a: \(=\dfrac{-a-2\sqrt{a}+a-2\sqrt{a}-4a-2\sqrt{a}+4}{a-4}:\dfrac{-2\sqrt{a}+2+\sqrt{a}}{\sqrt{a}\left(2-\sqrt{a}\right)}\)
\(=\dfrac{-4a-6\sqrt{a}+4}{a-4}\cdot\dfrac{-\sqrt{a}\left(\sqrt{a}-2\right)}{-\sqrt{a}+2}\)
\(=\dfrac{4a+6\sqrt{a}-4}{\sqrt{a}+2}\cdot\dfrac{\sqrt{a}}{2-\sqrt{a}}=\dfrac{\sqrt{a}\left(4a+6\sqrt{a}-4\right)}{4-a}\)
b: Để \(A=\sqrt{a}+2\) thì \(4a\sqrt{a}+6a-4\sqrt{a}=\left(\sqrt{a}+2\right)\left(4-a\right)=4\sqrt{a}-a\sqrt{a}+8-2a\)
=>\(5a\sqrt{a}+8a-8\sqrt{a}-8=0\)
=>\(5a\cdot\sqrt{a}+10a-2a-4\sqrt{a}-4\sqrt{a}-8=0\)
=>\(\left(\sqrt{a}+2\right)\left(5a-2\sqrt{a}-4\right)=0\)
=>\(5a-2\sqrt{a}-4=0\)
=>\(a=\dfrac{22+2\sqrt{21}}{25}\)