Odpowiedź:
zad 1
(x² - 4)/(x² - 4x + 4) = 0
założenie :
x² - 4x + 4 ≠ 0
a = 1 , b = - 4 , c = 4
Δ = b² - 4ac = (- 4)² - 4 * 1 * 4 = 16 - 16 = 0
x = - b/2a = 4/2 = 2
x ≠ 2
D: x ∈ R \ {2}
x² - 4 = 0
(x - 2)(x + 2) = 0
x - 2 = 0 ∨ x + 2 = 0
x = 2 nie należy do dziedziny
x = - 2
Odp: Równanie ma jedno rozwiązanie x = - 2
zad 2
cosα = 2/5
cos²α = (2/5)² = 4/25
1 - sin²α = 4/25
sin²α = 1 - 4/25 = 21/25
sinα = √(21/25) = √21/5
tgα = sinα/cosα = √21/5 : 2/5 = √21/5 * 5/2 = √21/2
tgα + 2sinαcosα = √21/2 + 2 * √21/5 * 2/5 = √21/2 + 4√21/25 =
= (25√21 + 2 * 4√21)/50 = (25√21 + 8√21)/50 = 23√21/50
zad 3
- x² + 2x + 3 ≤ 0
a = - 1 , b = 2 , c = 3
Δ = 2² - 4 * (- 1) * 3 = 4 + 12 = 16
√Δ = √16 = 4
x₁ = ( - b - √Δ)/2a = (- 2 - 4)/(- 2) = - 6/(- 2) = 6/2 = 3
x₂ = (- b + √Δ)/2a = (- 2 + 4)/(- 2) = 2/(- 2) = - 2/2 = - 1
a < 0 więc ramiona paraboli skierowane do dołu
x ∈ ( - ∞ , - 1 > ∪ < 3 , + ∞ )
