Đáp án:
$\begin{array}{l}
1)\\
\frac{1}{{\sqrt 2 - 1}} - \frac{{3\sqrt 6 - 3\sqrt {10} }}{{\sqrt 3 - \sqrt 5 }} + \frac{4}{{\sqrt 2 }}\\
= \frac{{\sqrt 2 + 1}}{{\left( {\sqrt 2 - 1} \right)\left( {\sqrt 2 + 1} \right)}} - \frac{{3\sqrt 2 \left( {\sqrt 3 - \sqrt 5 } \right)}}{{\sqrt 3 - \sqrt 5 }} + \frac{{2.\sqrt 2 .\sqrt 2 }}{{\sqrt 2 }}\\
= \frac{{\sqrt 2 + 1}}{{2 - 1}} - 3\sqrt 2 + 2\sqrt 2 \\
= \sqrt 2 + 1 - 3\sqrt 2 + 2\sqrt 2 \\
= 1\\
2)\\
a)a > 0\\
Q = \left( {\frac{{2\sqrt 2 }}{{\sqrt a }} + \frac{{\sqrt a }}{{2 + \sqrt a }}} \right):\frac{{1 - a}}{{a + 4\sqrt a + 4}}\\
= \frac{{2\sqrt 2 .\left( {2 + \sqrt a } \right) + \sqrt a .\sqrt a }}{{\sqrt a \left( {2 + \sqrt a } \right)}}:\frac{{1 - a}}{{{{\left( {\sqrt a + 2} \right)}^2}}}\\
= \frac{{4\sqrt 2 + 2\sqrt 2 .\sqrt a + a}}{{\sqrt a \left( {2 + \sqrt a } \right)}}.\frac{{{{\left( {\sqrt a + 2} \right)}^2}}}{{1 - a}}\\
= \frac{{\left( {a + 2\sqrt 2 a + 4\sqrt 2 } \right)\left( {a + \sqrt 2 } \right)}}{{\sqrt a \left( {1 - a} \right)}}\\
b)a > 0\\
Q < 0\\
\Rightarrow \frac{{\left( {a + 2\sqrt 2 a + 4\sqrt 2 } \right)\left( {a + \sqrt 2 } \right)}}{{\sqrt a \left( {1 - a} \right)}} < 0\\
\Rightarrow 1 - a < 0\left( {do:\left\{ \begin{array}{l}
\left( {a + 2\sqrt 2 a + 4\sqrt 2 } \right)\left( {a + \sqrt 2 } \right) > 0\\
\sqrt a > 0
\end{array} \right.} \right)\\
\Rightarrow a > 1
\end{array}$
