Question

Решить методом интервалов неравенств 6 и 8​

Answer 1

$6)\ \ \dfrac{y+\sqrt5}{(2y+9)(y-8)}<0\\\\y_1=-\sqrt5\ ,\ \ y_2=-4,5\ \ ,\ \ y_3=8\\\\znaki:\ \ ---(-4,5)+++(-\sqrt5)---(8)+++\\\\x\in (-\infty \, ;\, -4,5\, )\cup (\, -\sqrt5\, ;8)\\\\\\8)\ \ \dfrac{(5x-6)(2-3x)}{(4+x)(5-2x)}\geq 0\ \ \to \ \ \ \dfrac{(5x-6)(3x-2)}{(x+4)(2x-5)}\geq 0\\\\x_1=1,2\ \ ,\ \ x_2=\frac{2}{3}\ \ ,\ \ x_3=-4\ \ ,\ \ x_4=2,5\\\\znaki:\ \ +++(-4)---[\, \frac{2}{3}\, ]+++[1,2]---(2,5)+++\\\\x\in (-\infty \, ;\, -4\, )\cup \Big[\ \dfrac{2}{3}\, ;\, 1,2\ \Big]\cup (\, 2,5\, ;+\infty )$