a: ĐKXĐ: |x+1|<>|x-1|
=>x+1<>1-x
=>2x<>0
hay x<>0
Vậy: D=R\{0}
b: \(f\left(-x\right)=\dfrac{\left|-x+1\right|+\left|-x-1\right|}{\left|-x+1\right|-\left|-x-1\right|}=\dfrac{\left|x-1\right|+\left|x+1\right|}{\left|x-1\right|-\left|x+1\right|}\)
\(=-\dfrac{\left|x-1\right|+\left|x+1\right|}{\left|x+1\right|-\left|x-1\right|}=-f\left(x\right)\)