Đáp án:
$\begin{array}{l}
1)a)\left( {1 + \dfrac{{a + \sqrt a }}{{\sqrt a + 1}}} \right)\left( {1 - \dfrac{{a - \sqrt a }}{{\sqrt a - 1}}} \right)\\
= \left( {1 + \dfrac{{\sqrt a \left( {\sqrt a + 1} \right)}}{{\sqrt a + 1}}} \right)\left( {1 - \dfrac{{\sqrt a \left( {\sqrt a - 1} \right)}}{{\sqrt a - 1}}} \right)\\
= \left( {1 + \sqrt a } \right)\left( {1 - \sqrt a } \right)\\
= 1 - a\\
b)\dfrac{{a\sqrt b + b\sqrt a }}{{\sqrt {ab} }}:\dfrac{1}{{\sqrt a - \sqrt b }}\\
= \dfrac{{\sqrt {ab} \left( {\sqrt a + \sqrt b } \right)}}{{\sqrt {ab} }}.\left( {\sqrt a - \sqrt b } \right)\\
= \left( {\sqrt a + \sqrt b } \right)\left( {\sqrt a - \sqrt b } \right)\\
= a - b\\
c)\left( {\dfrac{{x\sqrt x - y\sqrt y }}{{\sqrt x - \sqrt y }} + \sqrt {xy} } \right).{\left( {\dfrac{{\sqrt x - \sqrt y }}{{x - y}}} \right)^2}\\
= \left( {\dfrac{{\left( {\sqrt x - \sqrt y } \right)\left( {x + \sqrt {xy} + y} \right)}}{{\sqrt x - \sqrt y }} + \sqrt {xy} } \right).{\left( {\dfrac{1}{{\sqrt x + \sqrt y }}} \right)^2}\\
= \left( {x + 2\sqrt {xy} + 1} \right).\dfrac{1}{{{{\left( {\sqrt x + \sqrt y } \right)}^2}}}\\
= 1\\
B2)\\
\left( {\dfrac{{\sqrt 6 - \sqrt 2 }}{{1 - \sqrt 3 }} - \dfrac{{\sqrt 5 - 5}}{{1 - \sqrt 5 }}} \right):\dfrac{1}{{\sqrt 2 - \sqrt 5 }}\\
= \left( {\dfrac{{\sqrt 2 \left( {\sqrt 3 - 1} \right)}}{{1 - \sqrt 3 }} - \dfrac{{\sqrt 5 \left( {1 - \sqrt 5 } \right)}}{{1 - \sqrt 5 }}} \right).\left( {\sqrt 2 - \sqrt 5 } \right)\\
= \left( { - \sqrt 2 - \sqrt 5 } \right)\left( {\sqrt 2 - \sqrt 5 } \right)\\
= - \left( {\sqrt 2 + \sqrt 5 } \right)\left( {\sqrt 2 - \sqrt 5 } \right)\\
= - \left( {2 - 5} \right)\\
= 3
\end{array}$
