Đáp án:
$\begin{array}{l}
a)Dkxd:a > 0;a \ne 1\\
\left( {\dfrac{{1 - a\sqrt a }}{{1 - \sqrt a }} + \sqrt a } \right){\left( {\dfrac{{1 - \sqrt a }}{{1 - a}}} \right)^2}\\
= \left( {\dfrac{{\left( {1 - \sqrt a } \right)\left( {1 + \sqrt a + a} \right)}}{{1 - \sqrt a }} + \sqrt a } \right){\left( {\dfrac{{1 - \sqrt a }}{{\left( {1 - \sqrt a } \right)\left( {1 + \sqrt a } \right)}}} \right)^2}\\
= \left( {1 + \sqrt a + a + \sqrt a } \right).\dfrac{1}{{{{\left( {1 + \sqrt a } \right)}^2}}}\\
= {\left( {1 + \sqrt a } \right)^2}.\dfrac{1}{{{{\left( {1 + \sqrt a } \right)}^2}}}\\
= 1\\
c)\left( {\dfrac{{\sqrt {14} - \sqrt 7 }}{{1 - \sqrt 2 }} + \dfrac{{\sqrt {15} - \sqrt 5 }}{{1 - \sqrt 3 }}} \right):\dfrac{1}{{\sqrt 7 - \sqrt 5 }}\\
= \left( {\dfrac{{\sqrt 7 \left( {\sqrt 2 - 1} \right)}}{{1 - \sqrt 2 }} + \dfrac{{\sqrt 5 \left( {\sqrt 3 - 1} \right)}}{{1 - \sqrt 3 }}} \right).\left( {\sqrt 7 - \sqrt 5 } \right)\\
= \left( { - \sqrt 7 - \sqrt 5 } \right).\left( {\sqrt 7 - \sqrt 5 } \right)\\
= - \left( {\sqrt 7 + \sqrt 5 } \right).\left( {\sqrt 7 - \sqrt 5 } \right)\\
= - \left( {7 - 5} \right)\\
= - 2
\end{array}$
