8. Let \( a_{k}=\frac{2^{k}}{3^{2^{k}}+1}, k \) be a natural number, and let \( A=a_{0}+a_{1}+\cdots+a_{9}, B=a_{0} a_{1} \cdots a_{9} \), then \( \frac{A}{B}=\) \qquad .
Since $\frac{1}{3^{2^{k}}-1}-\frac{1}{3^{2^{k}}+1}=\frac{2}{3^{2^{k+1}}-1} \Rightarrow \frac{2^{k}}{3^{2^{k}}+1}=\frac{2^{k}}{3^{2^{k}}-1}-\frac{2^{k+1}}{3^{2^{k+1}}-1}$,
thus $A=\sum_{k=0}^{9} a_{k}=\frac{1}{2}-\frac{2^{10}}{3^{2^{10}}-1}, B=\prod_{k=0}^{9} a_{k}=\frac{2^{46}}{3^{2^{10}}-1}$,
therefore $\frac{A}{B}=\frac{3^{2^{10}}-1}{2^{47}}-\frac{1}{2^{36}}=\frac{3^{2^{10}}-2^{11}-1}{2^{47}}$.