt x + 5}}{{\sqrt x + 5}}.\frac{1}{{\sqrt x + 1}} = \frac{1}{{\sqrt x + 1}}.\end{array}\) Chọn A.