Đáp án:

\(\begin{array}{l}
a)\left[ \begin{array}{l}
A = \dfrac{3}{{x - y}}\\
A =  - \dfrac{3}{{x - y}}
\end{array} \right.\\
b)\left[ \begin{array}{l}
B = 2a\sqrt 5 \\
B =  - 2a\sqrt 5 
\end{array} \right.\\
c)C =  - \sqrt {ab} \\
d)\left[ \begin{array}{l}
D =  - \dfrac{{2a + 3}}{b}\\
D = \dfrac{{2a + 3}}{b}
\end{array} \right.\\
e)\left[ \begin{array}{l}
E = 3x\\
E =  - 3x
\end{array} \right.
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
a)A = \dfrac{2}{{{x^2} - {y^2}}}.\dfrac{{3\left| {x + y} \right|}}{2}\\
 = \dfrac{{3\left| {x + y} \right|}}{{\left( {x - y} \right)\left( {x + y} \right)}}\\
 \to \left[ \begin{array}{l}
A = \dfrac{{3\left( {x + y} \right)}}{{\left( {x - y} \right)\left( {x + y} \right)}}\left( {DK:x \ge  - y} \right)\\
A =  - \dfrac{{3\left( {x + y} \right)}}{{\left( {x - y} \right)\left( {x + y} \right)}}\left( {DK:x <  - y} \right)
\end{array} \right.\\
 \to \left[ \begin{array}{l}
A = \dfrac{3}{{x - y}}\\
A =  - \dfrac{3}{{x - y}}
\end{array} \right.\\
b)B = \dfrac{2}{{2a - 1}}.\sqrt {5{a^2}{{\left( {2a - 1} \right)}^2}} \\
 = \dfrac{2}{{2a - 1}}.\left| {a.\left( {2a - 1} \right)} \right|.\sqrt 5 \\
 = \dfrac{{2a\sqrt 5 \left| {2a - 1} \right|}}{{2a - 1}}\\
 \to \left[ \begin{array}{l}
B = \dfrac{{2a\sqrt 5 \left( {2a - 1} \right)}}{{2a - 1}}\left( {DK:a \ge \dfrac{1}{2}} \right)\\
B =  - \dfrac{{2a\sqrt 5 \left( {2a - 1} \right)}}{{2a - 1}}\left( {DK:0 < a < \dfrac{1}{2}} \right)
\end{array} \right.\\
 \to \left[ \begin{array}{l}
B = 2a\sqrt 5 \\
B =  - 2a\sqrt 5 
\end{array} \right.\\
c)C = \left( {a - b} \right).\dfrac{{\sqrt {ab} }}{{\left| {a - b} \right|}}\\
 =  - \left( {a - b} \right).\dfrac{{\sqrt {ab} }}{{a - b}}\left( {do:a < b < 0} \right)\\
 =  - \sqrt {ab} \\
d)D = \sqrt {\dfrac{{{{\left( {3 + 2a} \right)}^2}}}{{{b^2}}}} \\
 = \left| {\dfrac{{3 + 2a}}{b}} \right| =  - \dfrac{{\left| {3 + 2a} \right|}}{b}\left( {do:b < 0} \right)\\
 \to \left[ \begin{array}{l}
D =  - \dfrac{{2a + 3}}{b}\left( {DK:a \ge  - \dfrac{3}{2}} \right)\\
D = \dfrac{{2a + 3}}{b}\left( {DK:a <  - \dfrac{3}{2}} \right)
\end{array} \right.\\
e)E = \dfrac{{3x}}{{x - 2}}.\sqrt {{{\left( {x - 2} \right)}^2}} \\
 = \dfrac{{3x}}{{x - 2}}.\left| {x - 2} \right|\\
 \to \left[ \begin{array}{l}
E = \dfrac{{3x}}{{x - 2}}.\left( {x - 2} \right)\left( {DK:x > 2} \right)\\
E =  - \dfrac{{3x}}{{x - 2}}.\left( {x - 2} \right)\left( {DK:x < 2} \right)
\end{array} \right.\\
 \to \left[ \begin{array}{l}
E = 3x\\
E =  - 3x
\end{array} \right.
\end{array}\)