Решите пожалуйста! 2x-1/1-x^2 меньше -1

ОДЗ 1-х²=0 (1-х)(1+х)=0 х=1 х=-1 [latex]- \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[/latex] x∈(-∞; -1)∨[0; 1)∨[2; +∞)