Question

нужен ответ только, даю 50 баллов

Answer 1

Ответ:

Объяснение:

$\displaystyle\\\frac{x}{x^2+2} \leq (2x)^{-1}~~~~~\Leftrightarrow ~~~\frac{x}{x^2+2} \leq \frac{1}{2x} \\\\\\\frac{x}{x^2+2}-\frac{1}{2x} \leq 0}~~~~~\Leftrightarrow ~~~\frac{2x^2-x^2-2}{2x(x^2+2)} \leq 0\\\\\\\frac{x^2-2}{2x(x^2+2)} \leq 0}~~~~~\Leftrightarrow ~~~\frac{(x-\sqrt{2})(x+\sqrt{2} ) }{x} \leq 0\\\\\\---[-\sqrt{2}]+++(0)---[ \sqrt{2} ]+++>x\\\\\\Otvet:~x\in\Big(-\infty;-\sqrt{2}\Big]\cup\Big(0;\sqrt{2}\Big]$