Đáp án:
$\begin{array}{l}
a)C = \left( {x - y} \right).\sqrt {\dfrac{{xy}}{{{{\left( {x - y} \right)}^2}}}} \left( {xy > 0} \right)\\
= \left( {x - y} \right).\dfrac{{\sqrt {xy} }}{{\left| {x - y} \right|}}\\
= \left[ \begin{array}{l}
\left( {x - y} \right).\dfrac{{\sqrt {xy} }}{{\left( {x - y} \right)}} = \sqrt {xy} \left( {khi:x > y} \right)\\
\left( {x - y} \right).\dfrac{{\sqrt {xy} }}{{y - x}} = - \sqrt {xy} \left( {khi:x < y} \right)
\end{array} \right.\\
b)D = \sqrt {\dfrac{{9 - 6x + {x^2}}}{{25{y^2}}}} = \dfrac{{\sqrt {{{\left( {x - 3} \right)}^2}} }}{{\sqrt {{{\left( {5y} \right)}^2}} }}\\
= \dfrac{{\left| {x - 3} \right|}}{{\left| {5y} \right|}}\\
= \left[ \begin{array}{l}
\dfrac{{x - 3}}{{5y}}\left( {khi:\left\{ \begin{array}{l}
x \ge 3\\
y > 0
\end{array} \right./\left\{ \begin{array}{l}
x \le 3\\
y < 0
\end{array} \right.} \right)\\
\dfrac{{3 - x}}{{5y}}\left( {khi:\left\{ \begin{array}{l}
x \le 3\\
y > 0
\end{array} \right./\left\{ \begin{array}{l}
x \ge 3\\
y < 0
\end{array} \right.} \right)
\end{array} \right.\\
e)E = \dfrac{3}{{2\left( {2x - 1} \right)}}.\sqrt {8{x^4}\left( {4x - 2x + 1} \right)} \\
= \dfrac{3}{{2\left( {2x - 1} \right)}}.2\sqrt 2 .{x^2}.\sqrt {2x + 1} \\
= \dfrac{{3\sqrt 2 {x^2}.\sqrt {2x + 1} }}{{2x - 1}}\\
d)D = \dfrac{2}{{2a - 1}}.\sqrt {5{a^2}.\left( {1 - 4a + 4{a^2}} \right)} \\
= \dfrac{2}{{2a - 1}}.\sqrt 5 .\left| a \right|.\sqrt {{{\left( {2a - 1} \right)}^2}} \\
= \dfrac{{2\sqrt 5 .\left| a \right|\left| {2a - 1} \right|}}{{2a - 1}}\\
= \left[ \begin{array}{l}
2\sqrt 5 a\left( {khi:a > \dfrac{1}{2}/a < 0} \right)\\
- 2\sqrt 5 a\left( {khi:0 \le a < \dfrac{1}{2}} \right)
\end{array} \right.
\end{array}$
