\(\dfrac{2}{x^2-1}<-1\\↔\dfrac{2}{x^2-1}+1<0\\↔\dfrac{x^2+1}{x^2-1}<0\\Vì\,\,x^2+1>0∀x\,\,mà\,\,M<0\\→x^2-1<0\\↔(x-1)(x+1)<0\\↔\left[\begin{array}{1}\begin{cases}x-1<0\\x+1>0\end{cases}\\\begin{cases}x-1>0\\x+1<0\end{cases}\end{array}\right.\\↔\left[\begin{array}{1}\begin{cases}x<1\\x>-1\end{cases}\\\begin{cases}x>1\\x<-1\end{cases}\end{array}\right.\\→-1<x<1\)