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Решите задачу:

$( \frac{1}{a^2}+ \frac{1}{b^2}+ \frac{2}{a+b}\cdot( \frac{1}{a}+ \frac{1}{b})): (\frac{a+b}{ab})^2=1$

$\boxed{1} \ \ \frac{1}{a}+ \frac{1}{b} = \frac{a+b}{ab}\\\\ \boxed{2} \ \ \frac{2}{a+b}\cdot \frac{a+b}{ab}= \frac{2}{ab}\\\\ \boxed{3} \ \ \frac{1}{a^2}+ \frac{1}{b^2}+ \frac{2}{ab}= \frac{b^2+a^2+2ab}{a^2b^2}= \frac{(a+b)^2}{a^2b^2}\\\\ \boxed{4} \ \ \frac{(a+b)^2}{a^2b^2}: (\frac{a+b}{ab})^2= \frac{(a+b)^2}{a^2b^2}\cdot \frac{a^2b^2}{(a+b)^2} =1$