Question

[Miyuki]

So sánh:

A= \(\frac{10^{2004}+1}{10^{2005}+1}\)và B=\(\frac{10^{2005}+1}{10^{2006}+1}\)

Answer 1

[Ninh Thế Quang Nhật]

\(B=\frac{10^{2005}+1}{10^{2006}+1}<\frac{10^{2005}+1+9}{10^{2006}+1+9}=\frac{10^{2005}+10}{10^{2006}+10}=\frac{10\left(10^{2004}+1\right)}{10\left(10^{2005}+1\right)}=\frac{10^{2004}+1}{10^{2005}+1}=A\)

\(\Rightarrow\)B < A