Question

100 балів! Допоможіть, будь ласка!!

Answer 1

$\displaystyle\bf\\g(x)=-6-2x\\\\\\1)\\\\\frac{1}{g(x)} \geq 0\\\\\\\frac{1}{-6-2x} \geq 0\\\\\\\frac{1}{-2\cdot(3+x)}\geq 0\\\\\\\frac{1}{x+3} \leq 0\\\\x+3\neq 0\\\\x+3 < 0\\\\x < -3\\\\x\in \Big(-\infty \ ; \ -3\Big)$

Наибольшее целое решение неравенства равно : - 2

$\displaystyle\bf\\2)\\\\\Big|g(x)\Big|\leq3\\\\\Big|-6-2x\Big|\leq 3\\\\\\\left \{ {{-6-2x\leq 3} \atop {-6-2x\geq -3 }} \right. \\\\\\\left \{ {{-2x\leq 3+6} \atop {-2x\geq -3+6 }} \right. \\\\\\\left \{ {{-2x\leq 9} \atop {-2x\geq 3}} \right. \\\\\\\left \{ {{x\geq -4,5} \atop {x\leq -1,5}} \right. \\\\\\x\in\Big[-4,5 \ ; \ -15\Big]\\\\\\Otvet \ : \ -4,5$

Answer 2

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