33 Let the line $y=m x$ have no common points with the curve $y=\frac{|x|-1}{|x-1|}$, then the range of values for $m$ is $\qquad$
When $33-1 \leqslant m1$, $y=1$.
When $0<x<1$, $y=-1$.
When $x \leqslant 0$, $y=\frac{-x-1}{-(x-1)}=1+\frac{2}{x-1}$ is a hyperbola, with its center at $(1,1)$.
From this, we can determine that when $m \geqslant 0$ or $m<-1$, the two lines must intersect. Therefore, we only need to consider $-1 \leqslant m<0$. By eliminating $y$ from $y=m x$ and $y=1+\frac{2}{x-1}$ and rearranging, we get $m x^{2}-(m+1) x-1=0$. Then, by the discriminant $\Delta=(m+1)^{2}+4 m<0$, we solve to get $-3-2 \sqrt{2}<m<-3+2 \sqrt{2}$. Hence, $-1 \leqslant m<-3+2 \sqrt{2}$.