Đáp án:
$B<0 \ \forall \ x\\ D<0 \ \forall \ x.$
Giải thích các bước giải:
$B=2x-5x^2-1\\ =-5\left(x^2-\dfrac{2}{5}x+\dfrac{1}{5}\right)\\ =-5\left(x^2-2.\dfrac{1}{5}x+\dfrac{1}{25}+\dfrac{4}{25}\right)\\ =-5\left(x^2-2.\dfrac{1}{5}x+\dfrac{1}{25}\right)-\dfrac{4}{5}\\ =-5\left(x-\dfrac{1}{5}\right)^2-\dfrac{4}{5}<0 \ \forall \ x\\ D=-x^2+x-1\\ =-\left(x^2-x+1\right)\\ =-\left(x^2-x+\dfrac{1}{4}+\dfrac{3}{4}\right)\\ =-\left(x^2-x+\dfrac{1}{4}\right)-\dfrac{3}{4}\\ =-\left(x-\dfrac{1}{2}\right)^2-\dfrac{3}{4}<0 \ \forall \ x.$
