Đáp án:
$\begin{array}{l}
75)\dfrac{{\sqrt x + \sqrt y }}{{\sqrt x - \sqrt y }}\\
= \dfrac{{{{\left( {\sqrt x + \sqrt y } \right)}^2}}}{{\left( {\sqrt x - \sqrt y } \right)\left( {\sqrt x + \sqrt y } \right)}}\\
= \dfrac{{x + 2\sqrt {xy} + y}}{{x - y}}\\
76)\dfrac{{\sqrt a + 1}}{{\sqrt a - 1}}\\
= \dfrac{{{{\left( {\sqrt a + 1} \right)}^2}}}{{a - 1}}\\
= \dfrac{{a + 2\sqrt a + 1}}{{a - 1}}\\
77)\dfrac{1}{{\sqrt 3 + \sqrt 2 + 1}}\\
= \dfrac{{\sqrt 3 + \sqrt 2 - 1}}{{{{\left( {\sqrt 3 + \sqrt 2 } \right)}^2} - 1}}\\
= \dfrac{{\sqrt 3 + \sqrt 2 - 1}}{{4 + 2\sqrt 6 }}\\
= \dfrac{{\left( {\sqrt 3 + \sqrt 2 - 1} \right)\left( {4 - 2\sqrt 6 } \right)}}{{{4^2} - {{\left( {2\sqrt 6 } \right)}^2}}}\\
= \dfrac{{4\sqrt 3 - 6\sqrt 2 + 4\sqrt 2 - 4\sqrt 3 - 4 + 2\sqrt 6 }}{{16 - 24}}\\
= \dfrac{{2\sqrt 6 - 2\sqrt 2 - 4}}{{ - 8}}\\
= \dfrac{{2 + \sqrt 2 - \sqrt 6 }}{4}
\end{array}$
