\(A=x^2+2x+4\)
\(=\left(x^2+2x+1\right)+3\)
\(=\left(x+1\right)^2+3\)
Vì \(\left(x+1\right)^2\ge0\)
\(\Rightarrow\left(x+1\right)^2+3\ge3\)
Hay \(A\ge3\)
Dấu \("="\) xảy ra khi: \(\left(x+1\right)^2=0\)
\(\Rightarrow x+1=0\)
\(\Rightarrow x=-1\)
Vậy min A=3 đạt tại x=-1