\(=\frac{2^{2019}}{2^{2010}.2^8}=\frac{2^{2019}}{2^{2018}}\)
vì \(2^{2019}>2^{2015}\)
\(=>\frac{2^{2019}}{2^{2018}}>\frac{2^{2015}}{2^{2018}}\)
Vậy \(\frac{2^{2019}}{2^{2010}.256}>\frac{2^{2015}}{2^{2018}}\)
Ta có:
\(\frac{2^{2019}}{2^{2010}\cdot256}=\frac{2^{2019}}{2^{2010}\cdot2^8}=\frac{2^{2019}}{2^{2018}}=\frac{2}{1}=2>1\)
\(\frac{2^{2015}}{2^{2018}}=\frac{1}{2^3}=\frac{1}{8}< 1\)
\(\Rightarrow\frac{2^{2015}}{2^{2018}}< \frac{2^{2019}}{2^{2010}\cdot256}\)