Решите номер 34.Есть вложение.

Решите задачу:

$\frac{(6-3x)^2}{3x^2+3x-18} = \frac{(3(2-x))^2}{3(x^2+x-6)} = \frac{9(2-x)^2}{3(x^2+3x-2x-6)} = \frac{3(2-x)^2}{x(x+3)-2(x+3)} = \\\ = \frac{3(2-x)^2}{(x+3)(x-2)} = \frac{3(x-2)}{x+3} =\frac{3x-6}{x+3}$

$\frac{2x^2+2x-24}{(6-2x)^2}= \frac{2(x^2+x-12)}{(2(3-x))^2}= \frac{2(x^2+4x-3x-12)}{4(3-x)^2}= \frac{x(x+4)-3(x+4)}{2(3-x)^2}= \\\ = \frac{(x+4)(x-3)}{2(x-3)^2}= \frac{x+4}{2(x-3)}= \frac{x+4}{2x-6}$