\(A=\left(\frac{x+1}{x-1}-\frac{x-1}{x+1}+\frac{x^2-4x-1}{x^2-1}\right).\frac{x+2003}{x}\)ĐK : \(x\ne0;\pm1\)
\(=\left(\frac{x^2+2x+1-x^2+2x-1+x^2-4x-1}{\left(x-1\right)\left(x+1\right)}\right).\frac{2003}{x}\)
\(=\frac{x^2-1}{\left(x-1\right)\left(x+1\right)}.\frac{2003}{x}=\frac{2003}{x}\)