\(B=\left(x^2+2x+1\right)+6=\left(x+1\right)^2+6\ge6\)
\(minB=6\Leftrightarrow x=-1\)
\(=x^2+2x+1+6\\ =\left(x+1\right)^2+6\\ Vì\left(x+1\right)^2\ge0\\ \Rightarrow\left(x+1\right)^2+6\ge0+6\\ \Rightarrow\left(x+1\right)^2+6\ge6\)
Dấu "=" \(\Leftrightarrow x+1=0\\ x=-1\)