\(C=\left(x+2\right)^2+3\ge3\forall x\)
Dấu '=' xảy ra khi x=-2
\(C=x^2+4x+7=\left(x^2+4x+4\right)+3=\left(x+2\right)^2+3\ge3\)
Dấu '=' xảy ra khi x=-2
Vậy \(C_{min}=3\Leftrightarrow x=-2\)
\(D=x^2+6x+15=\left(x^2+6x+9\right)+6=\left(x+3\right)^2+6\ge6\)
Dấu '=' xảy ra khi x=-3
Vậy\(D_{min}=6\Leftrightarrow x=-3\)