Задача 12. Решите систему неравенств \(\left\{ {\begin{array}{*{20}{c}} {4 \cdot {3^{x + 2}}-2 \cdot {5^{x + 2}} \leqslant {5^{x + 3}}-{3^{x + 3}}} \\ {\lg \left( {{x^2}-2x-2} \right) \leqslant 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \end{array}} \right.\)
Ответ
ОТВЕТ: \(\left[ {-1;\,1-\sqrt 3 } \right) \cup \left( {1 + \sqrt 3 ;\,\,3} \right].\)
Решение
Решим первое неравенство:
\(4 \cdot {3^{x + 2}}-2 \cdot {5^{x + 2}} \le {5^{x + 3}}-{3^{x + 3}}\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,4 \cdot {3^{x + 2}}-2 \cdot {5^{x + 2}} \le 5 \cdot {5^{x + 2}}-3 \cdot {3^{x + 2}}\,\,\,\,\,\, \Leftrightarrow \)
\( \Leftrightarrow \,\,\,\,\,7 \cdot {3^{x + 2}} \le 7 \cdot {5^{x + 2}}\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,{\left( {\dfrac{3}{5}} \right)^{x + 2}} \le 1\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,{\left( {\dfrac{3}{5}} \right)^{x + 2}} \le {\left( {\dfrac{3}{5}} \right)^0}\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,x + 2 \ge 0\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,x \ge -2.\)
Решим второе неравенство:
\(\lg \left( {{x^2}-2x-2} \right) \le 0\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\lg \left( {{x^2}-2x-2} \right) \le \lg 1\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{{x^2}-2x-2 \le 1,}\\{{x^2}-2x-2 > 0}\end{array}} \right.\,\,\,\,\,\, \Leftrightarrow \)
\( \Leftrightarrow \,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{{x^2}-2x-3 \le 0,}\\{{x^2}-2x-2 > 0\,}\end{array}} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left\{ \begin{array}{l}x\, \in \,\left[ {-1;3} \right],\\x\, \in \,\left( {-\infty ;1-\sqrt 3 } \right) \cup \left( {1 + \sqrt 3 ;\infty } \right)\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,x\, \in \left[ {-1;1-\sqrt 3 } \right) \cup \left( {1 + \sqrt 3 ;3} \right].\)
Найдём общее решение:
Таким образом, решение исходной системы неравенств: \(x \in \left[ {-1;1-\sqrt 3 } \right) \cup \left( {1 + \sqrt 3 ;3} \right].\)
Ответ: \(\left[ {-1;1-\sqrt 3 } \right) \cup \left( {1 + \sqrt 3 ;3} \right].\)