Решебник по алгебре 11 класс Никольский Параграф 11. Равносильность неравенств на множествах Задание 27

\[\boxed{\mathbf{27.}}\]

\[\textbf{а)}\log_{31}(8x - 9)^{2} > \log_{31}(9x - 11)^{2}\]

\[8x - 9 \neq 0\]

\[x \neq \frac{9}{8};\]

\[9x - 11 \neq 0\]

\[x \neq \frac{11}{9}.\]

\[(8x - 9)^{2} > (9x - 11)^{2}\]

\[(8x - 9)^{2} - (9x - 11)^{2} > 0\]

\[(17x - 20)( - x + 2) > 0\]

\[- 17\left( x - \frac{20}{17} \right)(x - 2) > 0\]

\[\frac{20}{17} < x < 2.\]

\[Решение\ неравенства:\]

\[x \in \left( \frac{20}{17};\frac{11}{9} \right) \cup \left( \frac{11}{9};2 \right).\]

\[Ответ:\ x \in \left( \frac{20}{17};\frac{11}{9} \right) \cup \left( \frac{11}{9};2 \right).\]

\[4x - 5 \neq 0\]

\[4x \neq 5\]

\[x \neq \frac{5}{4} \neq 1,25.\]

\[5x - 7 \neq 0\]

\[5x \neq 7\]

\[x \neq \frac{7}{5} \neq 1,4.\]

\[(4x - 5)^{2} < (5x - 7)^{2}\]

\[(4x - 5)^{2} - (5x - 7)^{2} < 0\]

\[( - x + 2)(9x - 12) < 0\]

\[- 9(x - 2)\left( x - \frac{4}{3} \right) < 0\]

\[x < \frac{4}{3};\ \ x > 2.\]

\[Решение\ неравенства:\]

\[x \in \left( \frac{4}{3};1,4 \right) \cup (1,4;2).\]

\[Ответ:\ x \in \left( \frac{4}{3};1,4 \right) \cup (1,4;2).\]