2:
a:
ĐKXĐ: a>0 và a<>1
\(P=\left(\dfrac{\sqrt{a}}{2}-\dfrac{1}{2\sqrt{a}}\right)^2\left(\dfrac{\sqrt{a}-1}{\sqrt{a}+1}-\dfrac{\sqrt{a}+1}{\sqrt{a}-1}\right)\)
\(=\left(\dfrac{a-1}{2\sqrt{a}}\right)^2\cdot\dfrac{\left(\sqrt{a}-1\right)^2-\left(\sqrt{a}+1\right)^2}{a-1}\)
\(=\dfrac{\left(a-1\right)^2}{4a}\cdot\dfrac{a-2\sqrt{a}+1-a-2\sqrt{a}-1}{a-1}\)
\(=\dfrac{\left(a-1\right)}{4a}\cdot\left(-4\sqrt{a}\right)=\dfrac{-a+1}{\sqrt{a}}\)
b: P<0
=>-a+1<0
=>-a<-1
=>a>1
c: P=-2
\(\Leftrightarrow-a+1=-2\sqrt{a}\)
=>\(a-1=2\sqrt{a}\)
=>\(a-2\sqrt{a}-1=0\)
=>\(a-2\sqrt{a}+1-2=0\)
=>\(\left(\sqrt{a}-1\right)^2=2\)
=>\(\left[{}\begin{matrix}\sqrt{a}-1=\sqrt{2}\\\sqrt{a}-1=-\sqrt{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\sqrt{a}=\sqrt{2}+1\\\sqrt{a}=1-\sqrt{2}\left(loại\right)\end{matrix}\right.\)
=>\(a=3+2\sqrt{2}\)
