Question

A=(\(\dfrac{x}{\left(x-2\right)\left(x+2\right)}\)+\(\dfrac{1}{x+2}\)-\(\dfrac{2}{x-2}\)):(1-\(\dfrac{x}{x+2}\))

Answer 1

A = (\(\dfrac{x}{\left(x-2\right)\left(x+2\right)}\) + \(\dfrac{1}{x+2}\) - \(\dfrac{2}{x-2}\)) : (1 - \(\dfrac{x}{x+2}\))

= (\(\dfrac{x}{\left(x-2\right)\left(x+2\right)}\) + \(\dfrac{x-2}{\left(x+2\right)\left(x-2\right)}\) - \(\dfrac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\)) : (\(\dfrac{x+2}{x+2}\) - \(\dfrac{x}{x+2}\))

= \(\dfrac{x+x-2-2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\) : \(\dfrac{x+2-x}{x+2}\)

= \(\dfrac{x+x-2-2x-4}{\left(x-2\right)\left(x+2\right)}\) : \(\dfrac{2}{x+2}\)

= \(\dfrac{-6}{\left(x-2\right)\left(x+2\right)}\) . \(\dfrac{x+2}{2}\)

= \(\dfrac{-3}{x-2}\)