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МатематикаГДЗ по алгебре 11 класс Никольский Параграф 11. Равносильность неравенств на множествах Задание 44
Задание 44
\[\boxed{\mathbf{44.}}\]
\[\textbf{а)}\log_{x}\frac{7x - 2}{2 - x} < 0\]
\[x > 0;\ \ x \neq 1\]
\[\frac{7x - 2}{2 - x} > 0\]
\[\frac{7\left( x - \frac{2}{7} \right)}{(x - 2)} < 0\]
\[\frac{2}{7} < x < 2.\]
\[M = \left( \frac{2}{7};1 \right) \cup (1;2).\]
\[\frac{\lg{(7x - 2) - \lg(2 - x)}}{\lg x} < 0\]
\[\frac{2}{7} < x < 1:\]
\[\lg{(7x - 2) - \lg(2 - x)} > 0\]
\[(7x - 2) - (2 - x) > 0\]
\[7x - 2 - 2 + x > 0\]
\[8x > 4\]
\[x > \frac{1}{2}.\]
\[1 < x < 2:\]
\[\lg{(7x - 2) - \lg(2 - x)} < 0\]
\[(7x - 2) - (2 - x) < 0\]
\[7x - 2 - 2 + x < 0\]
\[8x < 4\]
\[x < \frac{1}{2}.\]
\[Решение\ неравенства:\]
\[x \in \left( \frac{1}{2}\ ;1 \right).\]
\[Ответ:x \in \left( \frac{1}{2}\ ;1 \right).\]
\[\textbf{б)}\log_{x}\frac{3x - 1}{14x - 5} > 0\]
\[\frac{3x - 1}{14x - 5} > 0\]
\[\frac{3\left( x - \frac{1}{3} \right)}{14\left( x - \frac{5}{14} \right)} > 0\]
\[x < \frac{1}{3};\ \ x > \frac{5}{14}.\]
\[x > 0;\ \ x \neq 1.\]
\[M = \left( 0;\frac{1}{3} \right) \cup \left( \frac{5}{14};1 \right) \cup (1; + \infty).\]
\[\frac{\lg(3x - 1) - \lg(14x - 5)}{\lg x} > 0\]
\[0 < x < \frac{1}{3}:\]
\[\lg(3x - 1) - \lg(14x - 5) > 0\]
\[(3x - 1) - (14x - 5) > 0\]
\[3x - 1 - 14x + 5 > 0\]
\[11x < 4\]
\[x < \frac{4}{11}.\]
\[\frac{5}{14} < x < 1:\]
\[\lg(3x - 1) - \lg(14x - 5) < 0\]
\[(3x - 1) - (14x - 5) < 0\]
\[3x - 1 - 14x + 5 < 0\]
\[11x > 4\]
\[x > \frac{4}{11}.\]
\[Решение\ неравенства:\]
\[x \in \left( 0\ ;\frac{1}{3} \right) \cup \left( \frac{5}{14};1 \right) \cup (1; + \infty).\]
\[\textbf{в)}\log_{x}\frac{6x - 1}{13x - 7} < 0\]
\[x > 0;\ \ x \neq 1;\]
\[\frac{6x - 1}{13x - 7} > 0\]
\[\frac{6\left( x - \frac{1}{6} \right)}{13\left( x - \frac{7}{13} \right)} > 0\]
\[x < \frac{1}{6};\ \ \ x > \frac{7}{13}.\]
\[M = \left( 0;\frac{1}{6} \right) \cup \left( \frac{7}{13};1 \right) \cup (1; + \infty).\]
\[\frac{\lg(6x - 1) - \lg(13x - 7)}{\lg x} < 0\]
\[1)\ x > 1:\]
\[\lg(6x - 1) - \lg(13x - 7) < 0\]
\[6x - 1 - 13x + 7 < 0\]
\[7x > 6\]
\[x > \frac{6}{7}.\]
\[2)\ x < 1:\]
\[x < \frac{6}{7}.\]
\[Решение\ неравенства:\]
\[x \in \left( 0\ ;\frac{1}{6} \right) \cup \left( \frac{7}{13};\frac{6}{7} \right) \cup (1; + \infty).\]
\[\textbf{г)}\log_{x}\frac{16x - 11}{5x - 1} > 0\]
\[x > 0;\ \ x \neq 1\]
\[\frac{16x - 11}{5x - 1} > 0\]
\[\frac{16\left( x - \frac{11}{16} \right)}{5\left( x - \frac{1}{5} \right)} > 0\]
\[x < \frac{1}{5};\ \ \ x > \frac{11}{16}.\]
\[M = \left( 0;\frac{1}{5} \right) \cup \left( \frac{11}{16};1 \right) \cup (1; + \infty).\]
\[\frac{\lg(16x - 11) - \lg(5x - 1)}{\lg x} > 0\]
\[1)\ x > 1:\]
\[\lg(16x - 11) - \lg(5x - 1) > 0\]
\[16x - 11 - 5x + 1 > 0\]
\[11x > 10\]
\[x > \frac{10}{11}.\]
\[2)\ x < 1:\]
\[x < \frac{10}{11}.\]
\[Решение\ неравенства:\]
\[x \in \left( 0\ ;\frac{1}{5} \right) \cup \left( \frac{11}{16};\frac{10}{11} \right) \cup (1; + \infty).\]
