Giải thích các bước giải:
a.ĐKXĐ: $a\ge0, x\ne 1$
Ta có:
$A=(\dfrac{\sqrt{a}+1}{\sqrt{a} -1}-\dfrac{\sqrt{a} -1}{\sqrt{a} +1}+4\sqrt{a} )(\sqrt{a} -\dfrac{1}{\sqrt{a} })$
$\to A=\dfrac{(\sqrt{a}+1)^2-(\sqrt{a} -1)^2+4\sqrt{a}(\sqrt{a} -1)(\sqrt{a} +1)}{(\sqrt{a} -1)(\sqrt{a} +1)}\cdot \dfrac{a-1}{\sqrt{a} }$
$\to A=\dfrac{(a+2\sqrt{a}+1)-(a-2\sqrt{a} +1)+4\sqrt{a}(a-1)}{(\sqrt{a} -1)(\sqrt{a} +1)}\cdot \dfrac{a-1}{\sqrt{a} }$
$\to A=\dfrac{4\sqrt{a}+4\sqrt{a}(a-1)}{a-1}\cdot \dfrac{a-1}{\sqrt{a} }$
$\to A=\dfrac{4\sqrt{a}(a-1+1)}{a-1}\cdot \dfrac{a-1}{\sqrt{a} }$
$\to A=\dfrac{4a\sqrt{a}}{a-1}\cdot \dfrac{a-1}{\sqrt{a} }$
$\to A=4a$
b.Để $\sqrt{A}>A$
$\to \begin{cases}A\ge 0\\ A>A^2\end{cases}$
$\to \begin{cases}4a\ge 0\\ A^2-A<0\end{cases}$
$\to \begin{cases}a\ge 0\\ A(A-1)<0\end{cases}$
$\to \begin{cases}a\ge 0\\ 0<A<1\end{cases}$
$\to \begin{cases}a\ge 0\\ 0<4a<1\end{cases}$
$\to \begin{cases}a\ge 0\\ 0<a<\dfrac14\end{cases}$
$\to 0<a<\dfrac14$
