\(C=5^{2018}+\frac{1}{5^{2017}+1}=\left(5^{2017}+1\right)+\frac{1}{5^{2017}+1}\)
\(D=5^{2018}+\frac{1}{5^{2018}+1}=\left(5^{2017}+1\right)+\left(1+\frac{1}{5^{2017}+2}\right)\)
Do \(\frac{1}{5^{2017}+1}< 1+\frac{1}{5^{2017}+2}\)
Nên \(C< D\)
Ta có : C = \(\frac{5^{2018}+1}{5^{2017}+1}\)
=> \(\frac{C}{5}=\frac{5^{2018}+1}{5^{2018}+5}=1-\frac{4}{5^{2018}+5}\)
Lại có D = \(\frac{5^{2019}+1}{5^{2018}+1}\)
=> \(\frac{D}{5}=\frac{5^{2019}+1}{5^{2019}+5}=1-\frac{4}{5^{2019}+5}\)
Vì \(\frac{4}{5^{2018}+5}>\frac{4}{5^{2019}+5}\Rightarrow1-\frac{4}{5^{2018}+5}< 1-\frac{4}{5^{2019}+5}\Rightarrow\frac{C}{5}< \frac{D}{5}\Rightarrow C< D\)
Ta có : \(C=\frac{5^{2018}+1}{5^{2017}+1}\)
\(\Rightarrow\frac{1}{5}C=\frac{5^{2018}+1}{5^{2018}+5}=\frac{5^{2018}+5-4}{5^{2018}+5}=1-\frac{4}{5^{2018}+5}\)
Ta lại có : \(D=\frac{5^{2019}+1}{5^{2018}+1}\)
\(\Rightarrow\frac{1}{5}D=\frac{5^{2019}+1}{5^{2019}+5}=\frac{5^{2019}+5-4}{5^{2019}+5}=1-\frac{4}{5^{2019}+5}\)
Vì \(\frac{4}{5^{2018}+5}>\frac{4}{5^{2019}+5}\) nên \(1-\frac{4}{5^{2018}+5}< 1-\frac{4}{5^{2019}+5}\)\(\Rightarrow\frac{1}{5}C< \frac{1}{5}D\)
\(\Rightarrow C< D\)
Vậy \(C< D\).
