a)+)Ta có::\(B=\frac{7^{2014}+1}{7^{2015}+1}< 1\)
\(\Rightarrow B< \frac{7^{2014}+1+6}{7^{2015}+1+6}=\frac{7^{2014}+7}{7^{2015}+7}=\frac{7.\left(7^{2013}+1\right)}{7.\left(7^{2014}+1\right)}=\frac{7^{2013}+1}{7^{2014}+1}=A\)
\(\Rightarrow B< A\)
Vậy B<A
b)+)Ta có:\(A=\frac{2019^{2018}+1}{2019^{2019}+1}< 1\)
\(\Rightarrow A< \frac{2019^{2018}+1+2018}{2019^{2019}+1+2018}=\frac{2019^{2018}+2019}{2019^{2019}+2019}=\frac{2019.\left(2019^{2017}+1\right)}{2019.\left(2019^{2018}+1\right)}=B\)
\(\Rightarrow A< B\)
Vậy B<A
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