Odpowiedź:
zad 1
(x - 4)(x +2)(2x - 6) = 0
x - 4 = 0 ∨ x + 2 =0 ∨ 2x - 6 = 0
x = 4 ∨ x = - 2 ∨ 2x = 6
x = 4 ∨ x = - 2 ∨ x = 3
x = { - 2 , 3 , 4 }
zad 2
(2x-5)(6 - x)(x² + 1) = 0
Ponieważ x² + 1 > 0 dla x ∈ R , więc:
2x - 5 = 0 ∨ 6 - x = 0
2x = - 5 ∨ - x = - 6
x = - 5/2 ∨ x = 6
x = - 2,5 ∨ x = 6
zad 3
(4 - 2x)/(x + 2) = 0
założenie:
x + 2 ≠ 0
x ≠ - 2
D: x ∈ R \ { - 2}
(4 - 2x) = 0
- 2x = - 4
2x = 4
x = 4/2 = 2
