b) \(\dfrac{x+1}{x-3}-\dfrac{1-x}{x+3}-\dfrac{2x\left(1-x\right)}{9-x^2}\)
=\(\dfrac{x+1}{x-3}-\dfrac{1-x}{x+3}+\dfrac{2x\left(1-x\right)}{x^2-9}\)
=\(\dfrac{x+1}{x-3}-\dfrac{1-x}{x+3}+\dfrac{2x\left(1-x\right)}{\left(x-3\right)\left(x+3\right)}\)
=\(\dfrac{\left(x+1\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\dfrac{\left(1-x\right)\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}+\dfrac{2x\left(1-x\right)}{\left(x-3\right)\left(x+3\right)}\)
=\(\dfrac{x^2+3x+x+3-\left(x-3-x^2+3x\right)+2x-2x^2}{\left(x-3\right)\left(x+3\right)}\)
=\(\dfrac{x^2+4x+3-x+3+x^2-3x+2x-2x^2}{\left(x-3\right)\left(x+3\right)}\)
=\(\dfrac{2x+6}{\left(x-3\right)\left(x+3\right)}=\dfrac{2\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{2}{x-3}\)
