Đáp án:
$B=2x^2 +6x-5$
$= (\sqrt[]{2}x)^2 +2 . \sqrt[]{2}x . \dfrac{3\sqrt[]{2}}{2} + \dfrac{9}{2} -\dfrac{19}{2}$
$=(\sqrt[]{2}x +\dfrac{3\sqrt[]{2}}{2})^2 -\dfrac{19}{2}$
Vì $(\sqrt[]{2}x +\dfrac{3\sqrt[]{2}}{2})^2 ≥ 0$
Nên $(\sqrt[]{2}x +\dfrac{3\sqrt[]{2}}{2})^2 -\dfrac{19}{2} ≥ -\dfrac{19}{2}$
Dấu ''='' xảy ra khi $\sqrt[]{2}x +\dfrac{3\sqrt[]{2}}{2} =0 ⇔ x = -\dfrac{3}{2}$
Vậy Min B $=-\dfrac{19}{2}$ tại $x=-\dfrac{3}{2}$
