11. Find the real number $m$, such that the inequality $\frac{x^{2}-8 x+20}{m x^{2}+2(m+1) x+9 m+4}<0$ has the solution set as the set of all real numbers $\mathbf{R}$
11. Since $x^{2}-8 x+20>0$ always holds, the solution set of $m x^{2}+2(m+1) x+9 m+4<0$ is $\mathbf{R}$.
(1) When $m=0$, $2 x+4<0$ does not meet the requirement; (2) $m<0$, and $\Delta<0$, we get $m<-\frac{1}{2}$.