The minimum value of $\frac{x^2-2x+9}{x^2+2x+9}$  is 

$\begin{array}{r}\text{ Let }y=\frac{x^2-2x+9}{x^2+2x+9}\\ \Rightarrow (1-y)x^2-(2+2y)x+9(1-y)=0\\ \text{ Since }x\in R\text{, then }\mathrm{\Delta }\ge 0\end{array}$
$\begin{array}{r}\Rightarrow 8{\mathrm{y}}^2-20\mathrm{y}+8\le 0\\ \Rightarrow 2{\mathrm{y}}^2-5\mathrm{y}+2\le 0\\ \Rightarrow (2\mathrm{y}-1)(\mathrm{y}-2)\le 0\end{array}$
$\begin{array}{r}\Rightarrow \frac{1}{2}\le \mathrm{y}\le 2\\ \therefore \text{ minimum value of }\mathrm{y}=\frac{1}{2}\end{array}$