Найдите $$\frac{p(b)}{p(\frac{1}{b})}$$, если $$p(b) = (b + \frac{8}{b})(8b + \frac{1}{b})$$. При $$b \ne 0$$.

Вычислим $$p(\frac{1}{b})$$:

$$ p(\frac{1}{b}) = (\frac{1}{b} + \frac{8}{\frac{1}{b}})(8 \cdot \frac{1}{b} + \frac{1}{\frac{1}{b}}) = (\frac{1}{b} + 8b)(\frac{8}{b} + b) $$

Тогда:

$$ \frac{p(b)}{p(\frac{1}{b})} = \frac{(b + \frac{8}{b})(8b + \frac{1}{b})}{(\frac{1}{b} + 8b)(\frac{8}{b} + b)} = \frac{(b + \frac{8}{b})(8b + \frac{1}{b})}{(8b + \frac{1}{b})(b + \frac{8}{b})} = 1 $$

Ответ: 1