$a)\ A=4x^2-4x-3\\=4x^2-4x+1-4\\=(2x-1)^2-4\ge-4$
Dấu $=$ xảy ra khi
$2x-1=0\Leftrightarrow x=\dfrac{1}{2}$
Vậy $A_{min}=-4$ khi $x=\dfrac{1}{2}$
$b)\ B=3m^2-6m+1\\=3(m^2-2m+1)+1-3\\=3(m-1)^2-2\ge-2$
Dấu $=$ xảy ra khi
$m-1=0\Leftrightarrow m=1$
Vậy $B_{min}=-2$ khi $m=1$
$c)\ C=5a^2-4a-1\\=5(a^2-\dfrac{4}{5}a+\dfrac{4}{25})-1-\dfrac{4}{5}\\=5(a-\dfrac{2}{5})^2-\dfrac{9}{5}\ge\dfrac{-9}{5}$
Dấu $=$ xảy ra khi
$a-\dfrac{2}{5}=0\Leftrightarrow a=\dfrac{2}{5}$
Vậy $C_{min}=\dfrac{-9}{5}$ khi $a=\dfrac{2}{5}$
$d)\ D=(x-2)^2+(x+3)^2-2021\\=x^2-4x+4+x^2+6x+9-2021\\=2x^2+2x-2008\\=2(x^2+x+\dfrac{1}{4})-2008-\dfrac{1}{2}\\=2(x+\dfrac{1}{2})^2-\dfrac{4017}{2}\ge\dfrac{-4017}{2}$
Dấu $=$ xảy ra khi
$x+\dfrac{1}{2}=0\Leftrightarrow x=\dfrac{-1}{2}$
Vậy $D_{min}=\dfrac{-4017}{2}$ khi $x=\dfrac{-1}{2}$
