a)ĐKXĐ:
\(x+1\ne0\Leftrightarrow x\ne-1\)
\(x-1\ne0\Leftrightarrow x\ne1\)
b) \(A=\left(\dfrac{x}{x+1}+\dfrac{1}{x-1}\right):\left(\dfrac{2x+2}{x-1}-\dfrac{4x}{x^2-1}\right)\)
\(=\left[\dfrac{x\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}+\dfrac{x+1}{\left(x-1\right)\left(x+1\right)}\right]:\left[\dfrac{2\left(x+1\right)}{x-1}-\dfrac{4x}{x^2-1}\right]\)
\(=\left[\dfrac{x\left(x-1\right)+\left(x+1\right)}{x^2-1}\right]:\left[\dfrac{2\left(x+1\right)^2}{x^2-1}-\dfrac{4x}{x^2-1}\right]\)
\(=\left(\dfrac{x^2-x+x+1}{x^2-1}\right):\left(\dfrac{2\left(x^2+2x+1\right)-4x}{x^2-1}\right)\)
\(=\dfrac{x^2+1}{x^2-1}:\left(\dfrac{2x^2+4x+2-4x}{x^2-1}\right)\)
\(=\dfrac{x^2+1}{x^2-1}:\dfrac{2x^2+2}{x^2-1}\)
\(=\dfrac{x^2+1}{x^2-1}.\dfrac{x^2-1}{2x^2+2}\)
\(=\dfrac{x^2+1}{x^2-1}.\dfrac{x^2-1}{2\left(x^2+1\right)}\)
\(=\dfrac{1}{2}\)
Vậy với \(x\ne\pm1\) thì A không phụ thuộc vào biến x
