Đáp án:
$\begin{array}{l}
5)\\
a)Đkxđ:x \ge 0;x \ne 1\\
A = \left( {\frac{{x + 2}}{{x\sqrt x - 1}} + \frac{{\sqrt x }}{{x + \sqrt x + 1}} + \frac{1}{{1 - \sqrt x }}} \right):\frac{{\sqrt x - 1}}{2}\\
= \left( {\frac{{x + 2}}{{{{\left( {\sqrt x } \right)}^3} - 1}} + \frac{{\sqrt x }}{{x + \sqrt x + 1}} - \frac{1}{{\sqrt x - 1}}} \right).\frac{2}{{\sqrt x - 1}}\\
= \left( {\frac{{x + 2}}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}} + \frac{{\sqrt x }}{{x + \sqrt x + 1}} - \frac{1}{{\sqrt x - 1}}} \right).\frac{2}{{\sqrt x - 1}}\\
= \frac{{x + 2 + \sqrt x \left( {\sqrt x - 1} \right) - \left( {x + \sqrt x + 1} \right)}}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}.\frac{2}{{\sqrt x - 1}}\\
= \frac{{x - 2\sqrt x + 1}}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}.\frac{2}{{\sqrt x - 1}}\\
= \frac{{{{\left( {\sqrt x - 1} \right)}^2}.2}}{{{{\left( {\sqrt x - 1} \right)}^2}.\left( {x + \sqrt x + 1} \right)}}\\
= \frac{2}{{x + \sqrt x + 1}}\\
b)DKXD:x \ge 0;x \ne 1\\
Do\,0 < x + \sqrt x + 1 = x + 2.\frac{1}{2}.\sqrt x + \frac{1}{4} + \frac{3}{4} = {\left( {\sqrt x + \frac{1}{2}} \right)^2} + \frac{3}{4} \le \frac{3}{4} < 1\\
\Rightarrow 0 < x + \sqrt x + 1 < 1\\
\Rightarrow 0 < A = \frac{2}{{x + \sqrt x + 1}} < \frac{2}{1} = 2\\
Vậy\,0 < A < 2
\end{array}$
