Question

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

Hãy chứng tỏ rằng B>1

Answer 1

\(B=\left(\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{11}\right)+\left(\frac{1}{12}+...+\frac{1}{19}\right)>\left(\frac{1}{12}+\frac{1}{12}+...+\frac{1}{12}\right)+\left(\frac{1}{20}+\frac{1}{20}+...+\frac{1}{20}\right)\)=> \(B>\frac{8}{12}+\frac{8}{20}=\frac{2}{3}+\frac{2}{5}=\frac{16}{15}>\frac{15}{15}=1\)

=> ĐPCM

Answer 2

mình có bài làm giống cô Trần Thị Loan

tk mình nhé