a) Ta có: \(P=\left(\dfrac{1}{1-\sqrt{a}}-\dfrac{1}{1+\sqrt{a}}\right)\cdot\left(1-\dfrac{1}{\sqrt{a}}\right)\)
\(=\dfrac{1+\sqrt{a}-1+\sqrt{a}}{-\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\cdot\dfrac{\sqrt{a}-1}{\sqrt{2}}\)
\(=\dfrac{-2}{\sqrt{a}+1}\)
b) Ta có: \(P=\dfrac{-1}{2}\)
nên \(\dfrac{2}{\sqrt{a}+1}=\dfrac{1}{2}\)
\(\Leftrightarrow\sqrt{a}+1=4\)
\(\Leftrightarrow a=9\)(thỏa ĐK)