a) Điều kiện xác định $a\ge 0, a\ne 1$
$\begin{array}{l} \dfrac{4}{{\sqrt a - 1}} + \dfrac{3}{{\sqrt a + 1}} - \dfrac{{6\sqrt a + 2}}{{a - 1}}\\ = \dfrac{{4\left( {\sqrt a + 1} \right) + 3\left( {\sqrt a - 1} \right) - 6\sqrt a - 2}}{{\left( {\sqrt a - 1} \right)\left( {\sqrt a + 1} \right)}}\\ = \dfrac{{4\sqrt a + 3\sqrt a + 4 - 3 - 6\sqrt a - 2}}{{\left( {\sqrt a - 1} \right)\left( {\sqrt a + 1} \right)}}\\ = \dfrac{{\sqrt a - 1}}{{\left( {\sqrt a - 1} \right)\left( {\sqrt a + 1} \right)}} = \dfrac{1}{{\sqrt a + 1}}\\ b)a = 6 + 2\sqrt 5 = 5 + 1 + 2.1\sqrt 5 = {\left( {\sqrt 5 + 1} \right)^2}\\ P = \dfrac{1}{{\sqrt a + 1}} = \dfrac{1}{{\sqrt {{{\left( {\sqrt 5 + 1} \right)}^2}} + 1}} = \dfrac{1}{{\sqrt 5 + 2}} = \dfrac{{\sqrt 5 - 2}}{{5 - 4}} = \sqrt 5 - 2 \end{array}$
