6) 2/a²-9 - 1/a²+3a.

6) $$\frac{2}{a^2-9} - \frac{1}{a^2+3a} = \frac{2}{(a-3)(a+3)} - \frac{1}{a(a+3)} = \frac{2a - (a-3)}{a(a-3)(a+3)} = \frac{2a - a + 3}{a(a-3)(a+3)} = \frac{a+3}{a(a-3)(a+3)} = \frac{1}{a(a-3)}$$

Ответ: $$\frac{1}{a(a-3)}$$