k2 9-k2 27+k³ г) + :(3+k k²-3k 3-k 3-k).

г) Упростим выражение $$\frac{k^2}{3+k} + \frac{9-k^2}{k^2-3k} + \frac{27+k^3}{3-k}:\left(3+\frac{k^2}{3-k}\right)$$.

$$\frac{k^2}{3+k} + \frac{9-k^2}{k(k-3)} + \frac{27+k^3}{3-k}:\left(\frac{9-3k+k^2}{3-k}\right) = \frac{k^2}{3+k} - \frac{k^2-9}{k(k-3)} + \frac{(k+3)(k^2-3k+9)}{3-k}:\left(\frac{k^2-3k+9}{3-k}\right) = \frac{k^2}{3+k} - \frac{(k-3)(k+3)}{k(k-3)} + \frac{(k+3)(k^2-3k+9)(3-k)}{(3-k)(k^2-3k+9)} = \frac{k^2}{3+k} - \frac{(k+3)}{k} + k+3 = \frac{k^3 - (k+3)(k+3) + k(k+3)^2}{k(k+3)} = \frac{k^3 - (k^2+6k+9) + k(k^2+6k+9)}{k(k+3)} = \frac{k^3 - k^2 - 6k - 9 + k^3+6k^2+9k}{k(k+3)} = \frac{2k^3 + 5k^2 + 3k - 9}{k(k+3)} = \frac{(k+3)(2k^2-k-3)}{k(k+3)} = \frac{2k^2-k-3}{k}$$.

Ответ: $$\frac{2k^2-k-3}{k}$$.