Question

Cho B=\(\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+........+\frac{1}{19}\)          . hãy chứng minh B>1

Answer 1

\(B=\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{19}=\frac{1}{4}+\frac{1}{17}+\frac{1}{18}+\frac{1}{19}+\left(\frac{1}{5}+...+\frac{1}{8}\right)+\left(\frac{1}{9}+...+\frac{1}{16}\right)\)

\(\frac{1}{5}+...+\frac{1}{8}>\frac{1}{8}.4=\frac{1}{2}\)

\(\frac{1}{9}+...+\frac{1}{16}\frac{1}{2}+\frac{1}{2}=1\)

\(SuyraB>1\)

Answer 2

Ta có: \(B=\left(\frac{1}{4}+\frac{1}{19}\right).8\)

\(B=2\frac{8}{19}\)

=> B>1

Answer 3

B > 1 nha

chúc cậu học tốt

Answer 4

b=1nha

Answer 5

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