Question

So sánh A và B:

\(A=\frac{17^{2008}+1}{17^{2009}+1}\)

\(B=\frac{17^{2009}+1}{17^{2010}+1}\)

Answer 1

Ta có : 

\(17A=\frac{17^{2009}+17}{17^{2009}+1}=\frac{17^{1009}+1+16}{17^{2009}+1}=\frac{17^{2009}+1}{17^{2009}+1}+\frac{16}{17^{2009}+1}=1+\frac{16}{17^{2009}+1}\)

\(17B=\frac{17^{2010}+17}{17^{2010}+1}=\frac{17^{2010}+1+16}{17^{2010}+1}=\frac{17^{2010}+1}{17^{2010}+1}+\frac{16}{17^{2010}+1}=1+\frac{16}{17^{2010}+1}\)

Vì \(\frac{16}{17^{2009}+1}>\frac{16}{17^{2010}+1}\) nên \(17A>17B\)

\(\Rightarrow\)\(A>B\)

Vậy \(A>B\)

Chúc bạn học tốt ~