Question

So sánh : 2²⁰¹⁷ + 1 / 2²⁰¹⁸ + 1 và 2²⁰¹⁸ + 1 / 2²⁰¹⁹ + 1

Answer 1

Đặt \(A=\frac{2^{2017}+1}{2^{2018}+1}\Rightarrow2A=\frac{2^{2018}+2}{2^{2018}+1}=\frac{2^{2018}+1+1}{2^{2018}+1}=1+\frac{1}{2^{2018}+1}\)

\(B=\frac{2^{2018}+1}{2^{2019}+1}\Rightarrow2B=\frac{2^{2019}+2}{2^{2019}+1}=\frac{2^{2019}+1+1}{2^{2019}+1}=1+\frac{1}{2^{2019}+1}\)

Vì \(2^{2019}+1>2^{2018}+1\Rightarrow\frac{1}{2^{2019}+1}< \frac{1}{2^{2018}+1}\)

\(\Rightarrow2A>2B\Rightarrow A>B\)