a) Ta có: \(P=\left(\dfrac{1}{1-\sqrt{a}}-\dfrac{1}{1+\sqrt{a}}\right)\cdot\left(\dfrac{1}{\sqrt{a}}+1\right)\)
\(=\left(\dfrac{1+\sqrt{a}-\left(1-\sqrt{a}\right)}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)}\right)\cdot\left(\dfrac{1}{\sqrt{a}}+\dfrac{\sqrt{a}}{\sqrt{a}}\right)\)
\(=\dfrac{1+\sqrt{a}-1+\sqrt{a}}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)}\cdot\dfrac{1+\sqrt{a}}{\sqrt{a}}\)
\(=\dfrac{2\sqrt{a}}{\sqrt{a}\left(1-\sqrt{a}\right)}\)
\(=\dfrac{2}{1-\sqrt{a}}\)
b) Để \(P^2=P\) nên \(P^2-P=0\)
\(\Leftrightarrow P\left(P-1\right)=0\)
\(\Leftrightarrow P-1=0\)(Vì \(P\ne0\forall a\) thỏa mãn ĐKXĐ)
\(\Leftrightarrow P=1\)
\(\Leftrightarrow\dfrac{2}{1-\sqrt{a}}=1\)
\(\Leftrightarrow1-\sqrt{a}=2\)
\(\Leftrightarrow\sqrt{a}=-1\)(Vô lý)
Vậy: Không có giá trị nào của P để \(P^2=P\)
