Question

I = \(\dfrac{3}{8^3}\) + \(\dfrac{3}{8^4}\) và K = \(\dfrac{7}{8^3}\) + \(\dfrac{7}{8^4}\)

Answer 1

\(I=\dfrac{3}{8^3}+\dfrac{3}{8^4}=\dfrac{3.8}{8^4}+\dfrac{3}{8^4}=\dfrac{27}{8^4}\)

\(K=\dfrac{7}{8^3}+\dfrac{7}{8^4}=\dfrac{7.8}{8^4}+\dfrac{7}{8^4}=\dfrac{63}{8^4}\)

Vậy \(I< K\) vì \(\dfrac{27}{8^4}< \dfrac{63}{8^4}\)