What is the value of $\frac{1+x^2}{1-x}÷\frac{1-x^4}{x-1}×\frac{x(1-x)}{(1+x)}$?
Answer & explanation
Correct answer: option 3
$\frac{1+x^2}{1-x}÷\frac{1-x^4}{x-1}×\frac{x(1-x)}{(1+x)}$
\(\frac{1+x^2}{1-x}\) × \(\frac{-1(1-x)}{(1+x^2)(1- x^2)}\) × \(\frac{x(1-x)}{(1+x)}\)
= $\frac{-x}{(1+x)^2}$