Đáp án:
$\begin{array}{l}
A = \frac{{4{x^2} + 12x + 9}}{{4{x^2} - 9}}\\
a)Đkxđ:4{x^2} - 9 \ne 0\\
\Rightarrow \left( {2x - 3} \right)\left( {2x + 3} \right) \ne 0\\
\Rightarrow \left\{ \begin{array}{l}
2x - 3 \ne 0\\
2x + 3 \ne 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
x \ne \frac{3}{2}\\
x \ne - \frac{3}{2}
\end{array} \right.\\
b)A = \frac{{4{x^2} + 12x + 9}}{{4{x^2} - 9}}\\
= \frac{{{{\left( {2x} \right)}^2} + 2.2x.3 + {3^2}}}{{\left( {2x - 3} \right)\left( {2x + 3} \right)}}\\
= \frac{{{{\left( {2x + 3} \right)}^2}}}{{\left( {2x - 3} \right)\left( {2x + 3} \right)}}\\
= \frac{{2x + 3}}{{2x - 3}}\\
c)x = - 1\frac{1}{2} = - \frac{{2.1 + 1}}{2} = - \frac{3}{2}\left( {ktm} \right)\\
Vậy\,x = - 1\frac{1}{2}\,thì\,ko\,có\,giá\,trị\,của\,A.\\
d)\left\{ \begin{array}{l}
x \ne \frac{3}{2}\\
x \ne - \frac{3}{2}
\end{array} \right.\\
A = \frac{1}{3}\\
\Rightarrow \frac{{2x + 3}}{{2x - 3}} = \frac{1}{3}\\
\Rightarrow 3\left( {2x + 3} \right) = 2x - 3\\
\Rightarrow 6x - 2x = - 3 - 9\\
\Rightarrow 4x = - 12\\
\Rightarrow x = - 3\left( {tmdk} \right)\\
e)\left\{ \begin{array}{l}
x \ne \frac{3}{2}\\
x \ne - \frac{3}{2}
\end{array} \right.\\
A = \frac{{2x + 3}}{{2x - 3}} = \frac{{2x - 3 + 6}}{{2x - 3}} = 1 + \frac{6}{{2x - 3}}\\
A \in Z \Rightarrow \frac{6}{{2x - 3}} \in Z\\
\Rightarrow \left( {2x - 3} \right) \in Ư\left( 6 \right)\\
Mà:2x - 3\,là\,số\,lẻ\\
\Rightarrow 2x - 3 \in {\rm{\{ }} - 3; - 1;1;3\} \\
\Rightarrow 2x \in {\rm{\{ }}0; - 2;4;6\} \\
\Rightarrow x \in {\rm{\{ }}0; - 1;2;3\} \left( {tmdk} \right)
\end{array}$
