ớ chết, mk nhầm, lm lại nha

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

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

\(S< \frac{1}{30}.10+\frac{1}{40}.10+\frac{1}{50}.10\)

\(S< \frac{1}{3}+\frac{1}{4}+\frac{1}{5}< \frac{4}{5}\)

=> \(S< \frac{4}{5}\)

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

\(S< 30.\frac{1}{60}\)

\(S< \frac{1}{2}< \frac{4}{5}\)

\(S< \frac{4}{5}\)

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

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

\(\Rightarrow S< \frac{1}{30}.10+\frac{1}{40}.10+\frac{1}{50}.10\)

\(S< \frac{1}{3}+\frac{1}{4}+\frac{1}{5}< \frac{4}{5}\)

\(V\text{ậy}:S< \frac{4}{5}\)

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

\(S=\left(\frac{1}{31}+\frac{1}{32}+...+\frac{1}{40}\right)+\left(\frac{1}{41}+\frac{1}{42}+...+\frac{1}{49}\right)+\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{60}\right)\)

\(\frac{1}{40}.10+\frac{1}{50}.10+\frac{1}{60}.10< S< \frac{1}{30}.10+\frac{1}{40}.10+\frac{1}{50.10}\)

\(\frac{1}{4}+\frac{1}{5}+\frac{1}{6}< S< \frac{1}{3}+\frac{1}{4}+\frac{1}{5}\)

\(\frac{1}{4}+\frac{1}{5}+\frac{3}{20}< \frac{1}{4}+\frac{1}{5}+\frac{1}{6}< S< \frac{1}{3}+\frac{1}{4}+\frac{1}{5}< \frac{1}{3}+\frac{4}{15}+\frac{1}{5}\)

\(\frac{3}{5}< S< \frac{4}{5}\left(đpcm\right)\)

\(\hept{\begin{cases}A=-\frac{1}{2020}-\frac{3}{2019^2}-\frac{5}{2019^3}-\frac{7}{2019^4}^{ }\\B=-\frac{1}{2020}-\frac{7}{2019^2}-\frac{5}{2019^3}-\frac{3}{2019^4}\end{cases}}\)

=>\(A-B=-\frac{1}{2020}-\frac{3}{2019^2}-\frac{5}{2019^3}-\frac{7}{2019^4}+\frac{1}{2020}+\frac{7}{2019^2}+\frac{5}{2019^3}+\frac{3}{2019^4}\)

\(=>A-B=\left(-\frac{3}{2019^2}+\frac{7}{2019^2}\right)+\left(-\frac{7}{2019^4}+\frac{3}{2019^4}\right)\)

=>\(A-B=\frac{4}{2019^2}+-\frac{4}{2019^4}\)

=>\(A-B=\frac{2019^2.4}{2019^4}-\frac{4}{2019^4}\)

=>\(A>B\)

cách này mình tự nghĩ 

thank you \(v\text{er}y^{1000000000000}\)much

@Thùy Nguyễn Thị Bích

tao ko biết tao mới lên lớp 5 thôi mày 

Mình ko biết thông cảm nha .Năm nay mình mới lên lớp 5 thui à

            THẬT LÒNG XIN LỖI VÌ KO GIÚP ĐƯỢC GÌ

S=311​+321​+331​+...+601​

=(131+132+...+140)+(141+...+150)+(151+...+160)=(311​+321​+...+401​)+(411​+...+501​)+(511​+...+601​)

<(130+130+...+130)+(140+...+140)+(150+...+150)<(301​+301​+...+301​)+(401​+...+401​)+(501​+...+501​)

=1030+1040+1050=13+14+15=4760<4860=45=3010​+4010​+5010​=31​+

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

Vậy A > 3/5

Phần 2. 

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)

Tương tự : (1/41 + 1/42 + ...+ 1/50)  < 1/4 ;   (1/51 + 1/52+...+1/59+1/60) < 1/5

Mà S = (1/3 + 1/4 + 1/5) < 4/5 (Vì 1/3 + 1/5 < 3/5 hay 7/12 < 3/5 hay 35/60 < 36/60)

Vậy S <  4/5

Thank You