Đáp án:
$\begin{array}{l}
a)Dkxd:{x^2} - 4 \ne 0 \Rightarrow {x^2} \ne 4 \Rightarrow \left\{ \begin{array}{l}
x \ne 2\\
x \ne - 2
\end{array} \right.\\
b)A = \frac{{{x^2} - 4x + 4}}{{{x^2} - 4}} = \frac{{{{\left( {x - 2} \right)}^2}}}{{\left( {x - 2} \right)\left( {x + 2} \right)}} = \frac{{x - 2}}{{x + 2}}\\
C)\left| x \right| = 3 \Rightarrow \left[ \begin{array}{l}
x = 3\left( {tm} \right)\\
x = - 3\left( {tm} \right)
\end{array} \right.\\
+ Khi:x = 3 \Rightarrow A = \frac{{x - 2}}{{x + 2}} = \frac{{3 - 2}}{{3 + 2}} = \frac{1}{5}\\
+ Khi:x = - 3 \Rightarrow A = \frac{{x - 2}}{{x + 2}} = \frac{{ - 3 - 2}}{{ - 3 + 2}} = 5\\
d)Dk:x \ne 2;x \ne - 2\\
A < 2\\
\Rightarrow \frac{{x - 2}}{{x + 2}} < 2\\
\Rightarrow \frac{{x - 2}}{{x + 2}} - 2 < 0\\
\Rightarrow \frac{{x - 2 - 2x - 4}}{{x + 2}} < 0\\
\Rightarrow \frac{{ - x - 6}}{{x + 2}} < 0\\
\Rightarrow \frac{{x + 6}}{{x + 2}} > 0\\
\Rightarrow \left[ \begin{array}{l}
x > - 2\\
x < - 6
\end{array} \right.\\
Vậy\,x < - 6\,hoặc:x > - 2;x \ne 2
\end{array}$
