\(B=\dfrac{x}{x+2}-\dfrac{x^2+8}{x^2-4}+\dfrac{3}{x-2}\\ =\dfrac{x\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{x^2+8}{\left(x-2\right)\left(x+2\right)}+\dfrac{3\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\\ =\dfrac{x^2-2x-x^2-8+3x+6}{\left(x-2\right)\left(x+2\right)}\\ =\dfrac{x-2}{\left(x-2\right)\left(x+2\right)}\\ =\dfrac{1}{x+2}\)