Đáp án:
`a, x^2 - 2x + 7`
`= (x^2 - 2x + 1) + 6`
`= (x - 1)^2 + 6`
Vì `(x - 1)^2 ≥ 0 ∀ x`
`⇒ (x - 1)^2 ≥ 6 ∀ x`
Dấu "`=`" xảy ra khi:
`(x - 1)^2 = 0`
`⇔ x - 1 = 0`
`⇔ x = 1`
Vậy `MIN = 6 ↔ x = 1`
`b, x^2 + 6x + 1`
`= (x^2 + 6x + 9) - 8`
`= (x^2 + 2.x.3 + 3^2) - 8`
`= (x + 3)^2 - 8`
Vì `(x + 3)^2 ≥ 0 ∀ x`
`⇒ (x + 3)^2 ≥ - 8 ∀ x`
Dấu "`="` xảy ra khi:
`(x + 3)^2 = 0`
`⇔ x + 3 = 0`
`⇔ x = -3`
Vậy `MIN = -8 ↔ x = -3`
`c, 4x^2 - 4x + 2`
`= (4x^2 - 4x + 1) + 1`
`= [(2x)^2 - 2.2x.1 + 1^2] + 1`
`= (2x - 1)^2 + 1`
Vì `(2x - 1)^2 ≥ 0 ∀ x`
`⇒ (2x - 1)^2 + 1 ≥ 1 ∀ x`
Dấu "`=`" xảy ra khi:
`(2x - 1)^2 = 0`
`⇔ 2x - 1= 0`
`⇔ 2x = 1`
`⇔ x = 1/2`
Vậy `MIN = 1 ↔ x = 1/2`
`d, x^2 + 8x - 16`
`= x^2 + 8x + 16 - 32`
`= (x^2 + 2.x.4 + 4^2) - 32`
`= (x + 4)^2 - 32`
Vì `(x + 4)^2 ≥ 0 ∀ x`
`⇒ (x + 4)^2 - 32 ≥ - 32`
Dấu "`=`" xảy ra khi:
`(x + 4)^2 = 0`
`⇔ x +4 =0`
`⇔ x = -4`
Vậy `MIN = -32 ↔ x = -4`
