Задача 16А. Решите неравенство \(\left| {2x + 3} \right| \leqslant 4x\)
ОТВЕТ: \(\left[ {\frac{3}{2};\;\infty } \right).\)
\(\left| {2x + 3} \right| \le 4x.\) Неравенство вида \(\left| {f\left( x \right)} \right| \le g\left( x \right)\) равносильно двойному неравенству: \(-g\left( x \right) \le f\left( x \right) \le g\left( x \right)\;\;\;\; \Leftrightarrow \;\;\;\;\,\left\{ {\begin{array}{*{20}{c}}{f\left( x \right) \le g\left( x \right),\;\,}\\{f\left( x \right) \ge -g\left( x \right).}\end{array}} \right.\) \(\left| {2x + 3} \right| \le 4x\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,-4x \le 2x + 3 \le 4x\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{2x + 3 \le 4x\,\,\,}\\{2x + 3 \ge -4x}\end{array}} \right.\,\,\,\,\,\, \Leftrightarrow \) \( \Leftrightarrow \,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{x \ge \frac{3}{2},}\\{x \ge -\frac{1}{2}}\end{array}} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,x \in \left[ {\frac{3}{2};\, + \infty } \right).\) Ответ: \(\left[ {\frac{3}{2}; + \infty } \right)\).