Ta có \(A=\frac{10^{11}-1}{10^{12}-1}\)
=> \(10A=\frac{10^{12}-10}{10^{12}-1}\)
=>\(10A=\frac{\left(10^{12-1}\right)-9}{10^{12}-1}\)
=>\(10A=1-\frac{9}{10^{12}-1}\) ( 1 )
Ta có \(B=\frac{10^{10}+1}{10^{11}+1}\)
=>\(10B=\frac{10^{11}+10}{10^{11}+1}=\frac{\left(10^{11}+1\right)+9}{10^{11}+1}=1+\frac{9}{10^{11}+1}\) ( 2 )
Từ 1 và 2 => 10A < 10B => A < B