Question

помогите решиииить, пожалуйста.

Answer 1

$\frac{2x^2+2x-8+2x^2-2x-1}{x-1}\leq 0\Leftrightarrow \frac{4x^2-9}{x-1}\leq 0\\\frac{\left ( 2x-3 \right )\left ( 2x+3 \right )}{x-1}\leq 0\Rightarrow x\in \left ( -\infty ;-\frac{3}{2} \right ]\cup \left ( 1;\frac{3}{2} \right ]$

Answer 2

$\dfrac{(2x^2+2x-8)-(-2x^2+2x+1)}{x-1}\leq 0\ \ \ ,\ \ \ \ \ x\ne 1\\\\\\\dfrac{2x^2+2x-8+2x^2-2x-1}{x-1}\leq 0\\\\\\\dfrac{4x^2-9}{x-1}\leq 0\ \ \ ,\ \ \ \dfrac{(2x-3)(2x+3)}{x-1}\leq 0\\\\znaki:\ \ \ ---\Big[-\dfrac{3}{2}\, \Big]+++(1)---\Big[\, \dfrac{3}{2}\, \Big]+++\\\\\\x\in \Big(-\infty \, ;-\dfrac{3}{2}\ \Big]\cup \Big(\ 1\, ;\ \dfrac{3}{2}\ \Big]\\\\\\x\in (-\infty \, ;-1,5\, ]\cup (\ 1\, ;\ 1,5\, ]$