Đáp án:
Ta có:
`A = 16^2 - 8x + 3/2`
`= (4x)^2 - 2.4x.1 + 1 + 1/2`
`= (4x - 1)^2 + 1/2`
Vì `(4x - 1)^2 ≥ 0 ∀ x ∈ R`
`⇒ (4x - 1)^2 + 1/2 ≥ 1/2`
Suy ra: $Min_A$ `= 1/2` khi `(4x - 1)^2 = 0`
`⇔ 4x - 1 = 0`
`⇔ 4x = 1 ⇒ x = 1/4`
Vậy $Min_A$ `= 1/2` khi `x = 1/4`
`B = 4x^2 - 12x + 11`
`= 4x^2 - 12x + 9 + 2`
`= (2x)^2 - 2.2x.3 + 3^2 + 2`
`= (2x - 3)^2 +2`
Vì `(2x - 3)^2 ≥ 0 ∀ x ∈ R`
`⇒ (2x - 3)^2 + 2 ≥ 2`
Suy ra: $Min_B$ `= 2` , khi `(2x - 3)^2 = 0`
`⇔ 2x - 3 = 0 ⇒ 2x = 3`
`⇒ x = 3/2`
Vậy $Min_B$ `= 2` khi `x = 3/2`
`C = 25x^2 - 20x + 6`
`= 25x^2 - 20x + 4 + 2`
`= (5x)^2 - 2.5x.2 + 2^2 + 2`
`= (5x - 2)^2 + 2`
Vì `(5x - 2)^2 ≥ 0 ∀ x ∈ R`
`⇒ (5x - 2)^2 +2 ≥ 2`
Suy ra: $Min_C$ `= 2`, khi `(5x - 2)^2 = 0`
`⇔ 5x - 2 = 0 ⇒ 5x = 2`
`⇒ x = 2/5`
Vậy $Min_C$ `= 2`, khi `x = 2/5`
Học tốt
