$(x-3)(4x + 5) + 2019$
$= 4x^2 - 7x - 15 + 2019$
$= 4x^2 - 2.\dfrac{7}{4}.2x + \dfrac{49}{16} + \dfrac{32015}{16}$
$= \left(2x - \dfrac{7}{4}\right)^2 + \dfrac{32015}{16}$
Ta có:
$\left(2x - \dfrac{7}{4}\right)^2 \geq 0, \,\forall x$
$\to \left(2x - \dfrac{7}{4}\right)^2 + \dfrac{32015}{16} \geq \dfrac{32015}{16}$
Dấu = xảy ra $\Leftrightarrow 2x - \dfrac{7}{4} = 0 \Leftrightarrow x = \dfrac{7}{8}$
Vậy $\min (x-3)(4x + 5) + 2019 = \dfrac{32015}{16} \Leftrightarrow x = \dfrac{7}{8}$
