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(x - √2)(x + √2) ≥ 0
x nalezy (-oo, -√2> u <√2 ; +oo)
|x| = x^2 - 2
x = x^2 - 2 lub x = -(x^2 - 2)
x^2 - x - 2 = 0 x^2 + x - 2 = 0
Δ = 1 + 8 = 9 Δ = 1 + 8 = 9
√Δ = 3 √Δ = 3
x1 = (1 - 3)/2 = -1 x1 = (-1 - 3)/2 = -2
x2 = (1 + 3)/2 = 2 x2 = (-1 + 3)/2 = 1
Spośród liczb które otrzymałam czyli (-1) ; 2 ; (-2) i 1 do dziedziny należą liczby;
odp. x = 2 ; x = -2