$A = 2x^2+9x-13$
$ = 2.\bigg(x^2+\dfrac{9}{2}x-\dfrac{13}{2}\bigg)$
$ = 2\bigg(x^2+2.x.\dfrac{9}{4}+\dfrac{81}{16}-\dfrac{185}{16}\bigg)$
$ = 2\bigg(x+\dfrac{9}{4}\bigg)^2-\dfrac{185}{8} ≥ \dfrac{-185}{8}$
Dấu "=" xảy ra $⇔x=\dfrac{-9}{4}$
$B = x^2-4x+y^2-6y+2019$
$ = (x^2-4x+4)+(y^2-6y+9)+2004$
$ = (x-2)^2+(y-3)^2+2004 ≥ 2004$
Dấu "=" xảy ra $⇔x=2,y=3$
$C = \dfrac{2x^2-1}{x^2+3}$
$ = \dfrac{2.(x^2+3)-7}{x^2+3}$
$ = 2-\dfrac{7}{x^2+3}$
Vì $x^2+3 ≥ 3$
$\to \dfrac{-7}{x^2+3} ≥ \dfrac{-7}{3}$
$\to 2-\dfrac{7}{x^2+3} ≥ 2-\dfrac{7}{3} = -\dfrac{1}{3}$
Dấu "=" xảy ra $⇔x=0$
Phần d) thiếu đề thì phải.
