b)$\dfrac{2x-4}{x-1}$ + $\dfrac{2x+2}{x^2-1}$
=$\dfrac{2x-4}{x-1}$ + $\dfrac{2x+2}{x^2-1^2}$
=$\dfrac{2x-4}{x-1}$ + $\dfrac{2x+2}{(x-1)(x+1)}$
=$\dfrac{(x+1)(2x-4)}{(x-1)(x+1)}$+$\dfrac{2x+2}{(x-1)(x+1)}$
=$\dfrac{2x^2-4x+2x-4}{(x-1)(x+1)}$+$\dfrac{2x+2}{(x-1)(x+1)}$
=$\dfrac{2x^2-4x+2x-4+2x+2}{(x-1)(x+1)}$
=$\dfrac{2x^2-4+2}{(x-1)(x+1)}$
=$\dfrac{2x^2-2}{(x-1)(x+1)}$
= $\dfrac{2(x^2-1)}{(x-1)(x+1)}$
=$\dfrac{2(x^2-1²)}{(x-1)(x+1)}$
=$\dfrac{2(x-1)(x+1)}{(x-1)(x+1)}$
=`2`
