Bài 5:
a: ĐKXĐ: \(x\notin\left\{0;-1;1\right\}\)
b: Ta có:\(A=\left(\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}+\dfrac{x^2-4x-1}{x^2-1}\right)\cdot\dfrac{x+2003}{x}\)
\(=\dfrac{x^2+2x+1-x^2+2x-1+x^2-4x-1}{\left(x-1\right)\left(x+1\right)}\cdot\dfrac{x+2003}{x}\)
\(=\dfrac{x+2003}{x}\)