\[\frac{7b^2}{a^2-9} - \frac{7b}{a+3} = \frac{7b^2}{(a-3)(a+3)} - \frac{7b}{a+3} = \frac{7b^2 - 7b(a-3)}{(a-3)(a+3)} = \frac{7b^2 - 7ba + 21b}{(a-3)(a+3)}\]
\[\frac{7(6)^2 - 7(6)(5) + 21(6)}{(5-3)(5+3)} = \frac{7(36) - 7(30) + 126}{2 \cdot 8} = \frac{252 - 210 + 126}{16} = \frac{168}{16} = 10.5\]
Ответ: 10,5