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Vereinfache m"oglichst weitgehend:

(a)~~$\D\frac{1}{6a}+\frac{3a^2b+ab^2}{4a+4b}:\frac{3a^2b^2}{a+b}$ \qquad
(b)~~$\D\frac{\frac{1}{x-y}-\frac{1}{x+y}}{x^2+y^2}\cdot (x^2-y^2)-
	\frac{(x+y)^2-x^2-y^2}{x^3+xy^2}$

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}
\LSG{ (a):~~$\D\frac{a+b}{4ab}$ \qquad (b):~~$0$ }
