Докажите тождество:
1) $(\frac{a - 2b}{a^2 + 2ab} - \frac{1}{a^2 - 4b^2} : \frac{a + 2b}{(2b - a)^2}) : \frac{a^2 - 2ab}{a^2 + 4ab + 4b^2} = \frac{2b}{a^2}$;
2) $(\frac{2a}{a + 3} - \frac{4a}{a^2 + 6a + 9}) * \frac{a^2 - 9}{a + 1} - \frac{a^2 - 9a}{a + 3} = a$.
$(\frac{a - 2b}{a^2 + 2ab} - \frac{1}{a^2 - 4b^2} : \frac{a + 2b}{(2b - a)^2}) : \frac{a^2 - 2ab}{a^2 + 4ab + 4b^2} = (\frac{a - 2b}{a(a + 2b)} - \frac{1}{(a - 2b)(a + 2b)} : \frac{a + 2b}{(a - 2b)^2}) : \frac{a(a - 2b)}{(a + 2b)^2} = (\frac{a - 2b}{a(a + 2b)} - \frac{1}{(a - 2b)(a + 2b)} * \frac{(a - 2b)^2}{a + 2b}) * \frac{(a + 2b)^2}{a(a - 2b)} = (\frac{a - 2b}{a(a + 2b)} - \frac{1}{a + 2b} * \frac{a - 2b}{a + 2b}) * \frac{(a + 2b)^2}{a(a - 2b)} = (\frac{a - 2b}{a(a + 2b)} - \frac{a - 2b}{(a + 2b)^2}) * \frac{(a + 2b)^2}{a(a - 2b)} = \frac{(a - 2b)(a + 2b) - a(a - 2b)}{a(a + 2b)^2} * \frac{(a + 2b)^2}{a(a - 2b)} = \frac{a^2 - 4b^2 - a^2 + 2ab}{a} * \frac{1}{a(a - 2b)} = \frac{2ab - 4b^2}{a} * \frac{1}{a(a - 2b)} = \frac{2b(a - 2b)}{a} * \frac{1}{a(a - 2b)} = \frac{2b}{a} * \frac{1}{a} = \frac{2b}{a^2}$
$(\frac{2a}{a + 3} - \frac{4a}{a^2 + 6a + 9}) * \frac{a^2 - 9}{a + 1} - \frac{a^2 - 9a}{a + 3} = (\frac{2a}{a + 3} - \frac{4a}{(a + 3)^2}) * \frac{(a - 3)(a + 3)}{a + 1} - \frac{a(a - 9)}{a + 3} = \frac{2a(a + 3) - 4a}{(a + 3)^2} * \frac{(a - 3)(a + 3)}{a + 1} - \frac{a(a - 9)}{a + 3} = \frac{2a^2 + 6a - 4a}{a + 3} * \frac{a - 3}{a + 1} - \frac{a(a - 9)}{a + 3} = \frac{2a^2 + 2a}{a + 3} * \frac{a - 3}{a + 1} - \frac{a(a - 9)}{a + 3} = \frac{2a(a + 1)}{a + 3} * \frac{a - 3}{a + 1} - \frac{a(a - 9)}{a + 3} = \frac{2a}{a + 3} * \frac{a - 3}{1} - \frac{a(a - 9)}{a + 3} = \frac{2a(a - 3)}{a + 3} - \frac{a(a - 9)}{a + 3} = \frac{2a^2 - 6a - a^2 + 9a}{a + 3} = \frac{a^2 + 3a}{a + 3} = \frac{a(a + 3)}{a + 3} = a$
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