a, $A = \frac{x-1}{2\sqrt{x}}\left(\frac{x-\sqrt{x}}{\sqrt{x}+1}-\frac{x+\sqrt{x}}{\sqrt{x}-1}\right)$
$= \frac{x-1}{2\sqrt{x}}.\left[\frac{\left(x-\sqrt{x}\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}-\frac{\left(x+\sqrt{x}\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right]$
$= \frac{x-1}{2\sqrt{x}}.\frac{\left(x-\sqrt{x}\right)\left(\sqrt{x}-1\right)-\left(x+\sqrt{x}\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}$
$= \frac{x-1}{2\sqrt{x}}.\left[\frac{{x\sqrt{x}-2x+\sqrt{x}-x\sqrt{x}-2x-\sqrt{x}} }{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right]$
$=\frac{x-1}{2\sqrt{x}}.\frac{-4x}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}$
$=-\frac{x-1}{2\sqrt{x}}.\frac{4x}{x-1}$
$=-2\sqrt{x}$
Vậy $A=-2\sqrt{x}$
b, Để `A > -6 <=> -2\sqrtx > -6`
`<=>2sqrtx<6`
`<=> sqrtx < 3`
`=> 0 < x < 9`
Vậy `0 < x < 9 <=> A > -6`
