Ta có
\(\frac{7}{12}=\frac{3}{12}+\frac{4}{12}=\frac{1}{4}+\frac{1}{3}=\frac{20}{80}+\frac{20}{60}=\frac{1}{41}+\frac{1}{42}+\frac{1}{43}+...+\frac{1}{80}\)
\(>\frac{1}{42}+\frac{1}{43}+...+\frac{1}{80}\)
hay B>A
nhớ tick mình nha
Ta chia A làm hai phần mỗi phần 20 số hạng
\(C=\frac{1}{41}+\frac{1}{42}+...+\frac{1}{60}\)với \(D=\frac{1}{61}+\frac{1}{62}+...+\frac{1}{80}\)
Xét \(C=\frac{1}{41}+\frac{1}{42}+...+\frac{1}{61}\)\(>\frac{1}{61}+\frac{1}{61}+\frac{1}{61}+...+\frac{1}{61}\)\(=\frac{1}{61}.20=\frac{1}{3}\)
Xét \(D=\frac{1}{61}+\frac{1}{62}+...+\frac{1}{80}>\frac{1}{80}+\frac{1}{80}+...+\frac{1}{80}=\frac{1}{80}.20=\frac{1}{4}\)
Mà A = C + D > \(\frac{1}{3}+\frac{1}{4}=\frac{7}{12}\)
=> A > B