Решите пожалуйста! 2x-1/1-x^2 <-1

ОДЗ
1-х²=0
(1-х)(1+х)=0
х=1
х=-1

$- \frac{2(x-0.5)}{ x^{2} -1}\ \textless \ -1 \\ \frac{2(x-0.5)}{ x^{2} -1}\ \textgreater \ 1 \\ \frac{2(x-0.5)}{ x^{2} -1}-1\ \textgreater \ 0 \\ \frac{2x-1- x^{2} +1}{(x-1)(x+1)}\ \textgreater \ 0 \\ \frac{x(2- x)}{(x-1)(x+1)}\ \textgreater \ 0 \\ \frac{x(x-2)}{(x-1)(x+1)}\ \textless \ 0$

x∈(-∞; -1)∨[0; 1)∨[2; +∞)