Đáp án đúng: D
Giải chi tiết:Điều kiện: \(x \ge 0,\,\,x \ne 4.\)
\(\begin{array}{l}B = \frac{2}{{\sqrt x - 2}} + \frac{3}{{\sqrt x + 2}} - \frac{{x - 5\sqrt x + 2}}{{4 - x}}\\\,\,\,\,\, = \frac{2}{{\sqrt x - 2}} + \frac{3}{{\sqrt x + 2}} + \frac{{x - 5\sqrt x + 2}}{{x - 4}}\,\,\\\,\,\,\,\, = \frac{2}{{\sqrt x - 2}} + \frac{3}{{\sqrt x + 2}} + \frac{{x - 5\sqrt x + 2}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}\,\,\,\,\\\,\,\,\,\,\, = \frac{{2\left( {\sqrt x + 2} \right) + 3\left( {\sqrt x - 2} \right) + x - 5\sqrt x + 2}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}\,\,\,\,\\\,\,\,\,\,\, = \frac{{2\sqrt x + 4 + 3\sqrt x - 6 + x - 5\sqrt x + 2}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}\,\,\,\\\,\,\,\,\, = \frac{x}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}\,\,\, = \frac{x}{{x - 4}}\,\,\,.\end{array}\)
Vậy \(B = \frac{x}{{x - 4}}\) với \(x \ge 0;x \ne 4\).
Chọn D.
