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Najpierw założenia:
x - 2 ≠ 0, x ≠ 2
x - 3 ≠ 0, x ≠3
x² - 5x + 6 ≠ 0
Δ = 25 - 4 · 1 · 6 = 1, √Δ = 1
x₁ = 5 - 1/ 2 = 4/2 = 2
x₂ = 5 + 1/2 = 6/2 = 3
x² - 5x + 6 = (x - 2)(x - 3)
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Najpierw założenia:
x - 2 ≠ 0, x ≠ 2
x - 3 ≠ 0, x ≠3
x² - 5x + 6 ≠ 0
Δ = 25 - 4 · 1 · 6 = 1, √Δ = 1
x₁ = 5 - 1/ 2 = 4/2 = 2
x₂ = 5 + 1/2 = 6/2 = 3
x² - 5x + 6 = (x - 2)(x - 3)