Question

Không dùng máy tính cầm tay hãy so sánh

A = \(\dfrac{2017^{2017}+1}{2017^{2018}+1}\) và \(\dfrac{2017^{2016}+1}{2017^{2017}+1}\)

Answer 1

A = \(\dfrac{2017^{2017}+1}{2017^{2018}+1}\)< \(\dfrac{2017^{2017}+2017}{2017^{2018}+2017}\)= \(\dfrac{2017.\left(2017^{2016}+1\right)}{2017.\left(2017^{2017}+1\right)}\) = \(\dfrac{2017^{2016}+1}{2017^{2017}+1}\)= B

Vậy A < B

Answer 2

Ta có: \(2017A=\dfrac{2017^{2018}+2017}{2017^{2018}+1}=1+\dfrac{2016}{2017^{2018}+1}\)

\(2017B=\dfrac{2017^{2017}+2017}{2017^{2017}+1}=1+\dfrac{2016}{2017^{2017}+1}\)

Vì \(\dfrac{2016}{2017^{2018}+1}< \dfrac{2017}{2017^{2017}+1}\Rightarrow1+\dfrac{2016}{2017^{2018}+1}< 1+\dfrac{2016}{2017^{2017}+1}\)

\(\Rightarrow2017A< 2017B\Rightarrow A< B\)

Vậy A < B

Answer 3

Giải

Ta có: A =\(\dfrac{2017^{2017}+1}{2017^{2018}+1}< 1\)

\(\Rightarrow\) A< \(\dfrac{2017^{2017}+1+2016}{2017^{2018}+1+2016}\)=\(\dfrac{2017^{2017}+2017}{2017^{2018}+2017}=\dfrac{2017\left(2017^{2016}+1\right)}{2017\left(2017^{2017}+1\right)}=\dfrac{2017^{2016}+1}{2017^{2017}+1}=B\)Vậy A<B