Question

a) \(\dfrac{4}{x^2-2x+1}-\dfrac{6}{x^2-1}\)

Answer 1

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

Answer 2

\(=\dfrac{4}{\left(x-1\right)^2}-\dfrac{6}{\left(x-1\right)\left(x+1\right)}=\dfrac{4x+4-6x+6}{\left(x-1\right)^2\left(x+1\right)}=\dfrac{10-2x}{\left(x-1\right)^2\left(x+1\right)}\)