Ta có : \(A=\frac{x^2+2x+3}{\left(x+2\right)^2}\) . Đặt \(y=x+2\Rightarrow x=y-2\)
\(\Rightarrow x^2+2x+3=\left(y-2\right)^2+2\left(y-2\right)+3=y^2-2y+3\)
\(\Rightarrow A=\frac{y^2-2y+3}{y^2}=1-\frac{2}{y}+\frac{3}{y^2}\)
Đặt \(\frac{1}{y}=z\Rightarrow A=3z^2-2z+1=3\left(z-\frac{1}{3}\right)^2+\frac{2}{3}\ge\frac{2}{3}\)
Dấu đẳng thức xảy ra \(\Leftrightarrow z=\frac{1}{3}\Leftrightarrow y=3\Leftrightarrow x=1\)
Vậy Min A = \(\frac{2}{3}\Leftrightarrow x=1\)
Cách 2 : Ta có : \(A=\frac{x^2+2x+3}{\left(x+2\right)^2}=\frac{x^2+2x+3}{x^2+4x+4}=\frac{3\left(x^2+2x+3\right)}{3\left(x^2+4x+4\right)}=\frac{2\left(x^2+4x+4\right)+\left(x^2-2x+1\right)}{3\left(x^2+4x+4\right)}\)
\(=\frac{\left(x-1\right)^2}{3\left(x+2\right)^2}+\frac{2}{3}\ge\frac{2}{3}\). Dấu đẳng thức xảy ra khi x = 1.
Vậy Min A = 2/3 <=> x = 1
