Odpowiedź:
zad 3
(x⁵ - 3x⁴- 4x³ + 12x² + 4x- 12) : (x - 3) = 0
założenie:
x - 3 ≠ 0
x ≠ 3
D: x∈ R - {3}
(x⁵ - 3x⁴- 4x³ + 12x² + 4x- 12) : (x - 3) = 0
[x⁴(x -3) - 4x²(x - 3) + 4(x - 3)] : (x - 3) = 0 | : (x - 3)
x⁴(x - 3)/(x - 3) - 4x²(x - 3)/(x - 3) + 4(x - 3)/(x - 3) = 0
x⁴ - 4x² + 4 = 0
za x² wstawiam z
z²- 4x + 4= 0
a = 1 , b = - 4 , c = 4
Δ = b² - 4ac = (- 4)² - 4 * 1 * 4 = 16 - 16 = 0
z = - b/2a = 4/2 = 2
z = x²
x² = 2
x² - 2 = 0
(x - √2)(x + √2) = 0
x - √2 = 0 ∨ x + √2 = 0
x = √2 ∨ x = - √2
v - znaczy "lub"
