Răspuns:
Explicație pas cu pas:
{x+y = 1
{xy = 1
x = 1-y =>
(1-y)·y = 1 <=> y-y²-1 = 0 <=>
y²-y+1 = 0 ; a = 1 ; b = -1 ; c = 1 ; Δ = b²-4ac = (-1)²-4·1·1 = 1-4 = -3
√Δ = √(-3) = i√3 ; unde i = √-1
y₁,₂ = (-b±√Δ)/2a = (1±i√3)/2
y₁ = (1-i√3)/2 => x₁ = ²⁾1-(1-i√3)/2 => x₁ = (1+ i√3)/2
y₂ = (1+i√3)/2 => x₂ = ²⁾1-(1+i√3)/2 => x₂ = (1-i√3)/2
