a) Với \(0< a\ne1\), ta có:
\(P=\left(\dfrac{1}{1-\sqrt{a}}-\dfrac{1}{1+\sqrt{a}}\right)\cdot\left(\dfrac{1}{\sqrt{a}}+1\right)\)
\(=\left[\dfrac{1+\sqrt{a}}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)}-\dfrac{1-\sqrt{a}}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)}\right]\cdot\left(\dfrac{1}{\sqrt{a}}+\dfrac{\sqrt{a}}{\sqrt{a}}\right)\)
\(=\dfrac{1+\sqrt{a}-1+\sqrt{a}}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)}\cdot\dfrac{1+\sqrt{a}}{\sqrt{a}}\)
\(=\dfrac{2\sqrt{a}}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)}\cdot\dfrac{1+\sqrt{a}}{\sqrt{a}}\)
\(=\dfrac{2}{1-\sqrt{a}}\)
Vậy \(P=\dfrac{2}{1-\sqrt{a}}\).
b) Ta có: \(9+4\sqrt{2}=8+4\sqrt{2}+1=\left(\sqrt{8}+1\right)^2\).
Thay \(a=\left(\sqrt{8}+1\right)^2\) (TMĐK) vào P, ta có:
\(P=\dfrac{2}{1-\sqrt{\left(\sqrt{8}-1\right)^2}}=\dfrac{2}{1-\sqrt{8}-1}=\dfrac{2}{\sqrt{8}}=\dfrac{\sqrt{2}}{2}\)
Vậy \(P=\dfrac{\sqrt{2}}{2}\) khi \(a=9+4\sqrt{2}\).
c) Để \(P>\dfrac{1}{2}\) thì:
\(\dfrac{2}{1-\sqrt{a}}>\dfrac{1}{2}\Leftrightarrow\dfrac{4}{2\left(1-\sqrt{a}\right)}>\dfrac{1-\sqrt{a}}{2\left(1-\sqrt{a}\right)}\)
\(\Leftrightarrow\dfrac{4}{2\left(1-\sqrt{a}\right)}-\dfrac{1-\sqrt{a}}{2\left(1-\sqrt{a}\right)}>0\)
\(\Leftrightarrow\dfrac{3+\sqrt{a}}{2\left(1-\sqrt{a}\right)}>0\)
Vì \(a>0\Rightarrow\sqrt{a}>0\Rightarrow3+\sqrt{a}>0\)
Mà \(\dfrac{3+\sqrt{a}}{2\left(1-\sqrt{a}\right)}>0\Rightarrow1-\sqrt{a}>0\)
\(\Leftrightarrow1>\sqrt{a}\Leftrightarrow a< 1\)
Kết hợp điều kiện: \(0< a< 1\).
Vậy để \(P>\dfrac{1}{2}\) thì \(0< a< 1\).
