Question

Помогите решить неравенство (метод интервала)

Answer 1

$\frac{x^3-x^2+x-1}{x+8}\leq 0\\\\\frac{x^2(x-1)+(x-1)}{x+8}\leq 0\\\\\frac{(x-1)(x^2+1)}{x+8} \leq 0\\\\x^2+1>0\; pri\; x\in (-\infty ,+\infty )\; \; \to \; \; \frac{x-1}{x+8}\leq 0\\\\x-1=0\; ,\; \; x_1=1\; \; ;\; \; \; x+8=0\; ,\; \; x_2=-8\\\\ +++(-8)---[\, 1\, ]+++\\\\\underline {x\in (-8,1\, ]}$

Answer 2

ОДЗ

x≠-8

x^3-x^2+x-1≤0

(x^3+x)-(x^2+1)≤0

x*(x^2+1)-(x^2+1)≤0

(x^2+1)(x-1)≤0

Нули функции:

x^2+1=0 x-1=0

x^2≠-1 x=1

+ __ +

______-8_____________1__________

(-8;1]