Đáp án:
$\max(2 + x - x^2) =\dfrac94\Leftrightarrow x = \dfrac12$
Giải thích các bước giải:
Đặt $A = 2 + x - x^2$
$\to A = - \left(x^2 -2.\dfrac{1}{2}x +\dfrac{1}{4}\right) +\dfrac{9}{4}$
$\to A = -\left(x -\dfrac12\right)^2 +\dfrac94$
Ta có:
$\left(x -\dfrac12\right)^2 \geq 0,\,\,\forall x$
$\to - \left(x -\dfrac12\right)^2 \leq 0$
$\to -\left(x -\dfrac12\right)^2 +\dfrac94 \leq \dfrac94$
$\to A \leq \dfrac94$
Dấu $=$ xảy ra $\Leftrightarrow x - \dfrac12 = 0\Leftrightarrow x =\dfrac12$
Vậy $\max(2 + x - x^2) =\dfrac94\Leftrightarrow x = \dfrac12$
