Question

Решить неравенство , содержащие переменную под знаком модуля а) |2x + 3| < 9 б) |3x - 7| > 5 в) |5x + 2| > или равно 4,4 г) |2x - 9|

Answer 1

$1)\; \; |2x+3|\ \textless \ 9\; \; \to\; \; \; -9\ \textless \ 2x+3\ \textless \ 9\\\\-12\ \textless \ 2x\ \textless \ 6\\\\-6\ \textless \ x\ \textless \ 3\; \; \; \to \; \; x\in (-6,3)\\\\2)\; \; |3x-7|\ \textgreater \ 5\; \; \; \to \; \; \; \left [ {{3x-7\ \textgreater \ 5} \atop {3x-7\ \textless \ -5}} \right. \; \left [ {{x\ \textgreater \ 4} \atop {x\ \textless \ \frac{2}{3}}} \right. \; \to \\\\x\in (-\infty ,\frac{2}{3})\cup (4,+\infty )\\\\3)\; \; |5x+2| \geq 4,4\; \; \to \; \; \left [ {{5x+2 \geq 4,4} \atop {5x+2 \leq -4,4}} \right. \; \; \left [ {{x \geq 0,48} \atop {x \leq -1,28}} \right. \; \; \to$

$x\in (-\infty ;-1,28\, ]\cup [\, 0,48\; ; +\infty )$

$4)\; \; |2x-9| \leq 3,5\; \; \to \; \; -3,5 \leq 2x-9 \leq 3,5\\\\5,5 \leq 2x \leq 12,5\\\\2,75 \leq x \leq 6,25$

$x\in [\, 2,75\; ;\; 6,25\, ]$