a: ĐKXĐ: \(x\notin\left\{2;-2;0\right\}\)
\(A=\dfrac{-x^2-4x-4-4x^2+x^2-4x+4}{\left(x-2\right)\left(x+2\right)}:\dfrac{x\left(x-3\right)}{x^2\left(2-x\right)}\)
\(=\dfrac{-4x\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{-x\left(x-2\right)}{x-3}\)
\(=\dfrac{4x^2}{x-3}\)
b: Để A>0 thì x-3>0
hay x>3