Задача 40. Решите неравенство\(\left( {x-4} \right)\left( {{{\log }_5}\left( {125-25x} \right)-{{\log }_6}\left( {{x^2} + x-6} \right) + \frac{1}{{{{\log }_{5-x}}0,2}} + x + 6} \right) \le {x^2} + 2x-24\).
ОТВЕТ: \(\left[ {-7;\;-3} \right) \cup \left( {2;\;4} \right).\)
\(\left( {x-4} \right)\left( {{{\log }_5}\left( {125-25x} \right)-{{\log }_6}\left( {{x^2} + x-6} \right) + \dfrac{1}{{{{\log }_{5-x}}0,2}} + x + 6} \right) \le {x^2} + 2x-24.\) Найдём ОДЗ: \(\left\{ {\begin{array}{*{20}{c}}{125-25x > 0,}\\{{x^2} + x-6 > 0,}\\{5-x > 0,\,\,\;\,\,\,\,\,\,\,\,}\\{5-x \ne 1\;\,\,\,\,\;\,\,\,\,\,\,\,}\end{array}\,\,\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{x < 5,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\;\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{x\, \in \,\left( {-\infty ;-3} \right) \cup \left( {2;\infty } \right),}\\{x < 5,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\;\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{x \ne 4\;\,\,\,\,\,\,\,\,\,\,\,\,\;\,\;\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\end{array}} \right.} \right.\;\;\;\; \Leftrightarrow \;\;\;\;x\, \in \,\left( {-\infty ;-3} \right) \cup \left( {2;4} \right) \cup \left( {4;5} \right).\) Вернёмся к исходному неравенству: \(\left( {x-4} \right)\left( {{{\log }_5}\left( {125-25x} \right)-{{\log }_6}\left( {{x^2} + x-6} \right) + \dfrac{1}{{{{\log }_{5-x}}0,2}} + x + 6} \right) \le {x^2} + 2x-24\;\; \Leftrightarrow \) \( \Leftrightarrow \;\;\;\;\left( {x-4} \right)\left( {{{\log }_5}\left( {25\left( {5-x} \right)} \right)-{{\log }_6}\left( {{x^2} + x-6} \right)-{{\log }_5}\left( {5-x} \right) + x + 6} \right) \le {x^2} + 2x-24\;\;\;\; \Leftrightarrow \) \( \Leftrightarrow \;\;\;\left( {x-4} \right)\left( {2 + {{\log }_5}\left( {5-x} \right)-{{\log }_6}\left( {{x^2} + x-6} \right)-{{\log }_5}\left( {5-x} \right) + x + 6} \right) \le {x^2} + 2x-24\,\;\;\; \Leftrightarrow \) \( \Leftrightarrow \;\;\;\;\left( {x-4} \right)\left( {8 + x-{{\log }_6}\left( {{x^2} + x-6} \right)} \right)-\left( {x-4} \right)\left( {x + 6} \right) \le 0\;\;\;\; \Leftrightarrow \) \( \Leftrightarrow \;\;\;\;\left( {x-4} \right)\left( {8 + x-{{\log }_6}\left( {{x^2} + x-6} \right)-x-6} \right) \le 0\;\;\; \Leftrightarrow \;\;\;\left( {x-4} \right)\left( {2-{{\log }_6}\left( {{x^2} + x-6} \right)} \right) \le 0.\) Решим полученное неравенство методом интервалов: \(\left[ {\begin{array}{*{20}{c}}{x = 4,\;\;\;\;\;\,\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}\\{{{\log }_6}\left( {{x^2} + x-6} \right) = 2.}\end{array}} \right.\) Рассмотрим второе уравнение: \({\log _6}\left( {{x^2} + x-6} \right) = 2\;\;\;\; \Leftrightarrow \;\;\;\;{x^2} + x-6 = 36\;\;\;\; \Leftrightarrow \;\;\;\;{x^2} + x-42 = 0\;\;\;\; \Leftrightarrow \;\;\;\;\left[ {\begin{array}{*{20}{c}}{{x} = 6,\;\;}\\{{x} = -7.}\end{array}} \right.\) Таким образом, решением исходного неравенства является: \(x \in \left[ {-7;\;-3} \right) \cup \left( {2;\;4} \right).\) Ответ: \(\left[ {-7;\;-3} \right) \cup \left( {2;\;4} \right).\) 