áp dụng \(\frac{a}{b}< 1\Rightarrow\frac{a+m}{b+m}< 1\left(m\in N\right)\)
Ta có : \(B=\frac{9^{2009}+1}{9^{2010}+1}< 1\)
\(\Rightarrow B< \frac{9^{2009}+1+8}{9^{2010}+1+8}\)
\(\Rightarrow B< \frac{9.\left(9^{2008}+1\right)}{9.\left(9^{2009}+1\right)}=\frac{9^{2008}+1}{9^{2009}+1}\)
Vậy B < A
B = 92009 + 1/92010 + 1 < 1
=> B < 92009 + 1 + 8 / 92010 + 1 + 8 = 92009 + 9 / 92010 + 9 = 9 (92008 + 1 ) / 9 ( 92007 + 1) = A
=>B < A
#Hoq chắc _ Baccanngon
\(\frac{9^{2008}+1}{9^{2009}+1}=\frac{9^{2009}+9}{9^{2010}+9}>\frac{9^{2009}+1}{9^{2010}+1}\)
