№ 3. Упростить выражения: a) \(\\( \frac{2n^2 + 11n + 14}{n+3} - 2n + \frac{1}{n+3}\)) б) \(\\( \frac{a^2 - b^2}{a-b} - \frac{a^3 - b^3}{a^2 - b^2}\))

Решение:

1. \(\\(
   \frac{2n^2 + 11n + 14}{n+3} - 2n + \frac{1}{n+3}\)) = \(\\(
   \frac{2n^2 + 11n + 14 - 2n(n+3) + 1}{n+3}\)) = \(\\(
   \frac{2n^2 + 11n + 14 - 2n^2 - 6n + 1}{n+3}\)) = \(\\(
   \frac{5n + 15}{n+3}\)) = \(\\(
   \frac{5(n + 3)}{n+3}\)) = 5\)
2. \(\\(
   \frac{a^2 - b^2}{a-b} - \frac{a^3 - b^3}{a^2 - b^2}\)) = \(\\(
   \frac{(a-b)(a+b)}{a-b} - \frac{(a-b)(a^2+ab+b^2)}{(a-b)(a+b)}\)) = \(\\(
   (a+b) - \frac{a^2+ab+b^2}{a+b}\)) = \(\\(
   \frac{(a+b)^2 - (a^2+ab+b^2)}{a+b}\)) = \(\\(
   \frac{a^2+2ab+b^2 - a^2-ab-b^2}{a+b}\)) = \(\\(
   \frac{ab}{a+b}\))

Ответ: 5; \(\\(
\frac{ab}{a+b}\))