Đáp án:
$\begin{array}{l}
B = \left( {\sqrt {14 - 6\sqrt 5 } + 2\sqrt {9 + 4\sqrt 5 } } \right).\sqrt {{{\left( {\sqrt 5 - 7} \right)}^2}} \\
= \left( {\sqrt {9 - 2.3.\sqrt 5 + 5} + 2.\sqrt {5 + 2.2.\sqrt 5 + 4} } \right).\left| {\sqrt 5 - 7} \right|\\
= \left( {\sqrt {{{\left( {3 - \sqrt 5 } \right)}^2}} + 2.\sqrt {{{\left( {\sqrt 5 + 2} \right)}^2}} } \right).\left( {7 - \sqrt 5 } \right)\\
= \left( {3 - \sqrt 5 + 2\sqrt 5 + 4} \right).\left( {7 - \sqrt 5 } \right)\\
= \left( {7 + \sqrt 5 } \right).\left( {7 - \sqrt 5 } \right)\\
= {7^2} - 5\\
= 44\\
B2)\\
A = \left( {\dfrac{{\sqrt x - 2}}{{1 - x}} + \dfrac{{\sqrt x + 2}}{{1 + 2\sqrt x + x}}} \right):\dfrac{{2\sqrt x }}{{1 + \sqrt x }}\\
= \left( {\dfrac{{2 - \sqrt x }}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}} + \dfrac{{\sqrt x + 2}}{{{{\left( {\sqrt x + 1} \right)}^2}}}} \right).\dfrac{{1 + \sqrt x }}{{2\sqrt x }}\\
= \dfrac{{\left( {2 - \sqrt x } \right)\left( {\sqrt x + 1} \right) + \left( {\sqrt x + 2} \right)\left( {\sqrt x - 1} \right)}}{{\left( {\sqrt x - 1} \right){{\left( {\sqrt x + 1} \right)}^2}}}.\dfrac{{\sqrt x + 1}}{{2\sqrt x }}\\
= \dfrac{{2\sqrt x + 2 - x - \sqrt x + x - \sqrt x + 2\sqrt x - 2}}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right).2\sqrt x }}\\
= \dfrac{{4\sqrt x }}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right).2\sqrt x }}\\
= \dfrac{2}{{x - 1}}
\end{array}$
