Решебник по алгебре 8 класс Мерзляк ФГОС Задание 131

 

\[\boxed{\text{131\ (131).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

Пояснение.

Решение.

\[\frac{3^{\backslash 1 + a^{2}}}{1 - a^{2}} + \frac{3^{\backslash 1 - a^{2}}}{1 + a^{2}} + \frac{6}{1 + a^{4}} +\]

\[+ \frac{12}{1 + a^{8}} + \frac{24}{1 + a^{16}} = \frac{48}{1 - a^{32}}\]

\[Упростим\ левую\ часть\ \]

\[равенства:\]

\[\frac{6^{\backslash 1 + a^{4}}}{1 - a^{4}} + \frac{6^{\backslash 1 - a^{4}}}{1 + a^{4}} + \frac{12}{1 + a^{8}} +\]

\[+ \frac{24}{1 + a^{16}} = \frac{48}{1 - a^{32}}\]

\[\frac{12^{\backslash 1 + a^{8}}}{1 - a^{8}} + \frac{12^{\backslash 1 - a^{8}}}{1 + a^{8}} + \frac{24}{1 + a^{16}} =\]

\[= \frac{48}{1 - a^{32}}\]

\[\frac{24^{\backslash 1 + a^{16}}}{1 - a^{16}} + \frac{24^{\backslash 1 - a^{16}}}{1 + a^{16}} = \frac{48}{1 - a^{32}}\]

\[\frac{48}{1 - a^{32}} = \frac{48}{1 - a^{32}}\text{.\ }\]

\[\boxed{\text{131.}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[1)\ \frac{x}{3} + \frac{x - 1}{2} = 4\ \ \ \ \ \ \ \ \ \ \ \ | \cdot 6\]

\[2x + 3 \cdot (x - 1) - 4 \cdot 6 = 0\ \]

\[2x + 3x - 3 - 24 = 0\]

\[5x - 27 = 0\]

\[5x = 27\]

\[x = \frac{27}{5} = 5,4.\]

\[Ответ:x = 5,4.\]

\[2)\frac{x - 4}{2} - \frac{x - 1}{5} = 3\ \ \ \ \ \ \ | \cdot 10\]

\[5 \cdot (x - 4) - 2 \cdot (x - 1) -\]

\[- 3 \cdot 10 = 0\]

\[5x - 20 - 2x + 2 - 30 = 0\]

\[3x - 48 = 0\]

\[3x = 48\]

\[x = \frac{48}{3}\]

\[x = 16.\]

\[Ответ:x = 16.\ \]