Давай выполним все задания по порядку.
1) \[\frac{y^2-25}{y^2+12y+36} \cdot \frac{3y+18}{2y+10} = \frac{(y-5)(y+5)}{(y+6)^2} \cdot \frac{3(y+6)}{2(y+5)} = \frac{3(y-5)}{2(y+6)}\]
2) \[\frac{1-a^2}{4a+8b} \cdot \frac{a^2+10ab+4b^2}{3-3a} = \frac{(1-a)(1+a)}{4(a+2b)} \cdot \frac{a^2+10ab+4b^2}{3(1-a)} = \frac{(1+a)(a^2+10ab+4b^2)}{12(a+2b)}\]
3) \[\frac{x^2-10x+25}{3x+12} \cdot \frac{x^2-16}{2x-10} = \frac{(x-5)^2}{3(x+4)} \cdot \frac{(x-4)(x+4)}{2(x-5)} = \frac{(x-5)(x-4)}{6}\]
4) \[\frac{ax+ay}{x^2-2xy+y^2} \cdot \frac{x^2-xy}{7x+7y} = \frac{a(x+y)}{(x-y)^2} \cdot \frac{x(x-y)}{7(x+y)} = \frac{ax}{7(x-y)}\]
5) \(\(-\frac{7a^4b}{9bc^2}\)^2 = \frac{49a^8b^2}{81b^2c^4} = \frac{49a^8}{81c^4}\)
6) \(\( \frac{1}{2}mn^4c^6 \)^4 = \frac{1}{16}m^4n^{16}c^{24}\)
7) \(\( -0.21b^3a^8 \)^3 = -0.009261b^9a^{24}\)
Ответ: 1) \(\frac{3(y-5)}{2(y+6)}\); 2) \(\frac{(1+a)(a^2+10ab+4b^2)}{12(a+2b)}\); 3) \(\frac{(x-5)(x-4)}{6}\); 4) \(\frac{ax}{7(x-y)}\); 5) \(\frac{49a^8}{81c^4}\); 6) \(\frac{1}{16}m^4n^{16}c^{24}\); 7) \(-0.009261b^9a^{24}\)
Ты молодец! У тебя всё получится!