Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
DKXD:\,\,\,\left\{ \begin{array}{l}
a > 0\\
a \ne 1
\end{array} \right.\\
a,\\
Q = \left( {\dfrac{1}{{\sqrt a + 1}} - \dfrac{1}{{a + \sqrt a }}} \right):\dfrac{{\sqrt a - 1}}{{a + 2\sqrt a + 1}}\\
= \left( {\dfrac{1}{{\sqrt a + 1}} - \dfrac{1}{{\sqrt a \left( {\sqrt a + 1} \right)}}} \right):\dfrac{{\sqrt a - 1}}{{{{\left( {\sqrt a + 1} \right)}^2}}}\\
= \dfrac{{\sqrt a - 1}}{{\sqrt a \left( {\sqrt a + 1} \right)}}:\dfrac{{\sqrt a - 1}}{{{{\left( {\sqrt a + 1} \right)}^2}}}\\
= \dfrac{{\sqrt a - 1}}{{\sqrt a \left( {\sqrt a + 1} \right)}}.\dfrac{{{{\left( {\sqrt a + 1} \right)}^2}}}{{\sqrt a - 1}}\\
= \dfrac{{\sqrt a + 1}}{{\sqrt a }}\\
b,\\
Q = \dfrac{{\sqrt a + 1}}{{\sqrt a }} = 1 + \dfrac{1}{{\sqrt a }}\\
\dfrac{1}{{\sqrt a }} > 0,\,\,\,\forall a > 0,a \ne 1\\
\Rightarrow 1 + \dfrac{1}{{\sqrt a }} > 1,\,\,\,\forall a > 0,a \ne 1\\
\Rightarrow Q > 1
\end{array}\)
