Szczegółowe wyjaśnienie:
a) Df= R
b) x-1 =/ 0
x=/1
Df= R\{1}
c)6x+12>=0
6x>= -12
x>= -2
Df= <-2,oo)
d) 8-2x>0
-2x>-8/:(-2)
x<4
Df= (- oo, 4)
zad 3
[tex] {x}^{2} - 1 = (x - 1)(x + 1) \\ x1 = 1 \\ x2 = - 1 \\ \\ os \: symetri \: x = \frac{1 + ( - 1)}{2} = 0 \\ p = 0 \\ q = {0}^{2} - 1 = - 1 \\ a > 0 \\ wiec \: zw = < - 1. \infty )[/tex]
