Question

решите неравенство (х+2)(х-2)-2(х-1)≥23-2х

Answer 1

х^2-2х+2х-4-2х+2-23+2х>= 0
х^2-25>=0
х^2>=25
х>=5; -5 (-бесконечность;-5] и [5;+бесконечность)

Answer 2

$\textup{(x+2)(x-2)-2(x-1)} \geq \textup{23-2x} \\ \textup{x}^2-4-2\textup{x}+2 \geq 23-2\textup{x} \\ \textup{x}^2-2\textup{x}-2 \geq 23-2\textup{x} \\ \textup{x}^2-2\textup{x}+2\textup{x}-2-23 \geq 0 \\ \textup{x}^2-25 \geq 0 \\(\textup{x}+5)(\textup{x}-5) \geq 0$

_______+______$-5$______-______$5$______+_______

Ответ:  $\textup{A}) ~\textup{x}\in(-\infty;-5]\textup{ U }[5;+\infty)$