Question

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

Answer 1

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

Answer 2

Với `x \ne +-1` có:

`1/[x-1]-1/[x+1]+[4x+2]/[x^2-1]`

`=[x+1-x+1+4x+2]/[(x-1)(x+1)]`

`=[4x+4]/[(x-1)(x+1)]`

`=[4(x+1)]/[(x-1)(x+1)]`

`=4/[x-1]`