$\text{Giải thích các bước giải:}$
$D = 2x² - 6x + 12$
$= 2(x² - 3x + 6)$
$= 2[(x² - 3x + \dfrac{9}{4}) + \dfrac{15}{4}]$
$= 2[(x - \dfrac{3}{2})² + \dfrac{15}{4}]$
$\text{Do}$ $(x - \dfrac{3}{2})² ≥ 0 ∀ x$
$⇒ (x - \dfrac{3}{2})² + \dfrac{15}{4} ≥ \dfrac{15}{4}$
$⇒2[(x - \dfrac{3}{2})² + \dfrac{15}{4}] ≥ \dfrac{15}{2}$
$⇒ D ≥ \dfrac{15}{2}$
$D_{Min} = \dfrac{15}{2} ⇔ x = \dfrac{3}{2}$
$E = \dfrac{1}{3}x² + \dfrac{1}{9}x - 2$
$= \dfrac{1}{3}(x² + \dfrac{1}{3}x - 6)$
$= \dfrac{1}{3}[(x² + \dfrac{1}{3}x + \dfrac{1}{36}) - \dfrac{217}{36}]$
$= \dfrac{1}{3}[(x + \dfrac{1}{6})² - \dfrac{217}{36}]$
$\text{Do}$ $(x + \dfrac{1}{6})² ≥ 0 ∀ x$
$⇒ (x + \dfrac{1}{6})² - \dfrac{217}{36} ≥ -\dfrac{217}{36}$
$⇒ \dfrac{1}{3}[(x + \dfrac{1}{6})² - \dfrac{217}{36}] ≥ \dfrac{217}{108}$
$⇒ E ≥ \dfrac{217}{108}$
$E_{Min} = \dfrac{217}{108} ⇔ x = -\dfrac{1}{6}$
$\huge\text{Hok tốt !}$
