Đáp án:
$\begin{array}{l}
a)\dfrac{{{{\left( {\sqrt x - \sqrt y } \right)}^2} + 4\sqrt {xy} }}{{{{\left( {\sqrt x + \sqrt y } \right)}^2} - 4\sqrt {xy} }}.\dfrac{{x - y}}{{{{\left( {\sqrt x + \sqrt y } \right)}^2}}}\\
= \dfrac{{{{\left( {\sqrt x + \sqrt y } \right)}^2}}}{{{{\left( {\sqrt x - \sqrt y } \right)}^2}}}.\dfrac{{\left( {\sqrt x + \sqrt y } \right).\left( {\sqrt x - \sqrt y } \right)}}{{{{\left( {\sqrt x + \sqrt y } \right)}^2}}}\\
= \dfrac{{\sqrt x + \sqrt y }}{{\sqrt x - \sqrt y }}\\
b)\left( {3 + \dfrac{{a - 2\sqrt a }}{{\sqrt a - 2}}} \right)\left( {3 - \dfrac{{3a + \sqrt a }}{{3\sqrt a + 1}}} \right)\\
= \left( {3 + \dfrac{{\sqrt a \left( {\sqrt a - 2} \right)}}{{\sqrt a - 2}}} \right)\left( {3 - \dfrac{{\sqrt a \left( {3\sqrt a + 1} \right)}}{{3\sqrt a + 1}}} \right)\\
= \left( {3 + \sqrt a } \right)\left( {3 - \sqrt a } \right)\\
= 9 - a\\
c)\left( {\dfrac{{a\sqrt a + b\sqrt b }}{{\sqrt a + \sqrt b }} - \sqrt {ab} } \right){\left( {\dfrac{{\sqrt a + \sqrt b }}{{a - b}}} \right)^2}\\
= \left( {a - \sqrt {ab} + b - \sqrt {ab} } \right){\left( {\dfrac{1}{{\sqrt a - \sqrt b }}} \right)^2}\\
= {\left( {\sqrt a - \sqrt b } \right)^2}.\dfrac{1}{{{{\left( {\sqrt a - \sqrt b } \right)}^2}}}\\
= 1\\
d)\dfrac{{{{\left( {\sqrt x + \sqrt y } \right)}^2}}}{{x\sqrt y + y\sqrt x }}.\left( {\dfrac{{x - y}}{{\sqrt x - \sqrt y }} - \dfrac{{x\sqrt x - y\sqrt y }}{{x - y}}} \right)\\
= \dfrac{{\sqrt x + \sqrt y }}{{\sqrt {xy} }}.\left( {\sqrt x + \sqrt y - \dfrac{{x + \sqrt {xy} + y}}{{\sqrt x + \sqrt y }}} \right)\\
= \dfrac{{\sqrt x + \sqrt y }}{{\sqrt {xy} }}.\dfrac{{{{\left( {\sqrt x + \sqrt y } \right)}^2} - x - \sqrt {xy} - y}}{{\sqrt x + \sqrt y }}\\
= \dfrac{{\sqrt {xy} }}{{\sqrt {xy} }} = 1\\
e)\dfrac{{\sqrt x + 1}}{{x - 1}} - \dfrac{{x + 2}}{{x\sqrt x - 1}} - \dfrac{{\sqrt x + 1}}{{x + \sqrt x + 1}}\\
= \dfrac{1}{{\sqrt x - 1}} - \dfrac{{x + 2}}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}} - \dfrac{{\sqrt x + 1}}{{x + \sqrt x + 1}}\\
= \dfrac{{x + \sqrt x + 1 - x - 2 - \left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}\\
= \dfrac{{\sqrt x - 1 - x + 1}}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}\\
= \dfrac{{ - \sqrt x \left( {\sqrt x - 1} \right)}}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}\\
= \dfrac{{ - \sqrt x }}{{x + \sqrt x + 1}}
\end{array}$
