a) A= \(\dfrac{3x+15}{\left(x-3\right)\left(x+3\right)}\)+\(\dfrac{1}{x+3}\)-\(\dfrac{2}{x-3}\)

       =\(\dfrac{3x+15}{\left(x-3\right)\left(x+3\right)}\)+\(\dfrac{1\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}\)-\(\dfrac{2\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}\)

        =\(\dfrac{3x+15+x-3-2x-6}{\left(x-3\right)\left(x+3\right)}\)

        =\(\dfrac{2x+6}{\left(x-3\right)\left(x+3\right)}\)=\(\dfrac{2\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}\)=\(\dfrac{2}{x-3}\)

b) Để A=\(\dfrac{2}{3}\) thì \(\dfrac{2}{x-3}\)=\(\dfrac{2}{3}\)

Suy ra 3(x - 3) = 2.3 

            3x - 9   = 6 

             3x        = 15

                x        = 5    Vậy để A=\(\dfrac{2}{3}\) thì x = 5