C>1   vì c>1

a, Ta có: \(A=\frac{1}{11}+\frac{1}{12}+...+\frac{1}{50}=\left(\frac{1}{11}+\frac{1}{12}+...+\frac{1}{30}\right)+\left(\frac{1}{31}+\frac{1}{32}+...+\frac{1}{60}\right)\)

Nhận xét: \(\frac{1}{11}+\frac{1}{12}+....+\frac{1}{30}>\frac{1}{30}+\frac{1}{30}+...+\frac{1}{30}=\frac{20}{30}=\frac{2}{3}\)

\(\frac{1}{31}+\frac{1}{32}+...+\frac{1}{60}>\frac{1}{60}+\frac{1}{60}+...+\frac{1}{60}=\frac{20}{60}=\frac{1}{3}\)

\(\Rightarrow A>\frac{2}{3}+\frac{1}{3}=1>\frac{1}{2}\)

Vậy A > 1/2

b, Ta có: \(\frac{1}{50}>\frac{1}{100};\frac{1}{51}>\frac{1}{100};........;\frac{1}{99}>\frac{1}{100}\)

\(\Rightarrow B>\frac{1}{100}+\frac{1}{100}+...+\frac{1}{100}=\frac{50}{100}=\frac{1}{2}\)

Vậy B > 1/2

c, Ta có: \(C=\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+...+\frac{1}{100}=\frac{1}{10}+\left(\frac{1}{11}+\frac{1}{12}+...+\frac{1}{100}\right)\)

Nhận xét: \(\frac{1}{11}+\frac{1}{12}+...+\frac{1}{100}>\frac{1}{100}+\frac{1}{100}+...+\frac{1}{100}=\frac{90}{100}=\frac{9}{10}\)

\(\Rightarrow C>\frac{1}{10}+\frac{9}{10}=\frac{10}{10}=1\)

Vậy C > 1

Tớ đồng ý,bạn làm đúng rồi .......

Ta có:C=1/11+12+1/13+...+1/50

=(1/11+1/12+...+1/20)+(1/21+1/22+...+1/30)+(1/31+1/32+...+1/40)+(1/41+1/42+...+1/50)

=(1/20+1/20+...+1/20)+(1/30+1/30+...+1/30)+(1/40+1/40+...+1/40)+(1/50+1/50+...+1/50)

=1/20×10+1/30×10+1/40×10+1/50×10

Nhân với10 vì mỗi tổng đều có10 số hạng

=1/2+1/3+1/4+1/5

=30/60+20/60+15/60+12/60

=77/60 

Mik học trên lớp tương tự bài đấy

Hok tốt

\(\frac{1}{10.11}+\frac{1}{11.12}+\frac{1}{12.13}+.......+\frac{1}{49.50}\)

\(=\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+......+\frac{1}{50}\)

\(=\frac{1}{10}-\frac{1}{50}=\frac{2}{25}\)