Giải thích các bước giải:
$1)\\ A=x^2-2x+9\\ =x^2-2x+1+8\\ =(x-1)^2+8 >0 \ \forall \ x\\ B=4x^2+4x+2021\\ =4x^2+4x+1+2020\\ =(2x+1)^2+2020>0 \ \forall \ x\\ C=x^2-x+1\\ =x^2-x+\dfrac{1}{4}+\dfrac{3}{4}\\ =\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0 \ \forall \ x\\ D=9-6x+3x^2\\ =(\sqrt{3}x)^2-2.\sqrt{3}x.\sqrt{3}+3+6\\ =(\sqrt{3}x+\sqrt{3})^2+6>0 \ \forall \ x$
