Задача 8А. Решите неравенство \(\left| {2x + 5} \right| \geqslant 9\)
ОТВЕТ: \(\left( {-\infty ;\;-7} \right] \cup \left[ {2;\infty } \right).\)
\(\left| {2x + 5} \right| \ge 9.\) Неравенство вида \(\left| {f\left( x \right)} \right| \ge a\) при \(a > 0\) равносильно совокупности: \(\left[ {\begin{array}{*{20}{c}}{f\left( x \right) \ge a,\,\,}\\{f\left( x \right) \le -a.}\end{array}} \right.\) \(\left| {2x + 5} \right| \ge 9\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{2x + 5 \ge 9,\,}\\{2x + 5 \le -9}\end{array}} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{2x \ge 4,\,\,\,\,}\\{2x \le -14}\end{array}} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{x \ge 2,\,}\\{x \le -7}\end{array}} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,x \in \left( {-\infty ;\;-7} \right] \cup \left[ {2; + \infty } \right).\) Ответ: \(\left( {-\infty ;\;-7} \right] \cup \left[ {2; + \infty } \right).\)