bạn ơi đề sai ở chỗ dấu "  ,  "  phải không?? bạn hãy sửa đề đi 

Bạn Nguyễn Thị Bích Phương ơi, mình sửa lại đề rồi đó. Bạn giải giúp mình với.

\(A=\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+...+\frac{1}{20}\)

\(=\left(\frac{1}{6}+\frac{1}{7}+\frac{1}{8}\right)+\left(\frac{1}{9}+\frac{1}{10}+\frac{1}{11}\right)+\frac{1}{12}+\left(\frac{1}{13}+...+\frac{1}{16}\right)+\left(\frac{1}{17}+...+\frac{1}{20}\right)\)

\(>\left(\frac{1}{9}+\frac{1}{9}+\frac{1}{9}\right)+\left(\frac{1}{12}+\frac{1}{12}+\frac{1}{12}\right)+\frac{1}{12}+\left(\frac{1}{16}+...+\frac{1}{16}\right)+\left(\frac{1}{24}+...+\frac{1}{24}\right)\)

\(=\frac{1}{3}+\frac{1}{4}+\frac{1}{12}+\frac{1}{4}+\frac{1}{6}=1+\frac{1}{12}\)

\(B=\frac{1}{5}+\frac{1}{6}+...+\frac{1}{18}+\frac{1}{19}\)

\(=\left(\frac{1}{5}+...+\frac{1}{9}\right)+\left(\frac{1}{10}+...+\frac{1}{14}\right)+\left(\frac{1}{15}+...+\frac{1}{19}\right)\)

\(< \left(\frac{1}{5}+...+\frac{1}{5}\right)+\left(\frac{1}{10}+...+\frac{1}{10}\right)+\left(\frac{1}{15}+...+\frac{1}{15}\right)\)

\(=\frac{5}{5}+\frac{5}{10}+\frac{5}{15}=1+\frac{5}{6}\)

Nhóm lại từng nhóm rùi so sánh !!!

dạng này mik tưởng giảm tải mà

Đặt A = \(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+....+\frac{1}{17}+\frac{1}{18}+\frac{1}{19}\) 

\(A=\left(\frac{1}{5}+\frac{1}{6}+...+\frac{1}{9}\right)+\left(\frac{1}{10}+\frac{1}{11}+...+\frac{1}{14}\right)+\left(\frac{1}{15}+\frac{1}{16}+...+\frac{1}{19}\right)\) 

\(\Rightarrow A< \left(\frac{1}{5}+...+\frac{1}{5}\right)+\left(\frac{1}{10}+...+\frac{1}{10}\right)+\left(\frac{1}{15}+...+\frac{1}{15}\right)\)

\(\Rightarrow A< \frac{1}{5}\cdot5+\frac{1}{10}\cdot5+\frac{1}{15}\cdot5\)

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

\(\Rightarrow A< \frac{11}{6}< 2\) 

\(\Rightarrow A< 2\left(đpcm\right)\)

CMR A = 1/5 + 1/6 + 1/7 + ... + 1/17 < 2

A = 1/5 + 1/6 + 1/7 + ... + 1/17

Vì 1/6 < 1/7 < 1/8 < 1/9 < 1/5 và 1/10 < 1/11 < 1/12 < 1/13 < 1/14 <1/15 < 1/16 < 1/17 < 1/8 nên:

A = (1/5 + 1/6 + 1/7 + 1/8 + 1/9) + (1/10 + 1/11 + 1/12 + 1/13 + 1/14+ 1/15 + 1/16 + 1/17)

A < (1/5 + 1/5 + 1/5 + 1/5 + 1/5) + (1/8 + 1/8 + 1/8 + 1/8 + 1/8+1/8 + 1/8 + 1/8)

A < 1 + 1

A < 2

Vậy: A < 2 (đpcm)