\(ĐK:x\ne\pm1\)
\(\Leftrightarrow\frac{ax^2-x+ax-1+bx-b}{x^2-1}=\frac{a\left(x^2+1\right)}{x^2-1}\)
\(\Leftrightarrow\frac{ax^2+x\left(a-1+b\right)-b-1}{x^2-1}=\frac{ax^2+a}{x^2-1}\)
\(\Leftrightarrow\hept{\begin{cases}x\ne\pm1\\a+b-1=0\\-b-1=a\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x\ne\pm1\\a+b-1=0\\-b-1=a\end{cases}}\)
Giải ra :D