A=$\dfrac{2x}{x-3}$ -$\dfrac{1}{x+3}$+ $\dfrac{2x²-12}{9-x²}$
=$\dfrac{2x}{x-3}$ -$\dfrac{1}{x+3}$- $\dfrac{2x²-12}{x²-9}$
=$\dfrac{2x}{x-3}$ -$\dfrac{1}{x+3}$- $\dfrac{2x²-12}{(x-3)(x+3)}$
=$\dfrac{2x^2+6x}{(x-3)(x+3)}$ -$\dfrac{(x-3)}{(x+3)(x-3)}$- $\dfrac{2x²-12}{(x-3)(x+3)}$
=$\dfrac{2x^2+6x-x+3-2x^2+12}{(x-3)(x+3)}$
=$\dfrac{5x+15}{(x-3)(x+3)}$
=$\dfrac{5(x+3)}{(x-3)(x+3)}$
=$\dfrac{5}{(x-3}$
B=2x-$\dfrac{2}{5}$
=$\dfrac{2x.5}{5}$-$\dfrac{2}{5}$
=$\dfrac{10x}{5}$-$\dfrac{2}{5}$
=$\dfrac{10x-2}{5}$
