\(a,B=\dfrac{x^3-3x^2-x^3-3x^2-x-3+6x^2+x}{\left(x-3\right)\left(x+3\right)}:\dfrac{x-3-x-3}{x+3}\\ B=\dfrac{-3}{\left(x-3\right)\left(x+3\right)}\cdot\dfrac{x+3}{-6}=\dfrac{1}{2\left(x-3\right)}\\ b,B=-\dfrac{2}{3}\Rightarrow2\left(x-3\right)=-\dfrac{3}{2}\Rightarrow x-3=-\dfrac{3}{4}\Rightarrow x=\dfrac{9}{4}\)


