ĐKXĐ : \(x\ne1;-1\)
\(A=\left(\dfrac{x}{x+1}+\dfrac{x}{x-1}\right):\left(\dfrac{2x+2}{x-1}-\dfrac{4x}{x^2-1}\right)\)
\(\Leftrightarrow A=\left(\dfrac{x^2-x+x^2+x}{\left(x-1\right)\left(x+1\right)}\right)\left(\dfrac{\left(x-1\right)\left(x+1\right)}{2x^2+2x+2x+2-4x}\right)\)
\(\Leftrightarrow A=\dfrac{2x^2}{\left(x+1\right)\left(x-1\right)}.\dfrac{\left(x-1\right)\left(x+1\right)}{2\left(x^2+1\right)}=\dfrac{x^2}{x^2+1}\)
Ta thấy \(x^2>0\) \(\RightarrowĐPCM.\)