Question

Привидите дробь 2n/n-3 к знаменателю 9-n²

Answer 1

$9 - n ^{2} = (3 - n)(3 + n)$
поэтому
$\frac{2n}{n - 3} = \frac{ - 2n}{3 - n} = \frac{ - 2n(3 + n)}{(3 - n)(3 + n)} = \frac{ - 6n - 2 {n}^{2} }{9 - {n}^{2} }$

Answer 2

$\frac{2n}{n-3} =\frac{2n*(n+3)}{(n-3)(n+3)} =\frac{2n^2+6n}{n^2-3^2} =-\frac{2n^2+6n}{9-n^2}$