Đáp án:
$\begin{array}{l}
c)\dfrac{1}{{a - 3b}}.\sqrt {{a^6}{{\left( {a - 3b} \right)}^2}} \\
= \dfrac{1}{{a - 3b}}.\left| {{a^3}} \right|.\left| {a - 3b} \right|\\
= \left[ \begin{array}{l}
\dfrac{{{a^3}.\left( {a - 3b} \right)}}{{a - 3b}} = {a^3}\left( \begin{array}{l}
khi:\left\{ \begin{array}{l}
a \ge 0\\
a > 3b
\end{array} \right.\,\\
hoac\,\left\{ \begin{array}{l}
a \le 0\\
a < 3b
\end{array} \right.
\end{array} \right)\\
\dfrac{{{a^3}\left( {3b - a} \right)}}{{a - 3b}} = - {a^3}\left( {khi:\left\{ \begin{array}{l}
a \ge 0\\
a < 3b
\end{array} \right.} \right)
\end{array} \right.
\end{array}$
