Đáp án đúng: B Giải chi tiết:Điều kiện: \(x \ge 0,\,\,x \ne 1.\) \(\begin{array}{l}P = \frac{{x\sqrt x + 26\sqrt x - 19}}{{x + 2\sqrt x - 3}} - \frac{{2\sqrt x }}{{\sqrt x - 1}} + \frac{{\sqrt x - 3}}{{\sqrt x + 3}} = \frac{{x\sqrt x + 26\sqrt x - 19}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 3} \right)}} - \frac{{2\sqrt x }}{{\sqrt x - 1}} + \frac{{\sqrt x - 3}}{{\sqrt x + 3}}\\\,\,\,\, = \frac{{x\sqrt x + 26\sqrt x - 19 - 2\sqrt x \left( {\sqrt x + 3} \right) + \left( {\sqrt x - 3} \right)\left( {\sqrt x - 1} \right)}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 3} \right)}}\\\,\,\,\, = \frac{{x\sqrt x + 26\sqrt x - 19 - 2x - 6\sqrt x + x - 4\sqrt x + 3}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 3} \right)}}\\\,\,\,\, = \frac{{x\sqrt x - x + 16\sqrt x - 16}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 3} \right)}} = \frac{{x\left( {\sqrt x - 1} \right) + 16\left( {\sqrt x - 1} \right)}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 3} \right)}}\\\,\,\, = \frac{{\left( {x + 16} \right)\left( {\sqrt x - 1} \right)}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 3} \right)}} = \frac{{x + 16}}{{\sqrt x + 3}}.\end{array}\) Chọn B.