Đáp án:
$\begin{array}{l}
B = \left( {\dfrac{6}{{a - 1}} + \dfrac{{10 - 2\sqrt a }}{{a\sqrt a - a - \sqrt a + 1}}} \right).\dfrac{{{{\left( {\sqrt a - 1} \right)}^2}}}{{4\sqrt a }}\\
= \left( {\dfrac{6}{{a - 1}} + \dfrac{{10 - 2\sqrt a }}{{\left( {a - 1} \right)\left( {\sqrt a - 1} \right)}}} \right).\dfrac{{{{\left( {\sqrt a - 1} \right)}^2}}}{{4\sqrt a }}\\
= \dfrac{{6\left( {\sqrt a - 1} \right) + 10 - 2\sqrt a }}{{\left( {\sqrt a - 1} \right)\left( {a - 1} \right)}}.\dfrac{{{{\left( {\sqrt a - 1} \right)}^2}}}{{4\sqrt a }}\\
= \dfrac{{4\sqrt a + 4}}{{a - 1}}.\dfrac{{\left( {\sqrt a - 1} \right)}}{{4\sqrt a }}\\
= \dfrac{{4\left( {\sqrt a + 1} \right)\left( {\sqrt a - 1} \right)}}{{\left( {\sqrt a - 1} \right)\left( {\sqrt a + 1} \right).4\sqrt a }}\\
= \dfrac{1}{{\sqrt a }}
\end{array}$
