Đáp án:
$\max B =\dfrac{49}{12}\Leftrightarrow x = -\dfrac16$
Giải thích các bước giải:
$\quad B = (1-x)(3x+4)\qquad \left(-\dfrac43\leqslant x \leqslant 1\right)$
$\to B =\dfrac13(3 - 3x)(3x+4)$
$\to B \leqslant \dfrac13\left(\dfrac{3 - 3x + 3x + 4}{2}\right)^2\quad (BDT\ AM-GM)$
$\to B \leqslant \dfrac13\cdot\dfrac{49}{4}$
$\to B \leqslant \dfrac{49}{12}$
Dấu $=$ xảy ra $\Leftrightarrow 3 - 3x = 3x + 4 \Leftrightarrow -\dfrac16$
Vậy $\max B =\dfrac{49}{12}\Leftrightarrow x = -\dfrac16$
