\[\boxed{\text{131\ (131).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
Пояснение.
Решение.
\[\frac{3^{\backslash 1 + a^{2}}}{1 - a^{2}} + \frac{3^{\backslash 1 - a^{2}}}{1 + a^{2}} + \frac{6}{1 + a^{4}} +\]
\[+ \frac{12}{1 + a^{8}} + \frac{24}{1 + a^{16}} = \frac{48}{1 - a^{32}}\]
\[Упростим\ левую\ часть\ \]
\[равенства:\]
\[\frac{6^{\backslash 1 + a^{4}}}{1 - a^{4}} + \frac{6^{\backslash 1 - a^{4}}}{1 + a^{4}} + \frac{12}{1 + a^{8}} +\]
\[+ \frac{24}{1 + a^{16}} = \frac{48}{1 - a^{32}}\]
\[\frac{12^{\backslash 1 + a^{8}}}{1 - a^{8}} + \frac{12^{\backslash 1 - a^{8}}}{1 + a^{8}} + \frac{24}{1 + a^{16}} =\]
\[= \frac{48}{1 - a^{32}}\]
\[\frac{24^{\backslash 1 + a^{16}}}{1 - a^{16}} + \frac{24^{\backslash 1 - a^{16}}}{1 + a^{16}} = \frac{48}{1 - a^{32}}\]
\[\frac{48}{1 - a^{32}} = \frac{48}{1 - a^{32}}\text{.\ }\]