Răspuns:
Explicație pas cu pas:
E(x) = x²+4x+5
(x+2)² = x²+4x+4 =>
E(x) = x²+4x+4+1 = (x+2)²+1
(x+2)² ≥ 0 ; (∀) x ∈ R =>
E(x) ≥ 1 ; (∀) x ∈ R
E(a) + E(b) + E(c) = 3 <=>
a²+4a+5+b²+4b+5+c²+4c+5 = 3 <=>
(a+2)²+1 + (b+2)²+1 + (c+2)² + 1 = 3 <=>
(a+2)²+(b+2)²+(c+2)² = 0 =>
(a+2)² = 0 ; (b+2)² = 0 ; (c+2)² = 0 =>
a = b = c = -2
