Question

Решите методом интервалов:
1.1) (x+2)(x-7)>0
1.2) (x+5)(x-8) 0

2.1) $x^{2}$ -16x 0
2.3) ($x^{2}$-1)(x+3) 0

Answer 1

$1)\ \ (x+2)(x-7)>0\ \ ,\ \ \ \ +++(-2)---(7)+++\\\\x\in (-\infty ;-2)\cup (7;+\infty )\\\\2)\ \ (x+5)(x-8)>0\ \ ,\ \ \ +++(-5)---(8)+++\\\\x\in (-\infty ;-5)\cup (8;+\infty )\\\\b)\ \ (x+5)(x-8)<0\ \ \to \ \ \ x\in (-5\, ;\, 8)\\\\3)\ \ x^2-16x>0\ \ ,\ \ \ x(x-16)>0\ \ ,\\\\+++(0)---(16)+++\ \ \ \ x\in (-\infty ;0)\cup (16;+\infty )\\\\b)\ \ x^2-16x<0\ \ \ \to \ \ \ x\in (\, 0\, ;\, 16)\\\\4)\ \ (x^2-1)(x+3)>0\ \ ,\ \ \ (x-1)(x+1)(x+3)>0\ \ ,\\\\---(-3)+++(-1)---(1)+++$

$x\in (-3;-1)\cup (\, 1\, ;+\infty )\\\\b)\ \ (x^2-1)(x+3)<0\ \ \to \ \ \ x\in (-\infty ;-3)\cup (-1\, ;\, 1\, )$