Ta có:
\(9^{1005}=\left(3^2\right)^{1005}=3^{2010}\)
Vì \(2009< 2010\) \(\Rightarrow3^{2009}< 9^{1005}\)
Công thức : \(\left(a^m\right)^n=a^{m\cdot n}\) \(\left(m,n\ne0\right)\)
So sánh: \(3^{2009}\)và \(9^{1005}\)
Ta có: \(9^{1005}=\left(3^2\right)^{1005}=3^{2\cdot1005}=3^{2010}\)
Vì \(2009< 2010\) nên \(3^{2009}< 3^{2010}\) hay \(3^{2009}< 9^{1005}\)
\(3^{2009}=\left(3^2\right)^{1004}.3=9^{2008}.3>9^{1005}\)