Đáp án:
$\begin{array}{l}
A = \dfrac{{\sqrt {15} - \sqrt {12} }}{{\sqrt 5 - 2}} - \dfrac{{6 + 2\sqrt 6 }}{{\sqrt 3 + \sqrt 2 }}\\
= \dfrac{{\sqrt 3 .\sqrt 5 - \sqrt 4 .\sqrt 3 }}{{\sqrt 5 - 2}} - \dfrac{{\sqrt 2 .\sqrt 3 .\sqrt 6 + \sqrt 2 .\sqrt 6 }}{{\sqrt 3 + \sqrt 2 }}\\
= \dfrac{{\sqrt 3 \left( {\sqrt 5 - 2} \right)}}{{\sqrt 5 - 2}} - \dfrac{{\sqrt 2 .\sqrt 6 \left( {\sqrt 3 + \sqrt 2 } \right)}}{{\sqrt 3 + \sqrt 2 }}\\
= \sqrt 3 - \sqrt 2 .\sqrt 6 \\
= \sqrt 3 - 2\sqrt 3 \\
= - \sqrt 3 \\
B = \left( {\dfrac{{3\sqrt x - 1}}{{x - 1}} - \dfrac{1}{{\sqrt x - 1}}} \right):\dfrac{1}{{x + \sqrt x }}\\
= \dfrac{{3\sqrt x - 1 - \left( {\sqrt x + 1} \right)}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}.\sqrt x \left( {\sqrt x + 1} \right)\\
= \dfrac{{2\sqrt x - 2}}{{\sqrt x - 1}}.\sqrt x \\
= \dfrac{{2\left( {\sqrt x - 1} \right)}}{{\sqrt x - 1}}.\sqrt x \\
= 2\sqrt x
\end{array}$
