a) \(\dfrac{x-3}{x}-\dfrac{x}{x-3}+\dfrac{9}{x^2-3x}=\dfrac{\left(x-3\right)^2-x^2+9}{x\left(x-3\right)}=\dfrac{x^2-6x+9-x^2+9}{x\left(x-3\right)}=\dfrac{18-6x}{x\left(x-3\right)}=\dfrac{-6\left(x-3\right)}{x\left(x-3\right)}=\dfrac{-6}{x}\)b) \(\dfrac{1}{x-2}-\dfrac{6x}{x^3-8}+\dfrac{x-2}{x^2+2x+4}=\dfrac{x^2+2x+4+\left(x-2\right)^2-6x}{\left(x-2\right)\left(x^2+2x+4\right)}=\dfrac{x^2-4x+4+\left(x-2\right)^2}{\left(x-2\right)\left(x^2+2x+4\right)}=\dfrac{\left(x-2\right)^2+\left(x-2\right)^2}{\left(x-2\right)\left(x^2+2x+4\right)}=\dfrac{2\left(x-2\right)^2}{\left(x-2\right)\left(x^2+2x+4\right)}=\dfrac{2\left(x-2\right)}{x^2+2x+4}\)
\(\dfrac{x-3}{x}-\dfrac{x}{x-3}+\dfrac{9}{x^2-3x}=\dfrac{x-3}{x}-\dfrac{x}{x-3}+\dfrac{9}{x\left(x-3\right)}=\dfrac{\left(x-3\right).\left(x-3\right)}{x.\left(x-3\right)}-\dfrac{x.x}{\left(x-3\right).x}+\dfrac{9}{x.\left(x-3\right)}=\dfrac{x^2-6x+9}{x.\left(x-3\right)}-\dfrac{x^2}{x.\left(x-3\right)}+\dfrac{9}{x.\left(x-3\right)}=\dfrac{\left(x-3\right)^2}{x.\left(x-3\right)}-\dfrac{x^2}{x.\left(x-3\right)}+\dfrac{9}{x.\left(x-3\right)}=\dfrac{\left(x-3\right)^2-x^2+9}{x.\left(x-3\right)}\)