6.140. $\frac{z^{2}-z}{z^{2}-z+1}-\frac{z^{2}-z+2}{z^{2}-z-2}=1$.
Solution.

Domain of definition: $z^{2}-z-2 \neq 0 \Leftrightarrow\left\{\begin{array}{l}z \neq-1, \\ z \neq 2 .\end{array}\right.$

Let $z^{2}-z=y$. We have:

$\frac{y}{y+1}-\frac{y+2}{y-2}-1=0 \Leftrightarrow \frac{y(y-2)-(y+2)(y+1)-(y+1)(y-2)}{(y+1)(y-2)}=0 \Leftrightarrow$ $\Leftrightarrow \frac{y(y+4)}{(y+1)(y-2)}=0$, from which $y_{1}=0, y_{2}=-4$. Then $z^{2}-z=0, z_{1}=0$, $z_{2}=1$, or $z^{2}-z+4=0(D<0)$ and there are no solutions.

Answer: $z_{1}=0, z_{2}=1$