a)
[tex]x^{2}-5x > -6\\\\x^{2}-5x+6 > 0\\\\\Delta = b^{2}-4ac = (-5)^{2}-4\cdot1\cdot6 = 25 - 24 = 1\\\\\sqrt{\Delta} = 1\\\\x_1 = \frac{-b-\sqrt{\Delta}}{2a} = \frac{-(-5)-1}{2\cdot1} =\frac{4}{2} = 2\\\\x_2 = \frac{-b+\sqrt{\Delta}}{2a} = \frac{-(-5)+1}{2} = \frac{6}{2} = 3\\\\a > 0, \ ramiona \ wykresu \ paraboli \ skierowane \ do \ gory\\\\x \in (-\infty; 2) \ \cup \ (3;+\infty)[/tex]
d)
[tex]3x(x-2) < x\\\\3x^{2}-6x-x < 0\\\\3x^{2}-7x < 0\\\\M. \ xerowe:\\\\x(3x-7) = 0\\\\x = 0 \ \vee \ 3x-7 = 0\\\\x = 0 \ \vee \ x = \frac{7}{3}\\\\a > 0\\\\x \in (0;\frac{7}{3})[/tex]
g)
[tex]10x < (3x-1)(x+2)\\\\10x < 3x^{2}+6x - x - 2\\\\10x<3x^{2}+5x-2\\\\-3x^{2}+10x-5x+2 < 0\\\\-3x^{2} +5x + 2 < 0\\\\\Delta = 5^{2} - 4\cdot(-3)\cdot2 = 25+24 = 49\\\\\sqrt{\Delta} = \sqrt{49} = 7\\\\x_1 = \frac{-5+7}{-6} = \frac{-5+7}{-6} = \frac{2}{-6} = -\frac{1}{3}\\\\x_2 = \frac{-5-7}{-6} = \frac{-12}{-6} = 2\\\\a < 0, \ ramiona \ paraboli \ skierowane \ do \ dolu\\\\x \in (-\frac{1}{3};2)[/tex]
