Odpowiedź:
zad 14
a)
1/3x³ - x² = 0
x²(1/3x - 1) = 0
x² = 0 ∨ 1/3x - 1 = 0
x = 0 ∨ 1/3x = 1
x = 0 ∨ x = 1 : 1/3
x = 0 ∨ x = 1 * 3 = 3
b)
3x³ + 9x² = 0
3x²(x + 3) = 0
3x² = 0 ∨ x + 3 = 0
x = 0 ∨ x = - 3
c)
x⁴ - 2x³ = 0
x³(x - 2) = 0
x³ = 0 ∨ x - 2 = 0
x = 0 ∨ x = 2
d)
6x⁴ + x³ = 0
x³(6x + 1) = 0
x³ = 0 ∨ 6x + 1 = 0
x = 0 ∨ 6x = - 1
x = 0 ∨ x = - 1/6
zad 15
a)
(x + 1)(x + 2)(x + 3) ; (x² - 1)(x² - 9) = 0
x² - 1 = (x - 1)(x + 1)
x² - 9 = (x - 3)(x + 3)
x + 1 = 0 ∨ x + 2 = 0 ∨ x + 3 = 0 ∨ x - 1 = 0 ∨ x + 1 = 0 ∨ x - 3 = 0 ∨ x + 3 = 0
x = - 1 ∨ x = - 2 ∨ x = - 3 ∨ x = 1 ∨ x = - 1 ∨ x = 3 ∨ x = - 3
v - znaczy lub
jednocześnie obydwa równania spełniają x = {- 1 , - 3 }
b)
(x - 4)(2x + 1)(2x + 8) = 0 ; (4x² - 1)(x² - 16) = 0
4x² - 1 = (2x - 1)(2x + 1)
x² -16 = (x - 4)(x + 4)
x - 4 = 0 ∨ 2x + 1 = 0 ∨ 2x + 8 = 0 ∨ 2x - 1 = 0 ∨ 2x + 1 = 0 ∨ x - 4 = 0 ∨
∨ x + 4 = 0
x = 4 ∨ x = - 1/2 ∨ x = - 4 ∨ x = 1/2 ∨ x = - 1/2 ∨ x = 4 ∨ x = - 4
jednocześnie obydwa równania spełniają x = {- 4 , - 1/2 , 4 }
