\(\Leftrightarrow\frac{\left(ax-1\right)\left(x+1\right)}{x^2-1}+\frac{b\left(x-1\right)}{x^2-1}=\frac{a\left(x^2+1\right)}{x^2-1}\left(x\ne+-1\right)\)
\(\Rightarrow ax^2+ax-x-1+bx-b=ax^2+a\)
\(\Leftrightarrow ax+bx-a-b-x-1=0\)
\(\Leftrightarrow ax+bx-x-a-b+1=2\)
\(\Leftrightarrow x\left(a+b-1\right)-\left(a+b-1\right)=2\)
\(\Leftrightarrow\left(x-1\right)\left(a+b-1\right)=2\left(1\right)\)
-(1) vô nghiệm khi x=+-1
-(1) có nghiệm duy nhất \(x=\frac{2}{a+b-1}+1=\frac{a+b+1}{a+b-1}\).
Khi đó \(\left\{{}\begin{matrix}\frac{a+b+1}{a+b-1}\ne1\\\frac{a+b+1}{a+b-1}\ne-1\end{matrix}\right.\)
\(\Leftrightarrow a+b+1\ne-\left(a+b-1\right)\)
\(\Leftrightarrow2\left(a+b\right)\ne0\Leftrightarrow a\ne-b\)