Question

[Huycopper]

so sánh: A=2017⁹⁹/2017¹⁰⁰-2  và  B=2017¹⁰⁰/2017¹⁰¹-2

Answer 1

[Bùi Thế Hào]

\(A=\frac{2017^{99}}{2017^{100}-2}\) 

=> \(2017A=\frac{2017^{100}}{2017^{100}-2}=\frac{2017^{100}-2+2}{2017^{100}-2}=1+\frac{2}{2017^{100}-2}\)​

\(B=\frac{2017^{100}}{2017^{101}-2}\)

=>\(2017B=\frac{2017^{101}}{2017^{101}-2}=\frac{2017^{101}-2+2}{2017^{101}-2}=1+\frac{2}{2017^{101}-2}\)​

Do \(\frac{2}{2017^{100}-2}>\frac{2}{2017^{101}-2}\)

Nên 2017A > 2017B

Vậy A > B