Đáp án:
$\begin{array}{l}
a)\dfrac{{ - 138}}{{\sqrt 2 - 4\sqrt 3 }}\\
= \dfrac{{ - 138\left( {\sqrt 2 + 4\sqrt 3 } \right)}}{{2 - 48}}\\
= \dfrac{{ - 138}}{{ - 46}}\left( {\sqrt 2 + 4\sqrt 3 } \right)\\
= 3\sqrt 2 + 12\sqrt 3 \\
b)\dfrac{3}{{2\sqrt 7 - 3\sqrt 5 }} = \dfrac{{3\left( {2\sqrt 7 + 3\sqrt 5 } \right)}}{{28 - 45}}\\
= \dfrac{{6\sqrt 7 + 9\sqrt 5 }}{{ - 17}}\\
= \dfrac{{ - 6\sqrt 7 - 9\sqrt 5 }}{{17}}\\
c)A = \dfrac{1}{{3 - \sqrt 8 }} - \dfrac{1}{{\sqrt 8 - \sqrt 7 }} + \dfrac{1}{{\sqrt 7 - \sqrt 6 }}\\
- \dfrac{1}{{\sqrt 6 - \sqrt 5 }} + \dfrac{1}{{\sqrt 5 - 2}}\\
= \dfrac{{3 + \sqrt 8 }}{{9 - 8}} - \dfrac{{\sqrt 8 + \sqrt 7 }}{{8 - 7}} + \dfrac{{\sqrt 7 + \sqrt 6 }}{{7 - 6}}\\
- \dfrac{{\sqrt 6 + \sqrt 5 }}{{6 - 5}} + \dfrac{{\sqrt 5 + 2}}{{5 - 4}}\\
= 3 + \sqrt 8 - \sqrt 8 - \sqrt 7 + \sqrt 7 + \sqrt 6 - \sqrt 6 - \sqrt 5 + \sqrt 5 + 2\\
= 3 + 2 = 5\\
d)\dfrac{1}{{\sqrt x - 1}} = \dfrac{{\sqrt x + 1}}{{x - 1}}\\
e)\dfrac{{x - y}}{{\sqrt {{x^2} - {y^2}} }} = \dfrac{{\left( {x - y} \right)\left( {\sqrt {{x^2} - {y^2}} } \right)}}{{{x^2} - {y^2}}} = \dfrac{{\sqrt {{x^2} - {y^2}} }}{{x + y}}\\
g)\dfrac{{3\sqrt 5 - 2\sqrt 2 }}{{2\sqrt 5 - 3\sqrt 2 }} = \dfrac{{\left( {3\sqrt 5 - 2\sqrt 2 } \right)\left( {2\sqrt 5 + 3\sqrt 2 } \right)}}{{20 - 18}}\\
= \dfrac{{6.5 + 9\sqrt {10} - 4\sqrt {10} + 6.2}}{2}\\
= \dfrac{{42 + 5\sqrt {10} }}{2}
\end{array}$
