Câu $ a) $ Ct dài quá nên mình viết ra giấy nha.
$b$)
Để $B> A $ => $\frac{4}{3-\sqrt{x}} > 1 $
<=> $ \frac{4}{3-\sqrt{x}} -1 >0 $
<=> $\frac{4}{3-\sqrt{x}} - \frac{3-\sqrt{x}}{3-\sqrt{x}} > 0 $
<=> $\frac{4-3+\sqrt{x}}{3-\sqrt{x}} >0 $
<=> $\frac{1+\sqrt{x}}{3-\sqrt{x}} > 0$
=> $\left[\begin{array}{}\left\{{{\sqrt{x}+1>0}\atop{3-\sqrt{x}>0}}\right.\\\left\{{{\sqrt{x}+1<0}\atop{3-\sqrt{x}<0}}\right.\end{array}\right.$
$\Leftrightarrow\left[\begin{array}{}\left\{{{\sqrt{x}>-1}\atop{\sqrt{x}<3}}\right.\\\left\{{{\sqrt{x}<-1\text{ (loại)}}\atop{\sqrt{x}<3}}\right.\end{array}\right.$
$\Leftrightarrow \left\{{{x>1}\atop{x<9}}\right.$
Vậy $1<x<9 $ thì thoả mãn $B> A$
