ta co : A = 3/8^3+3/8^4+4/8^4
B=3/8^3+3/8^4+4/8^3
VI 4/8^4 <4/8^3 NEN A<B
có \(\frac{3}{8^3}+\frac{7}{8^4}=\frac{3}{8^3}+\frac{3}{8^4}+\frac{4}{8^4}\)
\(\frac{7}{8^3}+\frac{3}{8^4}=\frac{3}{8^3}+\frac{4}{8^3}+\frac{3}{8^4}\)
vì \(\frac{4}{8^4}
Ta có:
=> A= 3/8 ^ 3+3/8 ^ 4 + 4/8^4
B= 3/8 ^ 3 = 38 ^4 +4/8^3
Vậy:
A/8^4<4/8^3
Nên A <B