\(\left(\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}\right):\left(\dfrac{1}{x+1}-\dfrac{1}{1-x}+\dfrac{2}{x^2-1}\right)\left(đk:x\ne1\right)\)
\(=\dfrac{\left(x+1\right)^2-\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}:\dfrac{x-1+x+1+2}{\left(x-1\right)\left(x+1\right)}=\dfrac{x^2+2x+1-x^2+2x-1}{x^2-1}:\dfrac{2x+2}{x^2-1}=\dfrac{4x}{x^2-1}.\dfrac{x^2-1}{2\left(x+1\right)}=\dfrac{2x}{x+1}\)
Ta có: \(\left(\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}\right):\left(\dfrac{1}{x+1}+\dfrac{1}{x-1}+\dfrac{2}{x^2-1}\right)\)
\(=\dfrac{x^2+2x+1-x^2+2x-1}{\left(x-1\right)\left(x+1\right)}:\dfrac{x-1+x+1+2}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{4x}{\left(x-1\right)\left(x+1\right)}\cdot\dfrac{\left(x-1\right)\left(x+1\right)}{2x+2}\)
\(=\dfrac{4x}{2\left(x+1\right)}\)
\(=\dfrac{2x}{x+1}\)
