Đáp án:
$\begin{array}{l}
a)A = \dfrac{2}{{{x^2} - {y^2}}}\sqrt {\dfrac{{9\left( {{x^2} + 2xy + {y^2}} \right)}}{4}} \\
= \dfrac{2}{{\left( {x - y} \right)\left( {x + y} \right)}}.\dfrac{{3\sqrt {{{\left( {x + y} \right)}^2}} }}{2}\\
= \dfrac{{3\left| {x + y} \right|}}{{\left( {x - y} \right)\left( {x + y} \right)}}\\
= \left[ \begin{array}{l}
\dfrac{3}{{x - y}}\left( {khi:x + y > 0} \right)\\
\dfrac{3}{{y - x}}\left( {khi:x + y < 0} \right)
\end{array} \right.\\
b)\\
A = \dfrac{1}{{2a - 1}}\sqrt {25{a^4} - 100{a^5} + 100{a^6}} \\
= \dfrac{1}{{2a - 1}}.\sqrt {25{a^4}\left( {1 - 4a + 4{a^2}} \right)} \\
= \dfrac{1}{{2a - 1}}.5{a^2}.\sqrt {{{\left( {2a - 1} \right)}^2}} \\
= \dfrac{{5{a^2}\left| {2a - 1} \right|}}{{2a - 1}}\\
= \left[ \begin{array}{l}
5{a^2}\left( {khi:a > \dfrac{1}{2}} \right)\\
- 5{a^2}\left( {khi:a < \dfrac{1}{2}} \right)
\end{array} \right.
\end{array}$
