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Решите задачу:

$( \frac{x+2}{x^2+2x+4}- \frac{6x-13}{x^3-8})\cdot \frac{2x^2+4x+8}{3-x}\\\\ 1) \frac{x+2}{x^2+2x+4}- \frac{6x-13}{x^3-8}= \frac{x+2}{x^2+2x+4}- \frac{6x-13}{(x-2)(x^2+2x+4)}= \frac{x^2-6x+9}{(x-2)(x^2+2x+4)}=\\\\= \frac{(x-3)^2}{(x-2)(x^2+2x+4)}\\\\ 2) \frac{2x^2+4x+8}{3-x}\cdot \frac{(x-3)^2}{(x-2)(x^2+2x+4)}= \frac{2(x^2+2x+4)(3-x)}{(x-2)(x^2+2x+4)} = \frac{2(3-x)}{x-2}$

$\frac{2ab+4b-3a-6}{2b^2+2b}:( \frac{4b^2+21}{2b+2}-6 ) \\\\ 1) \frac{4b^2+21}{2b+2}-6= \frac{4b^2+21-12b-12}{2(b+1)}= \frac{4b^2-12b+9}{2(b+1)}= \frac{(2b-3)^2}{2(b+1)}\\\\ 2) \frac{2ab+4b-3a-6}{2b^2+2b}: \frac{(2b-3)^2}{2(b+1)}= \frac{(2b-3)(a+2)}{2b(b+1)}\cdot \frac{2(b+1)}{(2b-3)^2}= \frac{a+2}{b(2b-3)}$