68В. Решите неравенство \(\dfrac{{{{\log }_{1-x}}\left( {\left( {3x + 1} \right)\left( {1-2x + {x^2}} \right)} \right)}}{{{{\log }_{3x + 1}}\left( {1-x} \right)}} \le -1\).
ОТВЕТ: \(\dfrac{2}{3}\).
\(\dfrac{{{{\log }_{1-x}}\left( {\left( {3x + 1} \right)\left( {1-2x + {x^2}} \right)} \right)}}{{{{\log }_{3x + 1}}\left( {1-x} \right)}} \le -1.\) Запишем ОДЗ: \(\left\{ {\begin{array}{*{20}{c}}{1-x > 0,\,\,\,\,\,\,\,\,\;\;\;\;\;\;\;\,\,\,\,\,\,\;\;\,\,\,\,\,\,\,\,}\\{1-x \ne 1,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\;\;\;\;\;\;\;\;\;\;\,\,\,\,\,\,}\\{3x + 1 > 0,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\;\;\;\;\;\;\,}\\{3x + 1 \ne 1,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\;\;\;\;\;\;\,}\\{\left( {3x + 1} \right)\left( {1-2x + {x^2}} \right) > 0}\end{array}} \right.\;\;\;\; \Leftrightarrow \;\;\;\;\left\{ {\begin{array}{*{20}{c}}{x < 1,\,\,\,\,\,\,\,\,\;\;\;\;\,\;\;\;\;\,\,\,\,\,\;\,\,\,\,}\\{x \ne 0,\,\,\,\,\,\,\,\,\,\,\,\,\;\,\,\,\,\;\;\;\;\;\,\,\,\,\,\,}\\{x > -\dfrac{1}{3},\,\;\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\;\;\;\,}\\{3x \ne 0,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\;\;\;\,}\\{\left( {3x + 1} \right){{\left( {1-x} \right)}^2} > 0}\end{array}} \right.\;\;\;\; \Leftrightarrow \) \( \Leftrightarrow \;\;\;\;\left\{ {\begin{array}{*{20}{c}}{x < 1,\,\,\,\,\,\;\;\,\,\,\;\;\;\,\,\;\,\,\,\,}\\{x \ne 0,\,\,\,\,\,\,\,\,\,\,\,\,\;\;\;\,\,\,\,\,\,}\\{x > -\dfrac{1}{3},\,\,\,\,\,\,\,\,\,\,\,\,\,\;\;\,}\\{x \ne 0,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\;\,}\\{x > -\dfrac{1}{3},\;\;\;\;x \ne 1}\end{array}} \right.\;\;\;\;\; \Leftrightarrow \;\;\;\;\left\{ {\begin{array}{*{20}{c}}{x < 1,\;\;\,}\\{\begin{array}{*{20}{c}}{x \ne 0,\;\,}\\{x > -\dfrac{1}{3}}\end{array}}\end{array}} \right.\;\;\;\;\; \Leftrightarrow \;\;\;\;x\, \in \,\left( {-\dfrac{1}{3};0} \right) \cup \left( {0;1} \right).\) Воспользуемся свойством: \({\log _a}b = \dfrac{1}{{{{\log }_b}a}}.\) \(\dfrac{{{{\log }_{1-x}}\left( {\left( {3x + 1} \right)\left( {1-2x + {x^2}} \right)} \right)}}{{{{\log }_{3x + 1}}\left( {1-x} \right)}} \le -1\;\;\;\; \Leftrightarrow \;\;\;\;{\log _{1-x}}\left( {3x + 1} \right) \cdot {\log _{1-x}}\left( {\left( {3x + 1} \right){{\left( {1-x} \right)}^2}} \right) \le -1\;\;\;\; \Leftrightarrow \) \( \Leftrightarrow \;\;\;\;{\log _{1-x}}\left( {3x + 1} \right) \cdot \left( {{{\log }_{1-x}}\left( {3x + 1} \right) + 2{{\log }_{1-x}}\left| {1-x} \right|} \right) \le -1.\) Так как ОДЗ \(x\, \in \,\left( {-\dfrac{1}{3};0} \right) \cup \left( {0;1} \right),\) то \(\left| {1-x} \right| = 1-x.\) Тогда полученное неравенство примет вид: \({\log _{1-x}}\left( {3x + 1} \right) \cdot \left( {{{\log }_{1-x}}\left( {3x + 1} \right) + 2{{\log }_{1-x}}\left( {1-x} \right)} \right) \le -1\;\;\;\; \Leftrightarrow \) \( \Leftrightarrow \;\;\;\,{\log _{1-x}}\left( {3x + 1} \right) \cdot \left( {{{\log }_{1-x}}\left( {3x + 1} \right) + 2} \right) + 1 \le 0\;\;\;\; \Leftrightarrow \) \( \Leftrightarrow \;\;\;\;\log _{1-x}^2\left( {3x + 1} \right) + 2{\log _{1-x}}\left( {3x + 1} \right) + 1 \le 0\;\;\;\; \Leftrightarrow \;\;\;\;{\left( {{{\log }_{1-x}}\left( {3x + 1} \right) + 1} \right)^2} \le 0\;\;\;\; \Leftrightarrow \) \( \Leftrightarrow \;\;\;\;{\log _{1-x}}\left( {3x + 1} \right) + 1 = 0\;\;\;\; \Leftrightarrow \;\;\;\;3x + 1 = \dfrac{1}{{1-x}}\;\;\;\; \Leftrightarrow \;\;\;\;3x-3{x^2} + 1-x = 1\;\;\;\; \Leftrightarrow \) \( \Leftrightarrow \;\;\;\;3{x^2}-2x = 0\;\;\;\; \Leftrightarrow \;\;\;\;x\left( {3x-2} \right) = 0\,\;\;\; \Leftrightarrow \;\;\;\;\left[ {\begin{array}{*{20}{c}}{x = 0,\,}\\{x = \dfrac{2}{3}.}\end{array}} \right.\) Так как ОДЗ \(x\, \in \,\left( {-\dfrac{1}{3};0} \right) \cup \left( {0;1} \right),\) то решение исходного неравенства будет иметь вид: \(x = \dfrac{2}{3}.\) Ответ: \(\dfrac{2}{3}\).