Đáp án:
\[\dfrac{1}{x^2-x+2}\quad xac\quad dinh \Leftrightarrow x^2-x+2\ne 0\] \[\Leftrightarrow x^2-2.\dfrac{1}{2}.x+\Bigg(\dfrac{1}{2}\Bigg)^2+\dfrac{7}{4}\ne 0\] \[\Leftrightarrow {\Bigg(x-\dfrac{1}{2}\Bigg)}^2+\dfrac{7}{4}\ne 0\]
mà ${\Bigg(x-\dfrac{1}{2}\Bigg)}^2+\dfrac{7}{4}>0$
$→\dfrac{1}{x^2-x+2}$ luôn xác định.
\[\dfrac{-2}{-3-x^2}\quad xac\quad dinh \Leftrightarrow \dfrac{2}{x^2+3} \quad xac\quad dinh \]
$⇔x^2+3\ne 0 \quad (\quad luon\quad dung\quad )($ vì $x^2+3≥3)$
$→\dfrac{-2}{-3-x^2}$ luôn xác định.