15) $$\frac{29n^3(2n+4)}{n^2(4+2n)} = \frac{29n^3 \cdot 2(n+2)}{n^2 \cdot 2(2+n)} = 29n$$
$$\frac{29n^3(2n+4)}{n^2(4+2n)} = \frac{29n^3 \cdot 2(n+2)}{n^2 \cdot 2(2+n)} = 29n$$
Ответ: 29n