a) $\frac{x + 3}{x}$ - $\frac{x}{x - 3}$ + $\frac{9}{x^{2} – 3x}$ = $\frac{x + 3}{x}$ - $\frac{x}{x - 3}$ + $\frac{9}{x(x – 3)}$

= $\frac{(x + 3)(x – 3)}{x(x – 3)}$ - $\frac{x.x}{x(x – 3)}$ + $\frac{9}{x(x – 3)}$ = $\frac{x^{2} – 9 - x^{2} + 9}{x(x – 3)}$ = $\frac{0}{x(x – 3)}$ = 0;

b) $\frac{1}{x - 2}$ - $\frac{6x}{x^{3} - 8}$ + $\frac{x – 2}{x^{2} + 2x + 4}$ = $\frac{1}{x - 2}$ - $\frac{6x}{(x – 2)(x^{2} + 2x + 4)}$ + $\frac{x - 2}{x^{2} + 2x + 4}$

= $\frac{x^{2} + 2x + 4}{(x – 2)(x^{2} + 2x + 4)}$ - $\frac{6x}{(x – 2)(x^{2} + 2x + 4)}$ - $\frac{(x – 2)^{2}}{(x – 2)(x^{2} + 2x + 4)}$

= $\frac{x^{2} + 2x + 4 – 6x + x^{2} – 4x + 4}{(x – 2)(x^{2} + 2x + 4)}$ = $\frac{2x^{2} – 8x + 8}{(x – 2)(x^{2} + 2x + 4)}$ = $\frac{2(x – 2)}{x^{2} + 2x + 4}$.