Ta có: \(\frac{x+1}{x-3}-\frac{1-x}{x+3}-\frac{2x\left(1-x\right)}{9-x^2}\)
= \(\frac{\left(x+1\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\frac{\left(1-x\right)\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}+\frac{2x\left(1-x\right)}{\left(x-3\right)\left(x+3\right)}\)
= \(\frac{x^2+4x+3-x+3+x^2-3x+2x-2x^2}{\left(x-3\right)\left(x+3\right)}\)
= \(\frac{2x+6}{\left(x-3\right)\left(x+3\right)}\)
= \(\frac{2\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\frac{2}{x-3}\)