a, Ta có:
A= \(\dfrac{3}{8^3}+\dfrac{7}{8^4}=\dfrac{3}{8^3}+\dfrac{3}{8^4}+\dfrac{4}{8^4}\)
B= \(\dfrac{7}{8^3}+\dfrac{3}{8^4}=\dfrac{3}{8^3}+\dfrac{4}{8^3}+\dfrac{3}{8^4}\)
Vì \(\dfrac{4}{8^4}< \dfrac{4}{8^3}\) nên A < B.
b, Ta có:
\(\dfrac{20}{39}>\dfrac{14}{39}\)
\(\dfrac{22}{27}>\dfrac{22}{29}\)
\(\dfrac{18}{43}< \dfrac{18}{41}\)
\(\Rightarrow\)\(\dfrac{20}{39}+\dfrac{22}{27}+\dfrac{18}{43}>\dfrac{14}{39}+\dfrac{22}{29}+\dfrac{18}{41}\)
Hay A > B