Răspuns:
Explicație pas cu pas:
x²-6x+5 < 0
x²-6x+5 = 0 <=> x²-5x-x+5 = 0 <=>
x(x-5)-(x-5) = 0 <=> (x-5)(x-1) = 0 => x₁ = 1 ; x₂ = 5
=> Solutie : x ∈ (1 ; 5)
4x²+x+1 ≥ 0
4x²+x+1 = 0 ; a = 4 ; b = 1 ; c = 1
Δ = b²-4ac = 1²-4·4·1 = 1-16 < 0 =>
(∀) x ∈ R ; 4x²+x+1 > 0
=> Solutie : x ∈ R
