A<B

Tham khảo: https://hoc24.vn/cau-hoi/so-sanh-afrac720131720141va-bfrac720141720151.235038064198

\(\frac{A}{B}=\frac{7^{2013}+1}{7^{2014}+1}.\frac{7^{2015}+1}{7^{2014}+1}=\frac{7^{4028}+7^{2013}+7^{2015}+1}{7^{4028}+2.7^{2014}+1}=\)

\(=\frac{7^{4028}+7^{2013}\left(1+7^2\right)+1}{7^{4028}+2.7.7^{2013}+1}=\frac{7^{4028}+50.7^{2013}+1}{7^{4028}+14.7^{2013}+1}>1\)

\(\Rightarrow A>B\)

A/B sao lại nhân v bn

A/B thành A nhân với nghịch đảo của B mà

Đặt A= \(\frac{7^{2015}+1}{7^{2017}+1}\) 

B= \(\frac{7^{2017}+1}{7^{2019}+1}\)

Ta có  A= \(\frac{7^2\left(7^{2015}+1\right)}{7^2\left(7^{2017}+1\right)}\)

           = \(\frac{7^{2017}+49}{7^{2019}+49}\)

         = \(\frac{7^{2017}+1+48}{7^{2019}+1+48}\)

Vì \(\frac{7^{2017}+1+48}{7^{2019}+1+48}\)>\(\frac{7^{2017}+1}{7^{2019}+1}\)

=> A>B

K MK NHA !        

Bạn tham khảo nhé 
Ta có công thức : 

\(\frac{a}{b}< \frac{a+c}{b+c}\)\(\left(\frac{a}{b}< 1;a,b,c\inℕ^∗\right)\)

Áp dụng vào ta có : 

\(B=\frac{7^{2017}+1}{7^{2019}+1}< \frac{7^{2017}+1+48}{7^{2019}+1+48}=\frac{7^{2017}+49}{7^{2019}+49}=\frac{7^2\left(7^{2015}+1\right)}{7^2\left(7^{2017}+1\right)}=\frac{7^{2015}+1}{7^{2017}+1}=B\)

\(\Rightarrow\)\(B< A\) hay \(A>B\)

Vậy \(A>B\)

Chúc bạn học tốt ~

Cảm ơn mn nha

\(TA-CO':\)

\(A=\frac{4+\frac{7}{2014}-\frac{7}{2015}+\frac{7}{2012}-\frac{7}{2013}}{7+\frac{7}{2014}-\frac{7}{2015}+\frac{7}{2012}-\frac{7}{2013}}\)

\(A=\frac{4\left(\frac{1}{2014}-\frac{1}{2015}+\frac{1}{2012}-\frac{1}{2013}\right)}{7\left(\frac{1}{2014}-\frac{1}{2015}+\frac{1}{2012}-\frac{1}{2013}\right)}\)

\(A=\frac{4}{7}\)

\(B=\frac{1+2+...+2^{2013}}{2^{2015}-2}\)

ĐẶT \(C=1+2+...+2^{2013}\)

\(\Rightarrow2C=2+2^2+...+2^{2014}\)

\(\Rightarrow2C-C=\left(2+2^2+...+2^{2014}\right)-\left(1+2+...+2^{2013}\right)\)

\(\Rightarrow C=2^{2014}-2\)

\(\Rightarrow B=\frac{2^{2014}-1}{2^{2015}-2}\)

\(B=\frac{2^{2014}-1}{2\left(2^{2014}-1\right)}\)

\(B=\frac{1}{2}\)

\(\Rightarrow A-B=\frac{3}{7}-\frac{1}{2}=\frac{6}{14}-\frac{7}{14}\)

\(A-B=\frac{6-7}{14}=\frac{-1}{14}\)

VẬY, \(A-B=\frac{-1}{14}\)

\(2.\left(\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+...+\frac{1}{2013}+\frac{1}{2015}+\frac{1}{2017}\right)\)

\(=2.\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{2017}+\frac{1}{2018}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2018}\right)\)

\(=2.\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{2017}+\frac{1}{2018}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{1009}\right)\)

\(=2.\left(\frac{1}{1010}+\frac{1}{1011}+...+\frac{1}{2018}\right)-1\)