\(\frac{x}{x+1} < \frac{x}{x-1} \\ \frac{x}{x+1} - \frac{x}{x-1} <0 \\ x( \frac{1}{x+1} - \frac{1}{x+1})<0 \\ \frac{2x}{x^2-1} <0\)
Consider the function and define the domain of the function
\(y=\frac{2x}{x^2-1}\)
\(x^2-1
eq 0 \\ x^2
eq 1 \\ x
eq \pm 1\)
Function is zero
\(\frac{2x}{x^2-1}=0 \\ 2x=0 \\ x=0\)
Answer: \((-1;0)\cup(1;+\infty)\)
