Question

(х+3)(4х-1)/х+5<0 помогите решить неравенство методом интервала
/ это дробь

Answer 1

$\frac{(x+3)(4x-1)}{x+5}\ \textless \ 0\\\\ \frac{(x+3)*(4x-1)}{x+5}\ \textless \ 0\\\\ \frac{(x+3)*4*(x-\frac{1}{4})}{x+5}\ \textless \ 0\\\\ \frac{(x+3)*(x-\frac{1}{4})}{x+5}\ \textless \ 0\\\\ \frac{[x-(-3)]*[x-\frac{1}{4}]}{x-(-5)}\ \textless \ 0\\\\ -----(-5)+++++[-3]------[-\frac{1}{4}]+++++\ \textgreater \ x\\\\ x\in(-\infty;\ -5)\cup[\frac{1}{4};\ +\infty)$

Answer 2

(x+3)*(4x-1)/(x+5)<0   ОДЗ: x+5≠0   x≠-5
-∞____-____-5____+____-3____-____1/4____+____+∞
Ответ: x∈(-∞;-5)U(-3;1/4).