Odpowiedź:
(4x² - 9)/(x² + 5x +6) : (2x² - 7x + 6)/(x² -4)
założenie:
x² + 5x + 6 ≠ 0 ∧ x² - 4 ≠ 0
Obliczamy miejsca zerowe
x² + 5x + 6 = 0
a = 1 , b = 5 , c = 6
Δ = b² - 4ac = 5² - 4 * 1 * 6 = 25 - 24 = 1
√Δ = √1 = 1
x₁ = (- b - √Δ)/2a = ( - 5 - 1)/2 = - 6/2 = - 3
x₂ = ( - b + √Δ)/2a = (- 5 + 1)/2 = - 4/2 = - 2
x² - 4 = (x - 2)(x + 2)
x - 2 ≠ 0 ∧ x + 2 ≠ 0
x ≠ 2 ∧ x ≠ - 2
D: x ∈ R \ { - 3 , - 2 , 2 }
(4x² - 9)/(x² + 5x +6) : (2x² - 7x + 6)/(x² -4) =
= [(2x - 3)(2x + 3)]/[(x + 3)(x + 2) : (2x² - 7x + 6)/(x - 2)(x + 2) =
= [(2x - 3)(2x + 3)]/[(x + 3)(x + 2) * (x - 2)(x + 2)/(2x² - 7x + 6) =
= [(2x - 3)(2x + 3)]/(x + 3) * (x - 2)/(2x² - 7x + 6)
założenie:
2x² - 7x + 6 ≠ 0
Obliczamy miejsca zerowe
a = 2 , b = - 7 , c = 6
Δ = b² - 4ac = (- 7)² - 4 * 2 * 6 = 49 - 48 = 1
√Δ = √1 = 1
x₁ = (- b - √Δ)/2a = (7 - 1)/4 = 6/4 = 1 2/4 = 1 1/2
x₂ = ( - b + √Δ)/2a = (7 + 1)/4 = 8/4 = 2
D: x ∈ R \ { - 3 , - 2 , 1 1/2 , 2 }
[(2x - 3)(2x + 3)]/(x + 3) * (x - 2)/(2x² - 7x + 6) =
= [(2x - 3)(2x + 3)/(x + 3) * (x - 2)/[2(x - 1 1/2)(x - 2)] =
= [(2x - 3)(2x + 3)/(x + 3) * 1/[2(x - 1 1/2)] =
= [(2x - 3)(2x + 3)/(x + 3) * 1/(2x - 3) = (2x + 3)/x + 3) * 1 = (2x + 3)/(x + 3)