Cho C=\(10^{2010}+\frac{1}{10^{2010}}\)
Xét \(A_1=10^{2010}+\frac{1}{10^{2011}}\)và \(B^{ }_1=10^{2011}+\frac{1}{10^{2012}}\)
Ta có \(A_1-C=10^{2010}+\frac{1}{10^{2010}}-10^{2010}-\frac{1}{10^{2010}}\)
\(A_1-C=10.\left(\frac{1}{10^{2011}}-\frac{1}{10^{2010}}\right)\)
Giair tượng tự ta được \(B_1-C=10^{2010}.\left(9+\frac{1}{10^{2012}}-\frac{1}{10^{2010}}\right)\)
Ta thấy \(\frac{1}{10^{2012}}-\frac{1}{10^{2010}}