Question

ПОМОЩЬ НУЖНА !!!!!!!!!!!!!!!!!

Answer 1

8-n³=2³-n³=(2-n)(4+2n+n²)

$1) frac{8-n^3}{2+n}:(2+ frac{n^2}{2+n})= frac{8-n^3}{2+n}:( frac{2(2+n)+n^2}{2+n})= frac{8-n^3}{2+n}:( frac{4+2n+n^2}{2+n})= \ \ = frac{8-n^3}{2+n}cdot ( frac{2+n}{4+2n+n^2})= frac{(2-n)(4+2n+n^2)}{2+n}cdot ( frac{2+n}{4+2n+n^2})= 2-n$

$2) frac{n^2}{n-2}cdot frac{7-n^2}{n^2+2n}= frac{n^2}{n-2}cdot frac{7-n^2}{n(n+2)}= frac{n}{n-2}cdot frac{7-n^2}{n+2}=frac{(7-n^2)n}{n^2-4}\ \3)(2-n)-frac{(7-n^2)n}{n^2-4}= frac{(2-n)(n^2-4)-(7-n^2)n}{n^2-4} = frac{2n^2-n^3-8+4n-7n+n^3}{n^2-4} = \ \ = frac{2n^2-3n-8}{n^2-4}$

Answer 2

спасибо)