Đáp án:
$A<0\,\,∀\,x\in\mathbb R$
Giải thích các bước giải:
$A=-x^2+x-1$
$=-(x^2-x+1)$
$=-\left(x^2-2.x.\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{3}{4}\right)$
$=-\left(x-\dfrac{1}{2}\right)^2-\dfrac{3}{4}$
Ta có:
$\left(x-\dfrac{1}{2}\right)^2\ge 0$
$⇒-\left(x-\dfrac{1}{2}\right)^2\le 0$
$⇒-\left(x-\dfrac{1}{2}\right)^2-\dfrac{3}{4}\le-\dfrac{3}{4}$
$⇒A\le-\dfrac{3}{4}$
$⇒A<0\,\,∀\,x\in\mathbb R$
Vậy $A<0\,\,∀\,x\in\mathbb R$.
