a)
\(A=\left(\frac{\sqrt{a}+1}{\sqrt{a}-1}-\frac{\sqrt{a}-1}{\sqrt{a}+1}+4\sqrt{a}\right)\cdot\left(\sqrt{a}-\frac{1}{\sqrt{a}}\right)\\ =\left(\frac{\left(\sqrt{a}+1\right)^2-\left(\sqrt{a}-1\right)^2+4\sqrt{a}\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\right)\cdot\left(\frac{\left(\sqrt{a}\right)^2-1}{\sqrt{a}}\right)\\ =\left(\frac{a+2\sqrt{a}+1-a+2\sqrt{a}-1+4a\sqrt{a}-4\sqrt{a}}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\right)\cdot\left(\frac{a-1}{\sqrt{a}}\right)\)
\(=\frac{4a\sqrt{a}}{a-1}\cdot\frac{a-1}{\sqrt{a}}\\ =4a\)
b)
\(\sqrt{A}>A\Leftrightarrow A-\sqrt{A}< 0\\ \Leftrightarrow4a-\sqrt{4a}< 0\\ \Leftrightarrow4a-2\sqrt{a}< 0\\ \Leftrightarrow2\sqrt{a}\left(2\sqrt{a}-1\right)< 0\\ \Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}2\sqrt{a}< 0\\2\sqrt{a}-1>0\end{matrix}\right.\\\left\{{}\begin{matrix}2\sqrt{a}>0\\2\sqrt{a}-1< 0\end{matrix}\right.\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}a< 0\left(ktm\right)\\a>\frac{1}{4}\end{matrix}\right.\\\left\{{}\begin{matrix}a>0\\a< \frac{1}{4}\left(tm\right)\end{matrix}\right.\end{matrix}\right.\)
Vậy với \(0< a< \frac{1}{4}\)thì \(\sqrt{A}>A\)