f(x)=(2x-1)/(4x²-2)+(x-3)/(x²-6x+9)=(2x-1)/2(2x²-1)+(x-3)/(x-3)²=(2x-1)/2(2x²-1)+1/(x-3)
Wyznaczymy Df :
2x²-1≠0|:2 ∧ x-3≠0
x²-1/2≠0 ∧ x≠3
(x+√(1/2))(x-√(1/2))≠0
x≠-√2/2 ∧ x≠√2/2 ∧ x≠3
Df=R-{-√2/2,√2/2,3}
Odpowiedź:
4x^2 - 2 ≠ 0
4x^2 ≠ 2 /4
x^2 ≠ 1/2
x = √1/2
X^2 - 6x +9
Δ = 6^2 - 4(1*9) = 36 - 36 = 0
√Δ = 0
x1 = 6 -0/2 = 3
x2 = 6 + 0/2 = 3
Df: X ∈ (-∞,√1/2)∪ (√1/2,3)∪ (3,∞)
Szczegółowe wyjaśnienie:
