dễ dàng pt đc \(A=\frac{4\left(x^2+2x+5\right)^2+256}{x^2+2x+5}=4\left(x^2+2x+5\right)+\frac{256}{x^2+2x+5}\ge64\)
Dấu = xảy ra khi \(4\left(x^2+2x+5\right)=\frac{256}{x^2+2x+5}\Rightarrow x^2+2x+5=8\Leftrightarrow x^2+2x-3=0\)
\(\Rightarrow x=1,x=-3\)