Question

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Answer 1

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Answer 2

$\mathtt{(3a+1+\frac{1}{3a-1}):(3a-\frac{9a^2}{3a-1})=\frac{(3a+1)(3a-1)+1}{3a-1}:\frac{3a(3a-1)-9a^2}{3a-1}=}\\\\\mathtt{\frac{9a^2-1+1}{3a-1}:\frac{9a^2-3a-9a^2}{3a-1}=\frac{9a^2-1+1}{9a^2-3a-9a^2}=-3a}$