Đáp án:
`(x-1) ² + 2x + 3`
`= (x−1)(x−1)+2x+3`
`= x(x-1) - 1(x-1) + 2x + 3`
`= x^2 - x - x + 1 + 2x + 3`
`= x^2 - 2x + 2x + 4`
`= x^2 + 4 + (-2x+2x)`
`= x^2 + 4`
Do `x^2 ≥ 0`
`4 > 0`
`-> (x-1) ² + 2x + 3 > 0`
Vậy `(x-1) ² + 2x + 3` luôn luôn `> 0 ∀x`