Ta có: 
7/12 = 4/12 + 3/12 = 1/3 + 1/4 = 20/60 + 20/80 

1/41 + 1/42 + 1/43 +...+ 1/79 + 1/80 = (1/41 + 1/42 + 1/43 + ...+ 1/60) + (1/61 + 1/62 +...+ 1/79 + 1/80) 

Do 1/41> 1/42 > 1/43 > ...>1/59 > 1/60 
=> (1/41 + 1/42 + 1/43 + ...+ 1/60) > 1/60 + ...+ 1/60 = 20/60 

và 1/61> 1/62> ... >1/79> 1/80 
=> (1/61 + 1/62 +...+ 1/79 + 1/80) > 1/80 + ...+ 1/80 = 20/80 

Vậy: 1/41 + 1/42 + 1/43 +...+ 1/79 + 1/80 > 20/60 + 20/80 = 7/12 

=> 1/41 + 1/42 + 1/43 +...+ 1/79 + 1/80 > 7/12 

=> ĐPCM

Chứng minh 1/41 + 1/42 + 1/43 + ... + 1/79 + 1/80 > 7/12

Ta có:
7/12 = 4/12 + 3/12 = 1/3 + 1/4 = 20/60 + 20/80

1/41 + 1/42 + 1/43 +...+ 1/79 + 1/80 = (1/41 + 1/42 + 1/43 + ...+ 1/60) + (1/61 + 1/62 +...+ 1/79 + 1/80)

Do 1/41> 1/42 > 1/43 > ...>1/59 > 1/60
=> (1/41 + 1/42 + 1/43 + ...+ 1/60) > 1/60 + ...+ 1/60 = 20/60

và 1/61> 1/62> ... >1/79> 1/80
=> (1/61 + 1/62 +...+ 1/79 + 1/80) > 1/80 + ...+ 1/80 = 20/80

Vậy: 1/41 + 1/42 + 1/43 +...+ 1/79 + 1/80 > 20/60 + 20/80 = 7/12

=> 1/41 + 1/42 + 1/43 +...+ 1/79 + 1/80 > 7/12

=> ĐPCM

mình nghĩ chắc mình biết bài này 

mình chịu 

Đặt \(A=\frac{1}{41}+\frac{1}{42}+\frac{1}{43}+...+\frac{1}{79}+\frac{1}{80}\)

\(A>\frac{1}{80}+\frac{1}{80}+....+\frac{1}{80}\)

\(\frac{1}{80}+\frac{1}{80}+.....+\frac{1}{80}\\ =\frac{40}{80}=\frac{1}{2}\)

Vì \(\frac{1}{2}< \frac{5}{6}\\ =>A< \frac{5}{6}\)

\(A< \frac{1}{40}+\frac{1}{40}+.....+\frac{1}{40}\)

\(\frac{1}{40}+\frac{1}{40}+...+\frac{1}{40}\\ =\frac{40}{40}=1\)

Vì \(1>\frac{7}{12}\\ =>A>\frac{7}{12}\)

bài này đề có vấn để

Ta có:

1/41 + 1/42 + .....+1/60 < 1/40 . 20 = 1/2

1/61 + 1/62 +.......+1/80 < 1/60 . 20 = 1/3

=> 1/41 + 1/42 +.....+1/79 + 1/80 < 1/2 + 1/3 = 5/6

1/41 + 1/42 +...+1/60 > 1/60 . 20 = 1/3

1/61 + 1/62 +....+ 1/80 > 1/80 . 20 = 1/4

=> 1/41 + 1/42 +.......+ 1/79 + 1/80 > 1/3 + 1/4 = 7/12

KL: Vậy 7/12 < 1/41 + 1/42 +.....+ 1/80 < 5/6 (đpcm)

Bài 1:

Ta có: \(\frac{1}{51}>\frac{1}{100}\)

           \(\frac{1}{52}>\frac{1}{100}\)

......

             \(\frac{1}{99}>\frac{1}{100}\)

Công vế với vế lại ta được:

\(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{99}+\frac{1}{100}>\frac{1}{100}+\frac{1}{100}+...+\frac{1}{100}+\frac{1}{100}=\frac{50}{100}=\frac{1}{2}\)        (1)

Lại có: \(\frac{1}{51}< \frac{1}{50}\)

            \(\frac{1}{52}< \frac{1}{50}\)

.....

             \(\frac{1}{100}< \frac{1}{50}\)

Cộng vế với vế lại ta được:

\(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}< \frac{1}{50}+\frac{1}{50}+...+\frac{1}{50}=\frac{50}{50}=1\)             (2)

Từ (1)(2) => \(\frac{1}{2}< \frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}< 1\) (đpcm)

Bài 2:

Đặt S = 1/41 + 1/42 +...+ 1/80

S có 40 số hạng,chia thành 4 nhóm,mỗi nhóm có 10 số hạng

Ta có:S = \(\left(\frac{1}{41}+\frac{1}{42}+...+\frac{1}{50}\right)\) + \(\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{60}\right)\)+ \(\left(\frac{1}{61}+\frac{1}{62}+...+\frac{1}{70}\right)\)+ \(\left(\frac{1}{71}+\frac{1}{72}+...+\frac{1}{80}\right)\)

=> S > \(\left(\frac{1}{50}+\frac{1}{50}+...+\frac{1}{50}\right)+\left(\frac{1}{60}+\frac{1}{60}+...+\frac{1}{60}\right)+\left(\frac{1}{70}+\frac{1}{70}+...+\frac{1}{70}\right)+\left(\frac{1}{80}+\frac{1}{80}+...+\frac{1}{80}\right)\)

=> S > \(\frac{10}{50}+\frac{10}{60}+\frac{10}{70}+\frac{10}{80}\)

=> S > \(\frac{533}{840}>\frac{490}{840}=\frac{7}{12}\)

Vậy \(S=\frac{1}{41}+\frac{1}{42}+...+\frac{1}{80}>\frac{7}{12}\left(đpcm\right)\)