Đáp án:
$\begin{array}{l}
a)A = \left( {\frac{{x\sqrt x + 1}}{{x - 1}} - \frac{{x - 1}}{{\sqrt x - 1}}} \right):\left( {\sqrt x + \frac{{\sqrt x }}{{\sqrt x - 1}}} \right)\\
= \left[ {\frac{{{{\left( {\sqrt x } \right)}^3} + 1}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}} - \frac{{x - 1}}{{\sqrt x - 1}}} \right]:\\
\frac{{\sqrt x \left( {\sqrt x - 1} \right) + \sqrt x }}{{\sqrt x - 1}}\\
= \left[ {\frac{{\left( {\sqrt x + 1} \right)\left( {x - \sqrt x + 1} \right)}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}} - \frac{{x - 1}}{{\sqrt x - 1}}} \right].\frac{{\sqrt x - 1}}{{x - \sqrt x + \sqrt x }}\\
= \frac{{x - \sqrt x + 1 - x + 1}}{{\sqrt x - 1}}.\frac{{\sqrt x - 1}}{x}\\
= \frac{{2 - \sqrt x }}{x}\\
b)x > 0;x \ne 1\\
A > 0\\
\Rightarrow \frac{{2 - \sqrt x }}{x} > 0\\
\Rightarrow 2 - \sqrt x > 0\left( {do:x > 0} \right)\\
\Rightarrow \sqrt x < 2\\
\Rightarrow x < 4\\
Vậy\,0 < x < 4;x \ne 1
\end{array}$
