Найдите \(\displaystyle \frac {p(b)}{p( \frac{1}{b})}{\small,} \)если \(\displaystyle p(b) =\bigg (b + \frac {3}{b}\bigg)\bigg(3b + \frac {1}{b}\bigg){\small.}\) При \(\displaystyle b \,\cancel= \, 0{\small.}\)
По условию \(\displaystyle p(b) =\bigg (b + \frac {3}{b}\bigg)\bigg(3b + \frac {1}{b}\bigg){\small.}\)
Для того, чтобы найти \(\displaystyle p\left(\frac{1}{b}\right){\small,}\) нужно в данную формулу вместо \(\displaystyle \color{blue}b\) подставить \(\displaystyle \color{blue}{\frac{1}{b}}{\small:}\)
\(\displaystyle p(\color{blue}b) = \bigg (\color{blue}b + \frac {3}{\color{blue}b}\bigg)\bigg(3\color{blue}b + \frac {1}{\color{blue}b}\bigg){\small,}\)
\(\displaystyle p\left(\color{blue}{\frac{1}{b}}\right) =\bigg (\color{blue}{\frac{1}{b}} + \frac {\phantom 1 3\phantom 1}{\color{blue}{\frac{1}{b}}}\bigg)\bigg(3 \cdot \color{blue}{\frac{1}{b}} + \frac {\phantom 1 1 \phantom 1}{\color{blue}{\frac{1}{b}}}\bigg){\small.}\)
То есть
\(\displaystyle p\left(\frac{1}{b}\right) = \bigg (\frac{1}{b} + \frac {\phantom 1 3\phantom 1}{\frac{1}{b}}\bigg)\bigg(3 \cdot \frac{1}{b} + \frac {\phantom 1 1 \phantom 1}{\frac{1}{b}}\bigg)=\bigg (\frac{1}{b} + 3b\bigg)\bigg(\frac{3}{b} + b\bigg){\small.}\)
Подставим в заданное выражение \(\displaystyle \bigg (b + \frac {3}{b}\bigg)\bigg(3b + \frac {1}{b}\bigg)\) вместо \(\displaystyle p(b)\) и \(\displaystyle \bigg (\frac{1}{b} + 3b\bigg)\bigg(\frac{3}{b} + b\bigg)\) вместо \(\displaystyle p\left(\frac{1}{b}\right){\small.}\)
Тогда:
\(\displaystyle \frac {\color{red}{p(b)}}{\color{green}{p( \frac{1}{b})}}=\frac {\color{red}{\bigg (b + \frac {3}{b}\bigg)\bigg(3b + \frac {1}{b}\bigg)}}{\color{green}{\bigg (\frac{1}{b} + 3b\bigg)\bigg(\frac{3}{b} + b\bigg)}}{\small.}\)
В числителе и знаменателе дроби оказались одинаковые множители, которые можно сократить:
\(\displaystyle \frac {\cancel{\bigg (b + \frac {3}{b}\bigg)}\,\,\cancel{\bigg(3b + \frac {1}{b}\bigg)}}{\cancel{\bigg (\frac{1}{b} + 3b\bigg)} \,\, \cancel{\bigg(\frac{3}{b} + b\bigg)}}=1 {\small.}\)
Таким образом, верна следующая цепочка равенств:
\(\displaystyle \frac {p(b)}{p( \frac{1}{b})}=\frac {\bigg (b + \frac {3}{b}\bigg)\bigg(3b + \frac {1}{b}\bigg)}{\bigg (\frac{1}{b} + \frac {\phantom 1 3\phantom 1}{\frac{1}{b}}\bigg)\bigg(3 \cdot \frac{1}{b} + \frac {\phantom 1 1 \phantom 1}{\frac{1}{b}}\bigg)}=\frac {\bigg (b + \frac {3}{b}\bigg)\bigg(3b + \frac {1}{b}\bigg)}{\bigg (\frac{1}{b} + 3b\bigg)\bigg(\frac{3}{b} + b\bigg)}=1{\small.}\)
Ответ: \(\displaystyle 1 {\small.} \)