Ta có \(9^{50}=\left(3^2\right)^{50}=3^{100}\)
\(27^{33}=\left(3^3\right)^{33}=3^{99}\)
\(\Rightarrow3^{100}>3^{99}\)hay \(9^{50}>27^{33}\)
\(9^{50}=\left(3^3\right)^{50}=3^{3.50}=3^{100}\)
\(27^{33}=\left(3^3\right)^{33}=3^{3.33}=3^{99}\)
Ta có: \(3^{100}>3^{99}\)
Vậy: \(9^{50}>27^{33}\)