Đặt `A = (2x - 1)(x + 2)(2x + 1)(x + 3)`

`A = (4x^2 - 1)(x^2 + 5x + 6)`

`A = 4x^4 + 20x^3 + 24x^2 - x^2 - 5x - 6`

`A = 4x^4 + 20x^3 + 23x^2 - 5x - 6`

`A = 4x^4 + 20x^3 + 25x^2 - 2x^2 - 5x - 6`

`A = (2x^2 + 5x)^2 - (2x^2 + 5x) - 6`

`A = (2x^2 + 5x)^2 - 2 . (2x^2 + 5x) . 1/2 + 1/4 - 1/4 - 6`

`A = [(2x^2 + 5x) - 1/2]^2 - 25/4`

`A = (2x^2 + 5x - 1/2)^2 - 25/4`

Vì `(2x^2 + 5x - 1/2)^2 >= 0 AA x`

`to A >= -25/4`

Dấu "`=`" xảy ra khi `2x^2 + 5x - 1/2 = 0`

`4x^2 + 10x - 1 = 0`

`(2x + 5/2)^2 - 29/4 = 0`

`(2x + 5/2)^2 = 29/4`

`[(2x + 5/2 = sqrt29/2),(2x + 5/2 = -sqrt29/2):}`

`[(2x = (-5 + sqrt29)/2),(2x = (-5 - sqrt29)/2):}`

`[(x = (-5 + sqrt29)/4),(x = (-5 - sqrt29)/4):}`

Vậy `A_(min) = -25/4`khi `x = (-5 + sqrt29)/4` hoặc `x = (-5 - sqrt29)/4`