`a, A = x^2 - 8x + 38 = x^2 - 8x + 16 + 22 = (x - 4)^2 + 22 ≥ 22`
Dấu `=` xảy ra `↔ x - 4 = 0 ↔ x = 4`
Vậy `A_(min) = 22 ↔ x = 4`
`b, B = (x + 1)(2x - 1) = 2x^2 - x + 2x - 1 = 2x^2 + x - 1 = 2x^2 + x + 1/8 - 9/8`
`= (x√2 + 1/(√8))^2 - 8/9 ≥ -9/8`
Dấu `=` xảy ra `↔ x√2 + 1/(√8) = 0 ↔ x√2 = -1/(√8) ↔ x = -1/4`
Vậy `B_(min) = -9/8 ↔ x = -1/4`