Question

A=\(\left(\frac{x+1}{x-1}-\frac{x-1}{x+1}+\frac{\left(x-1\right)^2}{x^2-1}\right).\frac{x+2003}{x}.\)
a) ĐK? Rút gọn A?
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Answer 1

\(A=\left(\frac{x+1}{x-1}-\frac{x-1}{x+1}+\frac{\left(x-1\right)^2}{x^2-1}\right).\frac{x+2003}{x}\)ĐKXĐ: \(x\ne-1;0;1\)

\(A=\frac{\left(x+1\right)^2-\left(x-1\right)^2+\left(x-1\right)^2}{\left(x+1\right)\left(x-1\right)}.\frac{x+2003}{x}\)

\(A=\frac{\left(x+1\right)^2}{\left(x+1\right)\left(x-1\right)}.\frac{x+2003}{x}\)

\(A=\frac{x+1}{x-1}.\frac{x+2003}{x}\)

\(A=\frac{x^2+2004x+2003}{x^2-x}\)