Question

Решите систему неравенств (40) баллов

Answer 1

Ответ:

$\left\{\begin{array}{l}19x-2x^2-42>0\\(x^2-9)(x-4)(1-x)\leq 0\end{array}\right\ \ \left\{\begin{array}{l}2x^2-19x+42<0\\(x-3)(x+3)(x-4)(1-x)\leq 0\end{array}\right\\\\\\\left\{\begin{array}{l}2(x-3,5)(x-6)<0\\(x-3)(x+3)(x-4)(x-1)\geq 0\end{array}\right$

$2(x-3,5)(x-6)<0\ \ ,znaki:\ \ +++(3,5)---(6)+++\\\\x\in (\ 3,5\ ;\ 6\ )\\\\\\(x-3)(x+3)(x-4)(x-1)\geq 0\\\\znaki:\ +++[-3\ ]---[\ 1\ ]+++ [\ 3\ ]---[\ 4\ ]+++\\\\x\in (-\infty ;-3\ ]\cup [\ 1\ ;\ 3\ ]\cup [\ 4\ ;+\infty \, )\\\\\\\left\{\begin{array}{l}x\in (\ 3,5\ ;\ 6\ )\\x\in (-\infty ;-3\ ]\cup [\ 1\ ;\ 3\ ]\cup [\ 4\ ;+\infty \, )\end{array}\right\ \ \ \Rightarrow \ \ \ \ \boxed{\ x\in [\ 4\ ;\ 6\ )\ }$