Đáp án:
$\min B = -\dfrac 92 \Leftrightarrow x =\dfrac 12$
Giải thích các bước giải:
$B = 2x^2 - 2x - 4$
$\to B = 2\left(x^2 - x + \dfrac{1}{4}\right) - \dfrac{9}{2}$
$\to B = 2\left(x - \dfrac{1}{2}\right)^2 - \dfrac 92$
Ta có:
$\left(x - \dfrac{1}{2}\right)^2 \geq 0, \,\forall x$
$\to 2\left(x - \dfrac{1}{2}\right)^2 - \dfrac 92 \geq -\dfrac 92$
$\to B \geq -\dfrac 92$
Dấu = xảy ra $\Leftrightarrow x = \dfrac{1}{2}$
Vậy $\min B = -\dfrac 92 \Leftrightarrow x =\dfrac 12$
