a: Ta có: \(P=\left(\dfrac{\sqrt{a}}{2}-\dfrac{1}{2\sqrt{a}}\right)^2\cdot\left(\dfrac{\sqrt{a}-1}{\sqrt{a}+1}-\dfrac{\sqrt{a}+1}{\sqrt{a}-1}\right)\)
\(=\dfrac{\left(\sqrt{a}-1\right)^2\left(\sqrt{a}+1\right)^2}{4a}\cdot\dfrac{a-2\sqrt{a}+1-a-2\sqrt{a}-1}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\)
\(=\dfrac{\left(a-1\right)}{4a}\cdot\dfrac{-4\sqrt{a}}{1}\)
\(=\dfrac{-\left(a-1\right)}{\sqrt{a}}\)
b: Để P<0 thì a-1>0
hay a>1
c: Để P=-2 thì \(-\left(a-1\right)=-2\sqrt{a}\)
\(\Leftrightarrow a-2\sqrt{a}-1=0\)
\(\Leftrightarrow\left(\sqrt{a}-1\right)^2=2\)
\(\Leftrightarrow\sqrt{a}-1=\sqrt{2}\)
hay \(a=3+2\sqrt{2}\)
