Question

Решите систему неравенств:

{-x^2+2x+8≤0
6-2(x+1)>2x​

Answer 1

Ответ:

$\left\{\begin{array}{l}-x^2+2x+8\leq 0\\6-2(x+1)>2x\end{array}\right\ \ \left\{\begin{array}{l}x^2-2x-8\geq 0\\6-2x-2>2x\end{array}\right\ \ \left\{\begin{array}{l}(x-4)(x+2)\geq 0\\4x<4\end{array}\right\\\\\\\left\{\begin{array}{l}x\in (-\infty \, ;-2\ ]\cup [\ 4\ ;+\infty \, )\\x<1\end{array}\right\ \ \ \ \Rightarrow \ \ \ x\in (-\infty \ ;-2\ ]$