\(A=2x^2+4x=2x^2+4x+2-2=2\left(x^2+2x+1\right)-2=2\left(x+1\right)^2-2,tacó2\left(x+1\right)^2\ge0\Leftrightarrow2\left(x+1\right)^2-2\ge-2.VậyminA=-2khix+1=0\Rightarrow x=-1\)\(B=2x^2+4x+3=\left(2x^2+4x+2\right)+1=2\left(x^2+2x+1\right)+1=2\left(x+1\right)^2+1,vì2\left(x+1\right)^2\ge0\Leftrightarrow2\left(x+1\right)^2+1\ge1.VậyminB=1khix=-1\)