67В. Решите неравенство \(\frac{{{{\log }_{1-2x}}\left( {\left( {x + 1} \right)\left( {1-4x + 4{x^2}} \right)} \right)}}{{{{\log }_{x + 1}}\left( {1-2x} \right)}} \le -1\).
ОТВЕТ: \(-0,5.\)
\(\frac{{{{\log }_{1-2x}}\left( {\left( {x + 1} \right)\left( {1-4x + 4{x^2}} \right)} \right)}}{{{{\log }_{x + 1}}\left( {1-2x} \right)}} \le -1.\) Запишем ОДЗ: \(\left\{ {\begin{array}{*{20}{c}}{1-2x > 0,\,\,\,\,\,\,\,\,\;\;\;\;\;\;\;\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{1-2x \ne 1,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\;\;\;\;\;\;\;\;\,\,\,\,\,\,}\\{x + 1 > 0,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\;\;\;\;\;\;\;\;\,}\\{x + 1 \ne 1,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\;\;\;\;\;\;\;\;\,}\\{\left( {x + 1} \right)\left( {1-4x + 4{x^2}} \right) > 0}\end{array}} \right.\;\;\;\; \Leftrightarrow \;\;\;\;\left\{ {\begin{array}{*{20}{c}}{2x < 1,\,\,\,\,\,\,\,\,\;\;\;\;\;\;\;\,\,\,\,\,\;\,\,\,\,}\\{2x \ne 0,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\;\;\;\;\;\,\,\,\,\,\,}\\{x > -1,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\;\;\;\,}\\{x \ne 0,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\;\,\;\;\;\,}\\{\left( {x + 1} \right){{\left( {1-2x} \right)}^2} > 0}\end{array}} \right.\;\;\;\; \Leftrightarrow \) \( \Leftrightarrow \;\;\;\;\left\{ {\begin{array}{*{20}{c}}{x < \frac{1}{2},\,\,\,\,\,\,\,\,\;\,\;\;\,\,\;\,\,\,\,}\\{x \ne 0,\,\,\,\,\,\,\,\,\,\,\,\,\;\;\;\,\,\,\,\,\,}\\{x > -1,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\;\;\;\,}\\{x \ne 0,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\;\,}\\{x > -1,\;\;\;\;x \ne \frac{1}{2}}\end{array}} \right.\;\;\;\;\; \Leftrightarrow \;\;\;\;\left\{ {\begin{array}{*{20}{c}}{x < \frac{1}{2},}\\{\begin{array}{*{20}{c}}{x \ne 0,\,}\\{x > -1}\end{array}}\end{array}} \right.\;\;\;\;\; \Leftrightarrow \;\;\;\;x\, \in \,\left( {-1;0} \right) \cup \left( {0;\frac{1}{2}} \right).\) Воспользуемся свойством: \({\log _a}b = \frac{1}{{{{\log }_b}a}}.\) \(\frac{{{{\log }_{1-2x}}\left( {\left( {x + 1} \right)\left( {1-4x + 4{x^2}} \right)} \right)}}{{{{\log }_{x + 1}}\left( {1-2x} \right)}} \le -1\;\;\;\; \Leftrightarrow \;\;\;\;{\log _{1-2x}}\left( {x + 1} \right) \cdot {\log _{1-2x}}\left( {\left( {x + 1} \right){{\left( {1-2x} \right)}^2}} \right) \le -1\;\;\;\; \Leftrightarrow \) \( \Leftrightarrow \;\;\;\;{\log _{1-2x}}\left( {x + 1} \right) \cdot \left( {{{\log }_{1-2x}}\left( {x + 1} \right) + 2{{\log }_{1-2x}}\left| {1-2x} \right|} \right) \le -1.\) Так как ОДЗ \(x\, \in \,\,\left( {-1;0} \right) \cup \left( {0;\frac{1}{2}} \right),\) то \(\left| {1-2x} \right| = 1-2x.\) Тогда полученное неравенство примет вид: \({\log _{1-2x}}\left( {x + 1} \right) \cdot \left( {{{\log }_{1-2x}}\left( {x + 1} \right) + 2{{\log }_{1-2x}}\left( {1-2x} \right)} \right) \le -1\;\;\;\; \Leftrightarrow \) \( \Leftrightarrow \;\;\;\,{\log _{1-2x}}\left( {x + 1} \right) \cdot \left( {{{\log }_{1-2x}}\left( {x + 1} \right) + 2} \right) + 1 \le 0\;\;\;\; \Leftrightarrow \) \( \Leftrightarrow \;\;\;\;\log _{1-2x}^2\left( {x + 1} \right) + 2{\log _{1-2x}}\left( {x + 1} \right) + 1 \le 0\;\;\;\; \Leftrightarrow \;\;\;\;{\left( {{{\log }_{1-2x}}\left( {x + 1} \right) + 1} \right)^2} \le 0\;\;\;\; \Leftrightarrow \) \( \Leftrightarrow \;\;\;\;{\log _{1-2x}}\left( {x + 1} \right) + 1 = 0\;\;\;\; \Leftrightarrow \;\;\;\;x + 1 = \frac{1}{{1-2x}}\;\;\;\; \Leftrightarrow \;\;\;\;x-2{x^2} + 1-2x = 1\;\;\;\; \Leftrightarrow \) \( \Leftrightarrow \;\;\;\;2{x^2} + x = 0\;\;\;\; \Leftrightarrow \;\;\;\;x\left( {2x + 1} \right) = 0\,\;\;\; \Leftrightarrow \;\;\;\;\left[ {\begin{array}{*{20}{c}}{x = 0,\;\;\;\;\;}\\{x = -0,5.}\end{array}} \right.\) Так как ОДЗ \(x\, \in \,\left( {-1;0} \right) \cup \left( {0;\frac{1}{2}} \right),\) то решение исходного неравенства будет иметь вид: \(x = -0,5.\) Ответ: \(-0,5.\)