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x²-4 ≠ 0
(x-2)(x+2) ≠ 0
x≠2 v x≠-2
x ∈ R \ {-2,2}
x/(x+2) + 1/(x-2) = 4/(x²-4) // -1/(2-x) można zapisać jako + 1/(x-2)
x(x-2) / (x-2)(x+2) + 1*(x+2) / (x-2)(x+2) - 4 / (x-2)(x+2) = 0
(x² -2x + x + 2 - 4) / (x-2)(x+2) = 0
(x² - x - 2)/(x-2)(x+2) = 0 // a/b = 0 gdy a = 0
x² - x - 2 = 0
Δ = (-1)² - 4*1*(-2)
Δ = 1 + 8
Δ = 9
√Δ = 3
x₁ = (1-3) / 2
x₁ = -1
x₂ = (1+3)/2
x₂ = 2 odpada ( x ∉ Df)
spr
-1 / -1 + 2 - 1 / 2-(-1) = 4 / (-1)^2 -4
-1 - 1/3 = 4 / 1-4
-4/3 = -4/3
L=P
Odp rozwiązaniem jest x = -1