11. The solution set of the equation $\left(x^{2}-3 x-2\right)\left(x^{2}-3 x+1\right)=x-1$ is
11. $\{x \mid x=2 \pm \sqrt{3}$ or $x=1 \pm \sqrt{2}\}$ $=x$, i.e., $\left(x^{2}-3 x+1\right)^{2}-3\left(x^{2}-3 x+1\right)+1=x$, let $y=x^{2}-3 x+1$, by $\left\{\begin{array}{l}y=x^{2}-3 x+1 \\ x=y^{2}-3 y+1\end{array}\right.$
(1) - (2) gives: $y-x=(x+y)(x-y)-3(x-y)$,

i.e., $x=y$ or $x+y-3=-1$, substituting (1), we solve to get: $x=2 \pm \sqrt{3}$ and $x=1 \pm \sqrt{2}$.