\(\frac{x^2+2x+3}{x^2+2}=\frac{2\left(x^2+2x+3\right)}{2\left(x^2+2\right)}\)=\(\frac{\left(x^2+4x+4\right)+\left(x^2+2\right)}{2\left(x^2+2\right)}\)
=\(\frac{\left(x+2\right)^2}{2\left(x^2+2\right)}+\frac{1}{2}\ge\frac{1}{2}\)
dau = xay ra khi x=-2 .vay min E =1/2