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