a)
[tex]12 - x(3-x) > 0\\\\12-3x+x^{2} > 0\\\\x^{2}-3x+12 > 0\\\\a = 1, \ b = -3, \ c = 12\\\\\Delta = b^{2}-4ac = (-3)^{2}-4\cdot1\cdot12 = 9-48 = -39 < 0\\\\\Delta < 0, \ brak \ miejsc \ zerowych, \ wykres \ paraboli \ lezy \ nad \ osia \ Ox\\\\x \in R[/tex]
b)
[tex](5-2x)(3-4x) > 0\\\\\underline{M. \ zerowe}\\\\3-4x = 0 \ \vee \ 5-2x = \\\\-4x = -3 \ \vee \ -2x = -5\\\\x = \frac{3}{4} \ \vee \ x = \frac{5}{2}\\\\a > 0\\\\x \in (-\infty;\frac{3}{4}) \ \cup \ (\frac{5}{2};+\infty)[/tex]
c)
[tex](12x-3)(x+12) = 0\\\\\underline{M. \ zerowe}\\\\x+12 = 0 \ \vee \ 12x-3 = 0\\\\x = -12 \ \vee \ 12x = 3\\\\x = -12 \ \vee \ x = \frac{1}{4}\\\\a > 0\\\\x \in (-12;\frac{1}{4})[/tex]
