Ta có 10A = \(\frac{10\left(10^{14}-1\right)}{10^{15}-11}=\frac{10^{15}-10}{10^{15}-11}=\frac{10^{15}-11+1}{10^{15}-11}=1+\frac{1}{10^{15}-11}\)
Lại có 10B = \(\frac{10\left(10^{14}+1\right)}{10^{15}+9}=\frac{10^{15}+10}{10^{15}+9}=\frac{10^{15}+9+1}{10^{15}+9}=1+\frac{1}{10^{15}+9}\)
Nhận thấy 1015 - 11 < 1015 + 9
=> \(\frac{1}{10^{15}-11}>\frac{1}{10^{15}+9}\)
=> \(1+\frac{1}{10^{15}-11}>1+\frac{1}{10^{15}+9}\)
=> 10A > 10B
=> A > B