Ta có:\(\left(\frac{9}{11}-0,81\right)^{2005}\)=\(\left(\frac{9}{11}-\frac{81}{100}\right)^{2005}=\left(\frac{9}{1100}\right)^{2005}< \left(\frac{10}{1100}\right)^{2005}=\left(\frac{1}{110}\right)^{2005}\)
Mà \(\left(\frac{1}{110}\right)^{2005}< \left(\frac{1}{100}\right)^{2005}=\left[\left(\frac{1}{10}\right)^2\right]^{2005}=\left(\frac{1}{10}\right)^{4010}=\frac{1}{10^{4010}}\)
Vậy \(\left(\frac{9}{11}-0,81\right)^{2005}< \frac{1}{10^{4010}}\)