\(a,ĐK:x\ne\pm2\\ A=\dfrac{x+2-x+2+x^2+4x}{\left(x-2\right)\left(x+2\right)}=\dfrac{\left(x+2\right)^2}{\left(x-2\right)\left(x+2\right)}=\dfrac{x+2}{x-2}\\ b,x^2-2x=0\Leftrightarrow x\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\left(n\right)\\x=2\left(l\right)\end{matrix}\right.\Leftrightarrow=0\Leftrightarrow A=\dfrac{0+2}{0-2}=-1\\ c,A>0\Leftrightarrow\dfrac{x+2}{x-2}>0\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+2>0\\x-2>0\end{matrix}\right.\\\left\{{}\begin{matrix}x+2< 0\\x-2< 0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x>2\\x< -2\end{matrix}\right.\)
Vậy \(x\in\left\{x\in Z|\left(-\infty;-2\right)\cup\left(2;+\infty\right)\right\}\) hay \(x\in\left\{....;-4;-3;3;4;...\right\}\)
a: \(A=\dfrac{x+2-x+2+x^2+4x}{\left(x-2\right)\left(x+2\right)}=\dfrac{x^2+4x+4}{\left(x-2\right)\left(x+2\right)}=\dfrac{x+2}{x-2}\)
