Đáp án:
$\begin{array}{l}
Đkxđ:\left\{ \begin{array}{l}
x > 0\\
x \ne 1
\end{array} \right.\\
Q = \left( {\frac{1}{{\sqrt x + 1}} - \frac{1}{{x + \sqrt x }}} \right):\frac{{\sqrt x - 1}}{{x + 2\sqrt x + 1}}\\
= \left[ {\frac{1}{{\sqrt x + 1}} - \frac{1}{{\sqrt x \left( {\sqrt x + 1} \right)}}} \right].\frac{{{{\left( {\sqrt x + 1} \right)}^2}}}{{\sqrt x - 1}}\\
= \frac{{\sqrt x - 1}}{{\sqrt x \left( {\sqrt x + 1} \right)}}.\frac{{{{\left( {\sqrt x + 1} \right)}^2}}}{{\sqrt x - 1}}\\
= \frac{{\sqrt x + 1}}{{\sqrt x }}\\
= \frac{{\sqrt x }}{{\sqrt x }} + \frac{1}{{\sqrt x }}\\
= 1 + \frac{1}{{\sqrt x }}\\
b)Do:\sqrt x > 0\forall x > 0;x \ne 1\\
\Rightarrow \frac{1}{{\sqrt x }} > 0\forall x\\
\Rightarrow 1 + \frac{1}{{\sqrt x }} > 1\forall x\\
\Rightarrow Q > 1
\end{array}$
