Đáp án:
$A_{\max}=-\dfrac{3}{4}$
Giải thích các bước giải:
$A=x-x^2-1$
$=-\left(x^2-2.x.\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{3}{4}\right)$
$=-\left(x-\dfrac{1}{2}\right)^2-\dfrac{3}{4}$
Ta có:
$\left(x-\dfrac{1}{2}\right)^2\ge 0$
$⇒-\left(x-\dfrac{1}{2}\right)^2\le 0$
$⇒-\left(x-\dfrac{1}{2}\right)^2-\dfrac{3}{4}\le -\dfrac{3}{4}$
$⇒A\le-\dfrac{3}{4}⇒A_{\max}=-\dfrac{3}{4}$
Dấu "=" xảy ra khi:
$\left(x-\dfrac{1}{2}\right)^2= 0$
$⇒x=\dfrac{1}{2}$
Vậy $A_{\max}=-\dfrac{3}{4}$ khi $x=\dfrac{1}{2}$.
