a: ĐKXĐ: \(x\notin\left\{3;-3\right\}\)
b: \(B=\left(x+2\right)\left(x+3\right)\)
c: \(A=\dfrac{x}{x-3}-\dfrac{6x}{x^2-9}=\dfrac{x\left(x+3\right)-6x}{\left(x-3\right)\left(x+3\right)}=\dfrac{x^2+3x-6x}{\left(x-3\right)\left(x+3\right)}=\dfrac{x}{x+3}\)