Ta có\(10A=\frac{100^{100}+1}{100^{100}+10}=\frac{100^{100}+1}{100^{100}+1+9}=\frac{100^{100}+1}{1+9}\)
\(10B=\frac{100^{98}+1}{100^{98}+10}=\frac{100^{98}+1}{100^{98}+1+9}=\frac{100^{98}+1}{1+9}\)
Vì\(\frac{100^{100}+1}{1+9}>\frac{100^9+1}{1+9}\)
=>10A>10B
=>A>B