Ta có :
A=\(\frac{9}{a^{2013}}+\frac{7}{a^{2014}}\)
=\(\left(\frac{8}{a^{2013}}+\frac{1}{a^{2013}}\right)+\left(\frac{8}{a^{2014}}-\frac{1}{a^{2014}}\right)\)
=\(\left(\frac{8}{a^{2013}}+\frac{8}{a^{2014}}\right)+\left(\frac{1}{a^{2013}}-\frac{1}{a^{2014}}\right)\)
B=\(\frac{8}{a^{2014}}+\frac{8}{a^{2013}}\)
=\(\frac{8}{a^{2013}}+\frac{8}{a^{2014}}\)
Vì \(\frac{1}{a^{2013}}>\frac{1}{a^{2014}}\)nên\(\frac{1}{a^{2013}}-\frac{1}{a^{2014}}>0\)
=> \(\left(\frac{8}{a^{2013}}+\frac{8}{a^{2014}}\right)+\left(\frac{1}{a^{2013}}-\frac{1}{a^{2014}}\right)>\frac{8}{a^{2013}}+\frac{8}{a^{2014}}\)
Vậy \(A>B\)
Chúc em học tốt
#Thiên_Hy
\(A=\frac{9}{a^{2013}}+\frac{7}{a^{2014}}=\frac{8}{a^{2013}}+\frac{1}{a^{2013}}+\frac{7}{a^{2014}}\)
\(B=\frac{8}{a^{2014}}+\frac{8}{a^{2013}}=\frac{7}{a^{2014}}+\frac{1}{a^{2014}}+\frac{8}{a^{2013}}\)
Ta thấy :
\(\frac{8}{a^{2013}}=\frac{8}{a^{2013}}\)
\(\frac{7}{a^{2014}}=\frac{7}{a^{2014}}\)
\(\frac{1}{a^{2013}}>\frac{1}{a^{2014}}\left(a^{2013}< a^{2014}\right)\)
\(\Rightarrow A>B\)
