Đáp án:
$\begin{array}{l}
a)\left( {\dfrac{{1 - a\sqrt a }}{{1 - \sqrt a }} + \sqrt a } \right){\left( {\dfrac{{1 - \sqrt a }}{{1 - a}}} \right)^2}\\
= \left( {\dfrac{{\left( {1 - \sqrt a } \right)\left( {1 + \sqrt a + a} \right)}}{{1 - \sqrt a }} + \sqrt a } \right).{\left( {\dfrac{1}{{1 + \sqrt a }}} \right)^2}\\
= \left( {1 + \sqrt a + a + \sqrt a } \right).\dfrac{1}{{{{\left( {\sqrt a + 1} \right)}^2}}}\\
= {\left( {\sqrt a + 1} \right)^2}.\dfrac{1}{{{{\left( {\sqrt a + 1} \right)}^2}}}\\
= 1\\
b)\left( {\dfrac{1}{{a - \sqrt a }} + \dfrac{1}{{\sqrt a - 1}}} \right):\dfrac{{\sqrt a + 1}}{{a - 2\sqrt a + 1}}\\
= \dfrac{{1 + \sqrt a }}{{\sqrt a \left( {\sqrt a - 1} \right)}}.\dfrac{{{{\left( {\sqrt a - 1} \right)}^2}}}{{\sqrt a + 1}}\\
= \dfrac{{\sqrt a - 1}}{{\sqrt a }}
\end{array}$
