Question

\(\dfrac{3^{17}+1}{3^{20}+1}\) và \(\dfrac{3^{20}+1}{3^{23}+1}\)

Answer 1

Gọi \(A=\dfrac{3^{17}+1}{3^{20}+1}\) và \(B=\dfrac{3^{20}+1}{3^{23}+1}\)

Ta Có: \(A=\dfrac{3^{17}+1}{3^{20}+1}=\left(\dfrac{3^{17}+1}{3^{20}+1}\right).\dfrac{3^3}{3^3}=\dfrac{3^{20}+27}{3^{23}+27}\)

Ta lại có: \(1-A=1-\dfrac{3^{20}+27}{3^{23}+27}=\dfrac{3^{23}+27}{3^{23}+27}-\dfrac{3^{20}+27}{3^{23}+27}=\dfrac{3^{23}-3^{20}}{3^{23}+27}\)

\(1-B=1-\dfrac{3^{20}+1}{3^{23}+1}=\dfrac{3^{23}+1}{3^{23}+1}-\dfrac{3^{20}+1}{3^{23}+1}=\dfrac{3^{23}+3^{20}}{3^{23}+1}\)

Vì: \(\dfrac{3^{23}-3^{20}}{3^{23}+27}< \dfrac{3^{23}-3^{20}}{3^{23}+1}\Rightarrow A>B\)

Vậy \(\dfrac{3^{17}+1}{3^{20}+1}>\dfrac{3^{20}+1}{3^{23}+1}\)