41. Дано: $$p(b) = \left(b + \frac{2}{b}\right)\left(2b + \frac{1}{b}\right)$$Найти: $$\frac{p(b)}{p(\frac{1}{b})}$$Решение:
Найдем $$p\left(\frac{1}{b}\right)$$:$$\begin{aligned}p\left(\frac{1}{b}\right) &= \left(\frac{1}{b} + \frac{2}{\frac{1}{b}}\right)\left(2 \cdot \frac{1}{b} + \frac{1}{\frac{1}{b}}\right) \\&= \left(\frac{1}{b} + 2b\right)\left(\frac{2}{b} + b\right) \\&= \left(2b + \frac{1}{b}\right)\left(b + \frac{2}{b}\right) \\&= p(b)\end{aligned}$$
Найдем$$\frac{p(b)}{p(\frac{1}{b})}$$:$$\frac{p(b)}{p(\frac{1}{b})} = \frac{p(b)}{p(b)} = 1$$
Ответ: 1