70В. Решите неравенство \(\dfrac{{{{\log }_{{7^{x + 3}}}}49}}{{{{\log }_{{7^{x + 3}}}}\left( {-49x} \right)}} \le \dfrac{1}{{{{\log }_7}{{\log }_{\frac{1}{7}}}{7^x}}}\).
ОТВЕТ: \(\left[ {-49;\;-3} \right) \cup \left( {-3;\;-1} \right) \cup \left( {-\dfrac{1}{{49}};\;0} \right).\)
\(\dfrac{{{{\log }_{{7^{x + 3}}}}49}}{{{{\log }_{{7^{x + 3}}}}\left( {-49x} \right)}} \le \dfrac{1}{{{{\log }_7}{{\log }_{\frac{1}{7}}}{7^x}}}.\) Запишем ОДЗ: \(\left\{ {\begin{array}{*{20}{c}}{\begin{array}{*{20}{c}}{-49x > 0,\;\,}\\{-49x \ne 1\,,\,\;}\end{array}}\\{{7^{x + 3}} > 0,\;\;\;\,}\\{{7^{x + 3}} \ne 1,\,\,\;\;\;}\\{{7^x} > 0,\,\,\,\,\,\,\,\,\,\,}\\{{{\log }_{\frac{1}{7}}}{7^x} > 0}\end{array}} \right.\;\;\;\; \Leftrightarrow \;\;\;\;\left\{ {\begin{array}{*{20}{c}}{x < 0,\,\,\,\,\,\,\,\;\;\;\;\;\;\;\;\;}\\{x \ne -\dfrac{1}{{49}},\;\;\;\;\;\;\;\;}\\{x\, \in \,R,\,\,\,\,\;\;\;\;\;\;\;\;\;\,\,\,}\\{x \ne -3,\,\,\,\,\;\;\;\;\;\;\;\;\;}\\{x\, \in \,R,\,\;\;\;\;\;\;\;\;\;\,\,\,\,\,\,}\\{{{\log }_{\frac{1}{7}}}{7^x} > {{\log }_{\frac{1}{7}}}1}\end{array}} \right.\;\;\;\; \Leftrightarrow \;\;\;\;\left\{ {\begin{array}{*{20}{c}}{x < 0,\,\,\;\;\;\;}\\{x \ne -\dfrac{1}{{49}},}\\{x\, \in \,R,\,\,\,\,\;\;\,\,}\\{x \ne -3,\,\,\,\,\;}\\{x\, \in \,R,\,\,\;\;\,\,\,\,}\\{{7^x} < {7^0}\;\;\;\,}\end{array}} \right.\;\;\;\; \Leftrightarrow \) \( \Leftrightarrow \;\;\;\;\left\{ {\begin{array}{*{20}{c}}{x < 0,\,\,\;\;\;\;}\\{x \ne -\dfrac{1}{{49}},}\\{x \ne -3\;\;\;\;}\end{array}} \right.\;\;\;\; \Leftrightarrow \;\;\;\;x \in \left( {-\infty ;-3} \right) \cup \left( {-3;-\dfrac{1}{{49}}} \right) \cup \left( {-\dfrac{1}{{49}};0} \right).\) \(\dfrac{{{{\log }_{{7^{x + 3}}}}49}}{{{{\log }_{{7^{x + 3}}}}\left( {-49x} \right)}} \le \dfrac{1}{{{{\log }_7}{{\log }_{\frac{1}{7}}}{7^x}}}\;\;\;\; \Leftrightarrow \;\;\;\;\dfrac{{\dfrac{1}{{x + 3}}{{\log }_7}49}}{{\dfrac{1}{{x + 3}}{{\log }_7}\left( {-49x} \right)}} \le \dfrac{1}{{{{\log }_7}\left( {-x} \right)}}\;\;\;\; \Leftrightarrow \) \( \Leftrightarrow \;\,\;\;\,\dfrac{2}{{2 + {{\log }_7}\left( {-x} \right)}}-\dfrac{1}{{{{\log }_7}\left( {-x} \right)}} \le 0.\) Пусть \({\log _7}\left( {-x} \right) = t,\) тогда последнее неравенство примет вид: \(\dfrac{2}{{t + 2}}-\dfrac{1}{t} \le 0\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\dfrac{{t-2}}{{t\left( {t + 2} \right)}} \le 0\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{t < -2,\,\,\,\,}\\{0 < t \le 2.}\end{array}} \right.\) Вернёмся к прежней неизвестной: \(\left[ {\begin{array}{*{20}{c}}{{{\log }_7}\left( {-x} \right) < -2,\,\,\,\,}\\{0 < {{\log }_7}\left( {-x} \right) \le 2}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\left[ {\begin{array}{*{20}{c}}{{{\log }_7}\left( {-x} \right) < {{\log }_7}\dfrac{1}{{49}},\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{{{\log }_7}1 < {{\log }_7}\left( {-x} \right) \le {{\log }_7}49}\end{array}\,\,\,\,\,} \right. \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{0 < -x < \dfrac{1}{{49}},}\\{1 < -x \le 49\,\,\,\,}\end{array}} \right.\,\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\) \( \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{-\dfrac{1}{{49}} < x < 0,}\\{-49 \le x < -1}\end{array}} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,x \in \left[ {-49;-1} \right) \cup \left( {-\dfrac{1}{{49}};0} \right).\) Так как ОДЗ \(x \in \left( {-\infty ;-3} \right) \cup \left( {-3;-\dfrac{1}{{49}}} \right) \cup \left( {-\dfrac{1}{{49}};0} \right),\) то решение исходного неравенства будет иметь вид: \(x \in \left[ {-49;\;-3} \right) \cup \left( {-3;\;-1} \right) \cup \left( {-\dfrac{1}{{49}};\;0} \right).\) Ответ: \(\left[ {-49;\;-3} \right) \cup \left( {-3;\;-1} \right) \cup \left( {-\dfrac{1}{{49}};\;0} \right).\)