CMR 41/2*42/2....80/2=1*3*5...79
Những câu hỏi liên quan
CM: 41/2×42/2×43/2×.....×80/2=1×3×5×....×79
CMR 1/41+1/42+1/43+...+1/79+7/80 <7/12
Cmr 1/41+4/42+4/43+.........+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
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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
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CMR:
\(\frac{7}{12}< \frac{1}{41}+\frac{1}{42}+\frac{1}{43}+.........+\frac{1}{79}+\frac{1}{80}< \frac{5}{6}\)
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
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Xem thêm câu trả lời
CMR : a) 1/41 + 1/42 + 1/43 +...+ 1/80 < 5/6
b) 1/2^2 + 1/2^4 + 1/2^6 +...+ 1/2^200 < 1/3
Chứng minh rằng
a) 1/2 + 1/3 + 1/4 +...+1/63 > 2
b) 1/41 + 1/42 + 1/43 +...+1/79 + 1/80 > 7/12
Chứng tỏ rằng : 7/12 < 1/41 + 1/42 +...+1/79+1/80<5/6
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)
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1,Chứng minh rằng:
1/2<1/51+1/52+...+1/100<1
2,Chứng minh 1/41+1/42+1/43+...+1/79+1/80>7/12
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)
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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 \(S=\frac{1}{41}+\frac{1}{42}+...+\frac{1}{80}>\frac{7}{12}\left(đpcm\right)\)
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Hãy chứng tỏ rằng:
a) 1/41+1/42+1/43+...+1/79+1/80>7/12
b)11/15<1/21+1/22+1/23+...+1/59+1/60<3/2