Odpowiedź:
a)
(x - 1)/(2x - 3) = - 1/x
założenie:
2x - 3 ≠ 0 ∧ x ≠ 0
2x ≠ 3 ∧ x ≠ 0
x ≠ 3/2 ∧ x ≠ 0
x ≠ 1,5 ∧ x ≠ 0
D: x ∈ R \ {0 , 1,5 }
x(x - 1) = - (2x - 3)
x² - x = - 2x + 3
x² - x + 2x - 3 = 0
x² + x - 3 = 0
a = 1 , b = 1 , c = - 3
Δ = b² - 4ac = 1² - 4 * 1 * (- 3) = 1 + 12 = 13
√Δ = √13
x₁ = (- b - √Δ)/2a = (- 1 - √13)/2 = - (1 + √13)/2
x₂ = (- b + √Δ)/2a = (- 1 + √13)/2 = (√13 - 1)/2
b)
(x - 1)(x - 3)(x² - 16) = 0
(x - 1)(x - 3)(x - 4)(x + 4) = 0
x - 1 = 0 ∨ x - 3 = 0 ∨ x - 4 = 0 ∨ x + 4 = 0
x = 1 ∨ x = 3 ∨ x = 4 ∨ x = - 4
x₀ = { - 4 , 1 , 3 , 4 }
c)
(x + 1)(x² - 3)(x² + 4) = 0
Ponieważ x² + 4 > 0 dla x ∈ R , więc:
(x + 1)(x² - 3) = 0
(x + 1)(x - √3)(x + √3) = 0
x + 1 = 0 ∨ x - √3 = 0 ∨ x + √3 = 0
x = - 1 ∨ x = √3 ∨ x = - √3
x₀ = { - √3 , - 1 , √3 }
d)
x³ - 16x = 0
x(x² - 16) = 0
x(x - 4)(x + 4) = 0
x = 0 ∨ x - 4 = 0 ∨ x + 4 = 0
x = 0 ∨ x = 4 ∨ x = - 4
x₀ = { - 4 , 0 , 4 }
