Đặt \(x+2=a\Rightarrow x=a-2\)
\(A=\frac{\left(a-2\right)^2+2\left(a-2\right)+3}{a^2}=\frac{a^2-2a+3}{a^2}\)
\(=\frac{3}{a^2}-\frac{2}{a}+1=3t^2-2t+1\)(đặt \(t=\frac{1}{a}\))
\(=3\left(t-\frac{1}{3}\right)^2+\frac{2}{3}\ge\frac{2}{3}\)
Đẳng thức xảy ra khi \(t=\frac{1}{3}\Leftrightarrow a=3\Leftrightarrow x=1\)