ý bạn là: \(\frac{1}{31}+\frac{1}{32}+...+\frac{1}{60}\left(CMR:S< \frac{4}{5}\right)\)

Tổng S có 30 số hạng. Vậy tách S ra thành 3 nhóm, mỗi nhóm có 10 số.

\(S=\left(\frac{1}{31}+\frac{1}{32}+...+\frac{1}{40}\right)+\left(\frac{1}{41}+\frac{1}{42}+...+\frac{1}{50}\right)+\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{60}\right)\)

\(\Leftrightarrow S< \left(\frac{1}{30}+\frac{1}{30}+...+\frac{1}{30}\right)+\left(\frac{1}{40}+\frac{1}{40}+....+\frac{1}{40}\right)+\left(\frac{1}{50}+\frac{1}{50}+...+\frac{1}{50}\right)\)

\(\Leftrightarrow S< \frac{1}{3}+\frac{1}{4}+\frac{1}{5}=\frac{20}{60}+\frac{15}{60}+\frac{12}{60}=\frac{47}{60}< \frac{48}{60}=\frac{4}{5}\)

\(S=\frac{1}{31}+\frac{1}{32}+\frac{1}{33}+...+\frac{1}{60}\)

\(=\left(\frac{1}{31}+\frac{1}{32}+...+\frac{1}{40}\right)+\left(\frac{1}{41}+...+\frac{1}{50}\right)+\left(\frac{1}{51}+...+\frac{1}{60}\right)\)

\(< \left(\frac{1}{30}+\frac{1}{30}+...+\frac{1}{30}\right)+\left(\frac{1}{40}+...+\frac{1}{40}\right)+\left(\frac{1}{50}+...+\frac{1}{50}\right)\)

\(=\frac{10}{30}+\frac{10}{40}+\frac{10}{50}=\frac{1}{3}+\frac{1}{4}+\frac{1}{5}=\frac{47}{60}< \frac{48}{60}=\frac{4}{5}\)

ai trả lời đúng mình tặng coin

S = (1/31+1/32+1/33+...+1/40) + (1/41 + 1/42 + ...+ 1/50) + (1/51 + 1/52+...+1/59+1/60)

Mà : (1/31+1/32+1/33+...+1/40) > 1/40 x 10 = 1/4 (gồm 10 số hạng)

Tương tự : (1/41 + 1/42 + ...+ 1/50) > 1/5 ;   (1/51 + 1/52+...+1/59+1/60) > 1/6

S > 1/4 + 1/5 + 1/6.

Trong khi đó (1/4 + 1/5 + 1/6) > 3/5

=>S > 3/5                             (1)

S = (1/31+1/32+1/33+...+1/40) + (1/41 + 1/42 + ...+ 1/50) + (1/51 + 1/52+...+1/59+1/60)

Mà : (1/31+1/32+1/33+...+1/40) < 1/31 x 10 = 10/30 = 1/3 (gồm 10 số hạng)

=> S <  4/5                             (2)

Từ (1) và (2) => 3/5 <S<4/5

giúp mình nhé. ai nhanh mình tick cho

Lời giải:

$S=(\frac{1}{31}+\frac{1}{32}+...+\frac{1}{40})+(\frac{1}{41}+\frac{1}{42}+...+\frac{1}{50})+(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{60})$

$> \frac{10}{40}+\frac{10}{50}+\frac{10}{60}=\frac{37}{60}> \frac{36}{60}=\frac{3}{5}$

Mặt khác: 

$S=(\frac{1}{31}+\frac{1}{32}+...+\frac{1}{40})+(\frac{1}{41}+\frac{1}{42}+...+\frac{1}{50})+(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{60})$

$< \frac{10}{30}+\frac{10}{40}+\frac{1}{50}=\frac{47}{60}< \frac{48}{60}=\frac{4}{5}$