\(a,A=\dfrac{x^2+3x+2+x^2-3x+2-x^2-4x}{\left(x-2\right)\left(x+2\right)}\\ A=\dfrac{x^2-4x+4}{\left(x-2\right)\left(x+2\right)}=\dfrac{\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)}=\dfrac{x-2}{x+2}\\ b,x=4\Leftrightarrow A=\dfrac{4-2}{4+2}=\dfrac{2}{6}=\dfrac{1}{3}\\ c,A>0\Leftrightarrow\dfrac{x-2}{x+2}>0\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-2>0\\x+2>0\end{matrix}\right.\\\left\{{}\begin{matrix}x-2< 0\\x+2< 0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x>2\\x< -2\end{matrix}\right.\\ \Leftrightarrow x=\left(-\infty;-2\right)\cup\left(2;+\infty\right)\)
