Question

Pilne, rozwiąż 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

Odpowiedź:

a)

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

5x - 1 = 0   v   3 - x = 0

x = 0,2   v   x = - 3

X ∈ (- ∞, -3) ∪ (0,2, + ∞)

b)

6x³ + 10x² + 4x = 0

2x (3x² + 5x + 2) = 0

2x = 0   v   3x² + 5x + 2 = 0

x = 0   v   Δ = 1

x = 0   v   [tex]\sqrt{\Delta}} = 1[/tex]

x = 0   v   [tex]x = \frac{- 5 - 1}{6}\ \ \ \vee \ \ \ x = \frac{-5 +1}{6}[/tex]

x = 0   v   x = - 1   v   x = [tex]x = -\frac{2}{3}[/tex]

c)

2x - 1 /x = 2x / x-2

(2x - 1)(x - 2) = x * 2x

2x² - x - 4x + 2 = 2x²

- 5x = - 2

x = 0,4

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

[tex]\frac{1}{2} - \frac{2}{3}x - \frac{4}{3} < \frac{3}{4}x + 1[/tex]

[tex]-\frac{2}{3}x - \frac{3}{4}x < 1 - \frac{1}{2} + \frac{4}{3}[/tex]

[tex]-\frac{2*4}{3*4}x - \frac{3 * 3}{4 * 3}x < 1 - \frac{1 * 3}{2 * 3} + \frac{4 * 2}{3 * 2}[/tex]

[tex]- \frac{6}{12}x - \frac{9}{12}x < 1 - \frac{3}{6} + \frac{8}{6}[/tex]

[tex]- \frac{15}{12}x < \frac{11}{6}[/tex]

[tex]x < \frac{11}{6} * (- \frac{12}{15})[/tex]

[tex]x = - \frac{22}{15}[/tex]