Question

PILNE!! Na jutro równania



a) (5x - 1)(3 - x) >/ 0

b) 6x^3 + 10x^2 + 4x = 0

c) 2x - 1 /x = 2x / x-2

d) 1/2 - 2x +4 /3 < 3/4x + 1

Answer 1

a)

[tex](5x-1)(3-x) \geq 0\\\\\underline{M. \ zerowe}\\\\(5x-1)(3-x) = 0\\\\5x-1 = 0 \ \vee \ 3-x = 0\\\\5x = 1 \ \vee \ -x = -3\\\\x = \frac{1}{5} \ \vee \ x = 3[/tex]

a < 0, to parabola zwrócona jest ramionami do dołu, wówczas:

[tex]\boxed{x\in\langle\frac{1}{5};3\rangle}[/tex]

b)

[tex]6x^{3}+10x^{2}+4x = 0 \ \ \ /:2\\\\3x^{3}+5x^{2}+2x = 0\\\\x(3x^{2}+5x+2) = 0\\\\x_{o} = 0\\\\lub\\\\3x^{2}+5x+2 = 0\\\\a = 3, \ b = 5, \ c = 2\\\\\Delta = b^{2}-4ac} = 5^{2} -4\cdot3\cdot2 = 25 - 24 = 1\\\\\sqrt{\Delta} = \sqrt{1} = 1\\\\x_1 = \frac{-b-\sqrt{\Delta}}{2a} = \frac{-5-1}{2\cdot3} = \frac{-6}{6} = -1\\\\x_2 = \frac{-b+\sqrt{\Delta}}{2a} = \frac{-5+1}{6} = \frac{-4}{6} = -\frac{2}{3}[/tex]

[tex]\boxed{x \in \{-3, -\frac{2}{3},0\}}[/tex]

c)

[tex]\frac{2x-1}{x} = \frac{2x}{x-2}\\\\\underline{Dziedzina}\\\\x \neq 0 \ \wedge \ x \neq 2\\\\D = R \setminus\{0,2\}\\\\\\(2x-1)(x-2) = 2x\cdot x\\\\2x^{2}-4x-x+2 = 2x^{2}\\\\-5x=-2 \ \ \ /:(-5)\\\\\boxed{x = \frac{2}{5}}[/tex]