Ответ:
Решение:
- Приведем дроби к общему знаменателю. Общий знаменатель равен \( (x-3)(x+3)^2 \).
- Первую дробь умножим на \( (x+3)^2 \), вторую на \( (x+3) \), третью на \( (x-3) \).
- \(\begin{align*}\) \(\left\)\(\frac{x+3}{x-3} - \frac{x}{x+3} \right\) \(\cdot\) \(\frac{x+1}{x+3}\) &= \(\left\)( \(\frac{(x+3)(x+3)^2}{(x-3)(x+3)^2}\) - \(\frac{x(x-3)(x+3)}{(x-3)(x+3)^2}\) \(\right\)) \(\cdot\) \(\frac{x+1}{x+3}\) \\ &= \(\frac{(x+3)^3 - x(x-3)(x+3)}{(x-3)(x+3)^2}\) \(\cdot\) \(\frac{x+1}{x+3}\) \\ &= \(\frac{(x^3 + 9x^2 + 27x + 27) - x(x^2-9)}{(x-3)(x+3)^2}\) \(\cdot\) \(\frac{x+1}{x+3}\) \\ &= \(\frac{x^3 + 9x^2 + 27x + 27 - x^3 + 9x}{(x-3)(x+3)^2}\) \(\cdot\) \(\frac{x+1}{x+3}\) \\ &= \(\frac{9x^2 + 36x + 27}{(x-3)(x+3)^2}\) \(\cdot\) \(\frac{x+1}{x+3}\) \\ &= \(\frac{9(x^2 + 4x + 3)}{(x-3)(x+3)^2}\) \(\cdot\) \(\frac{x+1}{x+3}\) \\ &= \(\frac{9(x+1)(x+3)}{(x-3)(x+3)^2}\) \(\cdot\) \(\frac{x+1}{x+3}\) \\ &= \(\frac{9(x+1)^2}{(x-3)(x+3)^2}\) \\ &= \(\frac{9(x+1)^2}{(x-3)(x^2+6x+9)}\) \\ &= \(\frac{9(x^2+2x+1)}{x^3+6x^2+9x-3x^2-18x-27}\) \\ &= \(\frac{9x^2+18x+9}{x^3+3x^2-9x-27}\) \(\end{align*}\)
Ответ: \( \frac{9(x+1)^2}{(x-3)(x+3)^2} \)
