Question

Answer 1

\(\dfrac{6}{1+x}-\dfrac{4}{1-x}-\dfrac{100x}{x^2-1}\\ =\dfrac{6}{x+1}+\dfrac{4}{x-1}-\dfrac{100x}{\left(x-1\right)\left(x+1\right)}\)

\(=\dfrac{6\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}+\dfrac{4\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}-\dfrac{100x}{\left(x-1\right)\left(x+1\right)}\)

\(=\dfrac{6x-6+4x+4-100x}{\left(x-1\right)\left(x+1\right)}\\ =\dfrac{-90x-2}{\left(x-1\right)\left(x+1\right)}\)