Question

[ke giau mat]

\(A=\frac{17^{2008}+1}{17^{2009}+1}\) và ​\(B=\frac{17^{2009}+1}{17^{2010}+1}\)​​​​

So sánh

Answer 1

[Trần Duy Vương]

Ta có: B=\(\frac{17^{2009}+1}{17^{2010}+1}\)<1 ( Vì 17²⁰⁰⁹+1< 17²⁰¹⁰+1 )

 Nên    B=\(\frac{17^{2009}+1}{17^{2010}+1}\)<\(\frac{17^{2009}+1+16}{17^{2010}+1+16}\)

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

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

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

Vậy A>B