Đáp án:

$\begin{array}{l}
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)\left( {\dfrac{{1 - a\sqrt a }}{{1 - \sqrt a }} + \sqrt a } \right){\left( {\dfrac{{1 - \sqrt a }}{{1 - a}}} \right)^2}\\
 = \left( {\dfrac{{\left( {1 - \sqrt a } \right)\left( {1 + \sqrt a  + a} \right)}}{{1 - \sqrt a }} + \sqrt a } \right).{\left( {\dfrac{1}{{1 + \sqrt a }}} \right)^2}\\
 = \left( {1 + \sqrt a  + a + \sqrt a } \right)\dfrac{1}{{{{\left( {1 + \sqrt a } \right)}^2}}}\\
 = \left( {a + 2\sqrt a  + 1} \right).\dfrac{1}{{a + 2\sqrt a  + 1}}\\
 = 1\\
c)\left( {\dfrac{1}{{a + \sqrt a }} + \dfrac{1}{{a - \sqrt a }}} \right):\dfrac{{\sqrt a  + 1}}{{a - 2\sqrt a  + 1}}\\
 = \dfrac{{\sqrt a  - 1 + \sqrt a  + 1}}{{\sqrt a \left( {\sqrt a  + 1} \right)\left( {\sqrt a  - 1} \right)}}.\dfrac{{{{\left( {\sqrt a  - 1} \right)}^2}}}{{\sqrt a  + 1}}\\
 = \dfrac{{2\sqrt a }}{{\sqrt a \left( {\sqrt a  + 1} \right)}}.\dfrac{{\sqrt a  - 1}}{{\sqrt a  + 1}}\\
 = \dfrac{{2\left( {\sqrt a  - 1} \right)}}{{{{\left( {\sqrt a  + 1} \right)}^2}}}\\
d)\dfrac{{{{\left( {\sqrt a  - \sqrt b } \right)}^2} + 4\sqrt {ab} }}{{\sqrt a  + \sqrt b }} - \dfrac{{a\sqrt b  - b\sqrt a }}{{\sqrt {ab} }}\\
 = \dfrac{{{{\left( {\sqrt a  + \sqrt b } \right)}^2}}}{{\sqrt a  + \sqrt b }} - \dfrac{{\sqrt {ab} \left( {\sqrt a  - \sqrt b } \right)}}{{\sqrt {ab} }}\\
 = \sqrt a  + \sqrt b  - \left( {\sqrt a  - \sqrt b } \right)\\
 = 2\sqrt b \\
e)\dfrac{{x + \sqrt {xy} }}{{\sqrt x  + \sqrt y }} - \dfrac{{x - y}}{{\sqrt x  - \sqrt y }}\\
 = \dfrac{{\sqrt x \left( {\sqrt x  + \sqrt y } \right)}}{{\sqrt x  + \sqrt y }} - \dfrac{{\left( {\sqrt x  - \sqrt y } \right)\left( {\sqrt x  + \sqrt y } \right)}}{{\sqrt x  - \sqrt y }}\\
 = \sqrt x  - \left( {\sqrt x  + \sqrt y } \right)\\
 =  - \sqrt y \\
g)\dfrac{{{{\left( {\sqrt a  + 1} \right)}^2} - 4\sqrt a }}{{\sqrt a  - 1}} + \dfrac{{a + \sqrt a }}{{\sqrt a }}\\
 = \dfrac{{{{\left( {\sqrt a  - 1} \right)}^2}}}{{\sqrt a  - 1}} + \dfrac{{\sqrt a \left( {\sqrt a  + 1} \right)}}{{\sqrt a }}\\
 = \sqrt a  - 1 + \sqrt a  + 1\\
 = 2\sqrt a \\
B2)\\
a)\sqrt {6 - 2\sqrt 5 }  - \sqrt {{{\left( {1 - 3\sqrt 5 } \right)}^2}} \\
 = \sqrt {{{\left( {\sqrt 5  - 1} \right)}^2}}  - \left( {3\sqrt 5  - 1} \right)\\
 = \sqrt 5  - 1 - 3\sqrt 5  + 1\\
 =  - 2\sqrt 5 \\
b)\left( {\sqrt {10}  + \dfrac{{5 - 2\sqrt 5 }}{{2 - \sqrt 5 }}} \right)\left( {\dfrac{{10 - \sqrt {10} }}{{\sqrt {10}  - 1}} + \sqrt 5 } \right)\\
 = \left( {\sqrt {10}  - \sqrt 5 } \right)\left( {\sqrt {10}  + \sqrt 5 } \right)\\
 = 10 - 5\\
 = 5
\end{array}$