Odpowiedź :
[tex]\frac{x-3}{x^{2}+x-20} = 0[/tex]
Z:
x² + x - 20 ≠ 0
M. zerowe:
Δ = b² - 4ac = 1² - 4·1·(-20) = 1 + 80 = 81
√Δ = √81 = 9
x₁ = (-1 - 9)/2 = -10/2 = -5
x₂ = (-1 + 9)/2 = 8/2 = 4
D = R \ {-5; 4}
x - 3 = 0
x = 3
[tex]\frac{x-3}{x^{2}+x-20} = 0[/tex]
Z:
x² + x - 20 ≠ 0
M. zerowe:
Δ = b² - 4ac = 1² - 4·1·(-20) = 1 + 80 = 81
√Δ = √81 = 9
x₁ = (-1 - 9)/2 = -10/2 = -5
x₂ = (-1 + 9)/2 = 8/2 = 4
D = R \ {-5; 4}
x - 3 = 0
x = 3