Question

So sánh A và B

A =\(\frac{2^{2017}+1}{2^{2018}+1}\) và B = \(\frac{2^{2018}+1}{2^{2019}+1}\)

Answer 1

Ta có: \(2A=\frac{2^{2018}+2}{2^{2018}+1}=1+\frac{1}{2^{2018}+1}\)

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

Vì \(\frac{1}{2^{2018}+1}>\frac{1}{2^{2019}+1}\Rightarrow1+\frac{1}{2^{2018}+1}>1+\frac{1}{2^{2019}+1}\)

\(\Rightarrow2A>2B\)

\(\Rightarrow A>B\)

Vậy A > B

Answer 2

Ta có:

2A = \(\frac{2\left(2^{2017}+1\right)}{2^{2018}+1}\) = \(\frac{2^{2018}+2}{2^{2018}+1}\) = \(\frac{2^{2018}+1+1}{2^{2018}+1}\) = \(\frac{2^{2018}+1}{2^{2018}+1}\) + \(\frac{1}{2^{2018}+1}\) = 1 + \(\frac{1}{2^{2018}+1}\)

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

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

Vậy A > B