Đáp án:
$\begin{array}{l}
1)\\
a)A = \dfrac{{\sqrt 2 }}{{\sqrt 2 - 2}} + \dfrac{{\sqrt {32} + 4}}{{\sqrt 2 + 1}}\\
= \dfrac{1}{{1 - \sqrt 2 }} + \dfrac{{4\sqrt 2 + 4}}{{\sqrt 2 + 1}}\\
= \dfrac{{1 + \sqrt 2 }}{{\left( {1 - \sqrt 2 } \right)\left( {1 + \sqrt 2 } \right)}} + \dfrac{{4\left( {\sqrt 2 + 1} \right)}}{{\sqrt 2 + 1}}\\
= \dfrac{{1 + \sqrt 2 }}{{1 - 2}} + 4\\
= - 1 - \sqrt 2 + 4\\
= 3 - \sqrt 2 \\
b)a \ge 0;a \ne 1\\
B = \left( {1 + \dfrac{{a + \sqrt a }}{{\sqrt a + 1}}} \right).\left( {1 - \dfrac{{a - \sqrt a }}{{\sqrt a - 1}}} \right)\\
= \left( {1 + \dfrac{{\sqrt a \left( {\sqrt a + 1} \right)}}{{\sqrt a + 1}}} \right).\left( {1 - \dfrac{{\sqrt a \left( {\sqrt a - 1} \right)}}{{\sqrt a - 1}}} \right)\\
= \left( {1 + \sqrt a } \right)\left( {1 - \sqrt a } \right)\\
= 1 - a\\
A > B\\
\Rightarrow 3 - \sqrt 2 > 1 - a\\
\Rightarrow a > \sqrt 2 - 2\\
Do:\sqrt 2 - 2 < 0\\
\Rightarrow a > \sqrt 2 - 2\left( {luôn\,đúng} \right)\\
Vậy\,A > B
\end{array}$