Đáp án:

a) \(A = \dfrac{{ - 5x - 4}}{{\left( {x - 3} \right)\left( {x + 2} \right)}}\)

Giải thích các bước giải:

\(\begin{array}{l}
a)DK:x \ne \left\{ { - 2;3} \right\}\\
A = \dfrac{{3\left( {x - 3} \right) - 4x\left( {x + 2} \right) + 4{x^2} + 5}}{{\left( {x - 3} \right)\left( {x + 2} \right)}}\\
 = \dfrac{{3x - 9 - 4{x^2} - 8x + 4{x^2} + 5}}{{\left( {x - 3} \right)\left( {x + 2} \right)}}\\
 = \dfrac{{ - 5x - 4}}{{\left( {x - 3} \right)\left( {x + 2} \right)}}\\
b)3x + 3 = 0\\
 \to x =  - 1\\
Thay:x =  - 1\\
 \to A = \dfrac{{ - 5.\left( { - 1} \right) - 4}}{{\left( { - 1 - 3} \right)\left( { - 1 + 2} \right)}} =  - \dfrac{1}{4}\\
c)A =  - 3\\
 \to \dfrac{{ - 5x - 4}}{{\left( {x - 3} \right)\left( {x + 2} \right)}} =  - 3\\
 \to  - 5x - 4 =  - 3\left( {{x^2} - x - 6} \right)\\
 \to  - 3{x^2} + 3x + 18 =  - 5x - 4\\
 \to 3{x^2} - 8x - 22 = 0\\
 \to 3{x^2} - 2.x\sqrt 3 .\dfrac{4}{{\sqrt 3 }} + \dfrac{{16}}{3} - \dfrac{{82}}{3} = 0\\
 \to {\left( {x\sqrt 3  - \dfrac{4}{{\sqrt 3 }}} \right)^2} = \dfrac{{82}}{3}\\
 \to \left[ \begin{array}{l}
x\sqrt 3  - \dfrac{4}{{\sqrt 3 }} = \sqrt {\dfrac{{82}}{3}} \\
x\sqrt 3  - \dfrac{4}{{\sqrt 3 }} =  - \sqrt {\dfrac{{82}}{3}} 
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x = \dfrac{{4 + \sqrt {82} }}{3}\\
x = \dfrac{{4 - \sqrt {82} }}{3}
\end{array} \right.\\
d)A = \dfrac{{ - 5x - 4}}{{\left( {x - 3} \right)\left( {x + 2} \right)}}\\
 =  - \dfrac{{5\left( {x - 3} \right) + 19}}{{\left( {x - 3} \right)\left( {x + 2} \right)}}\\
 =  - \dfrac{5}{{x + 2}} - \dfrac{{19}}{{{x^2} - x - 6}}\\
A \in Z\\
 \to \left\{ \begin{array}{l}
\dfrac{5}{{x + 2}} \in Z\\
\dfrac{{19}}{{{x^2} - x - 6}} \in Z
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
x + 2 \in U\left( 5 \right)\\
{x^2} - x - 6 \in U\left( {19} \right)
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
\left[ \begin{array}{l}
x + 2 = 5\\
x + 2 =  - 5\\
x + 2 = 1\\
x + 2 =  - 1
\end{array} \right.\\
\left[ \begin{array}{l}
{x^2} - x - 6 = 19\\
{x^2} - x - 6 =  - 19\left( {vonghiem} \right)\\
{x^2} - x - 6 = 1\\
{x^2} - x - 6 =  - 1
\end{array} \right.
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
\left[ \begin{array}{l}
x = 3\\
x =  - 7\\
x =  - 1\\
x =  - 3
\end{array} \right.\\
\left[ \begin{array}{l}
x = \dfrac{{1 \pm \sqrt {101} }}{2}\\
x = \dfrac{{1 \pm \sqrt {29} }}{2}\\
x = \dfrac{{1 \pm \sqrt {21} }}{2}
\end{array} \right.
\end{array} \right.\\
 \to x \in \emptyset 
\end{array}\)