Найдите значение выражения: x^2-4 / 6x^3+48 : x^2-2x / 7x^2-14x+28 при х=-0,5 x^9-1 /...

Решите задачу:

$\frac{x^2-4}{6x^3+48}: \frac{x^2-2x}{7x^2-14x+28}= \frac{(x-2)(x+2)}{6(x+2)(x^2-2x+4)}* \frac{7(x^2-2x+4)}{x(x-2)}= \frac{7}{6x} \\ \\ x=-0.5 \\ \\ \frac{7}{6*(-0.5)}=- \frac{7}{3}=-2 \frac{1}{3}$

$\frac{x^9-1}{3x^2+3x+3}: \frac{2x^6+2x^3+2}{10x-10}= \frac{(x^3-1)(x^6+x^3+1)}{3(x^2+x+1)}* \frac{10(x-1)}{2(x^6+x^3+1)}= \\ \\ \frac{(x-1)(x^2+x+1)(x^6+x^3+1)}{3(x^2+x+1)}* \frac{10(x-1)}{2(x^6+x^3+1)}= \frac{10(x-1)^2}{6}= \\ \\ x=-0.2 \\ \\ \frac{10(x^2-2x+1)}{6}= \frac{10((-0.2)^2-(2*(-0.2))+1)}{6}= \frac{10(0.04+0.4+1)}{6}= \frac{14.4}{6}=2.4$