Sửa đề: \(A=\left(\dfrac{x^2+1}{2x}-1\right)\left(\dfrac{1}{x-1}+\dfrac{1}{x+1}\right)\)
a: TXĐ: \(x\notin\left\{0;1;-1\right\}\)
b: \(A=\dfrac{x^2-2x+1}{2x}\cdot\dfrac{x+1+x-1}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}=\dfrac{x-1}{x+1}\)
c: Để A=0 thì x-1=0
hay x=1(loại)