\(A=\frac{3^{100}+1}{3^{99}+1}=\frac{\left(3^{99}+1\right)\times3-2}{3^{99}+1}=3-\frac{2}{3^{99}+1}\)
\(B=\frac{3^{99}+1}{3^{98}+1}=\frac{\left(3^{98}+1\right)\times3-2}{3^{98}+1}=3-\frac{2}{3^{98}+1}\)
Do 398 + 1 < 399 + 1
=> \(\frac{2}{3^{98}+1}>\frac{2}{3^{99}+1}\)
=> A > B