Ta có:
\(B=\frac{20^{10}-1}{20^{10}-3}>1\)
=> Ta có: \(B=\frac{20^{10}-1}{20^{10}-3}>\frac{20^{10}-1-2}{20^{10}-3-2}=\frac{20^{11}+1}{20^{10}+1}=A\)
\(\Rightarrow A< B\)
\(A=\frac{20^{10}+1}{20^{10}-1}=\frac{20^{10}-1+2}{20^{10}-1}=1+\frac{2}{20^{10}-1}\)
\(B=\frac{20^{10}-1}{20^{10}-3}=\frac{20^{10}-3+2}{20^{10}-3}=1+\frac{2}{20^{10}-3}\)
vì \(\frac{2}{20^{10}-1}\) < \(\frac{2}{20^{10}-3}\)
=> 1 + \(\frac{2}{20^{10}-1}\) < 1+\(\frac{2}{20^{10}-3}\)
hay A< B