\(A=\dfrac{x^2+6x+9-2x^2+6+x^2-3x}{\left(x-3\right)\left(x+3\right)}\cdot\dfrac{2\left(x-3\right)\left(x+3\right)}{6\left(x-2\right)}\)
\(=\dfrac{3x+15}{\left(x-3\right)\left(x+3\right)}\cdot\dfrac{\left(x-3\right)\left(x+3\right)}{3\left(x-2\right)}=\dfrac{x+5}{x-2}\)