7. 8 1) (1+3b)/b * (1/(2-6b) + a^2/(27b^3-1)) : (1+3b)/(1+3b+9b^2);

1) $$(\frac{1+3b}{b} \cdot (\frac{1}{2-6b} + \frac{a^2}{27b^3-1}) : \frac{1+3b}{1+3b+9b^2}) = (\frac{1+3b}{b} \cdot (\frac{1}{2(1-3b)} + \frac{a^2}{(3b-1)(9b^2+3b+1)}) : \frac{1+3b}{1+3b+9b^2}) = (\frac{1+3b}{b} \cdot (\frac{-1}{2(3b-1)} + \frac{a^2}{(3b-1)(9b^2+3b+1)}) : \frac{1+3b}{1+3b+9b^2}) = (\frac{1+3b}{b} \cdot \frac{-1(9b^2+3b+1) + 2a^2}{2(3b-1)(9b^2+3b+1)}) : \frac{1+3b}{1+3b+9b^2}) = (\frac{1+3b}{b} \cdot \frac{-9b^2-3b-1 + 2a^2}{2(3b-1)(9b^2+3b+1)}) : \frac{1+3b}{1+3b+9b^2}) = (\frac{1+3b}{b} \cdot \frac{2a^2-9b^2-3b-1}{2(3b-1)(9b^2+3b+1)}) \cdot \frac{1+3b+9b^2}{1+3b} = (\frac{1+3b}{b} \cdot \frac{2a^2-9b^2-3b-1}{2(3b-1)(9b^2+3b+1)}) \cdot \frac{1+3b+9b^2}{1+3b} = \frac{(1+3b)(2a^2-9b^2-3b-1)(1+3b+9b^2)}{b \cdot 2(3b-1)(9b^2+3b+1)(1+3b)} = \frac{(2a^2-9b^2-3b-1)(1+3b+9b^2)}{b \cdot 2(3b-1)(9b^2+3b+1)}$$.

Ответ: $$\frac{(2a^2-9b^2-3b-1)(1+3b+9b^2)}{b \cdot 2(3b-1)(9b^2+3b+1)}$$