a: \(A=\dfrac{-\left(x-3\right)}{x+3}\cdot\dfrac{\left(x+3\right)^2}{\left(x-3\right)\left(x+3\right)}+\dfrac{x}{x+3}\)
\(=-1+\dfrac{x}{x+3}=\dfrac{-x-3+x}{x+3}=\dfrac{-3}{x+3}\)
b: Khi x=1 thì \(A=\dfrac{-3}{1+3}=\dfrac{-3}{4}\)
Khi x=-1 thì \(A=\dfrac{-3}{-1+3}=\dfrac{-3}{2}\)