Đáp án:
$\begin{array}{l}
2)\dfrac{1}{{2 + \sqrt 3 }} - \dfrac{1}{{2 - \sqrt 3 }}\\
= \dfrac{{2 - \sqrt 3 - 2 - \sqrt 3 }}{{\left( {2 + \sqrt 3 } \right)\left( {2 - \sqrt 3 } \right)}}\\
= \dfrac{{ - 2\sqrt 3 }}{{{2^2} - 3}}\\
= - 2\sqrt 3 \\
3)\\
A = 1 + \dfrac{{3m}}{{m - 2}}\sqrt {{m^2} - 4m + 4} \\
= 1 + \dfrac{{3m}}{{m - 2}}.\sqrt {{{\left( {m - 2} \right)}^2}} \\
= 1 + \dfrac{{3m\left| {m - 2} \right|}}{{m - 2}}\\
= \left[ \begin{array}{l}
1 + 3m\left( {khi:m > 2} \right)\\
1 - 3m\left( {khi:m < 2} \right)
\end{array} \right.\\
B = \left( {1 + \dfrac{{a + \sqrt a }}{{\sqrt a + 1}}} \right)\left( {1 + \dfrac{{a - \sqrt a }}{{1 - \sqrt a }}} \right)\\
= \left( {1 + \sqrt a } \right)\left( {1 - \sqrt a } \right)\\
= 1 - a\\
C = \left( {\dfrac{{\sqrt x + 1}}{{\sqrt x - 1}} - \dfrac{{\sqrt x - 1}}{{\sqrt x + 1}}} \right):\dfrac{{2\sqrt x }}{{\sqrt x - 1}}\\
= \dfrac{{{{\left( {\sqrt x + 1} \right)}^2} - {{\left( {\sqrt x - 1} \right)}^2}}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}.\dfrac{{\sqrt x - 1}}{{2\sqrt x }}\\
= \dfrac{{x + 2\sqrt x + 1 - x + 2\sqrt x - 1}}{{\sqrt x + 1}}.\dfrac{1}{{2\sqrt x }}\\
= \dfrac{{4\sqrt x }}{{\sqrt x + 1}}.\dfrac{1}{{2\sqrt x }}\\
= \dfrac{2}{{\sqrt x + 1}}
\end{array}$
