Задача 12А. Решите неравенство \({\log _{0,5}}\left( {{x^2} + 0,5x} \right) \le 1\)
Ответ
ОТВЕТ: \(\,\left( {-\infty ;-1} \right] \cup \left[ {\dfrac{1}{2};\infty } \right).\)
Решение
\({\log _{\dfrac{1}{2}}}\left( {{x^2} + 0,5x} \right) \le 1\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,{\log _{\dfrac{1}{2}}}\left( {{x^2} + 0,5x} \right) \le {\log _{\dfrac{1}{2}}}\dfrac{1}{2}\,\,\,\,\,\,\, \Leftrightarrow \)
\( \Leftrightarrow \,\,\,\,\,\,\,{x^2} + 0,5x \ge \dfrac{1}{2}\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,2{x^2} + x-1 \ge 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,x\, \in \,\left( {-\infty ;-1} \right] \cup \left[ {\dfrac{1}{2};\infty } \right).\)
Ответ: \(\,\left( {-\infty ;-1} \right] \cup \left[ {\dfrac{1}{2};\infty } \right).\)