Question

\(\frac{4}{x-4}-\frac{x}{x+4}+\frac{32}{16-x^2}\)

Answer 1

\(\frac{4}{x-4}-\frac{x}{x+4}+\frac{32}{16-x^2}.\)

\(=\frac{4\left(x+4\right)}{\left(x-4\right)\left(x+4\right)}-\frac{x\left(x-4\right)}{\left(x-4\right)\left(x+4\right)}-\frac{32}{\left(x-4\right)\left(x+4\right)}\)

\(=\frac{4x+16-x^2+4x-32}{\left(x-4\right)\left(x+4\right)}\)

\(=\frac{-x^2+8x-16}{\left(x-4\right)\left(x+4\right)}=\frac{-\left(x-4\right)^2}{\left(x-4\right)\left(x+4\right)}=\frac{-\left(x-4\right)}{x+4}\)