Hướng dẫn trả lời:
Ta có: `-4x^2 - 4x - 2`
`= -4x^2 - 4x - 1 - 1`
`= - (4x^2 + 4x + 1) - 1`
`= - [(2x)^2 + 2cdot2x + 1^2] - 1`
`= - (2x + 1)^2 - 1`
Vì `(2x + 1)^2 ≥ 0 ∀ x ∈ \mathbb{Z}` nên `- (2x + 1)^2 ≤ 0 ∀ x ∈ \mathbb{Z}`
`→ - (2x + 1)^2 - 1 ≤ - 1 < 0 ∀ x ∈ \mathbb{Z}`
Hay `-4x^2 - 4x - 2 < 0 ∀ x ∈ \mathbb{Z}`
`→ đpcm.`
Vậy `-4x^2 - 4x - 2 < 0 ∀ x ∈ \mathbb{Z}`.