Chứng minh :
\(\frac{1}{2}< \frac{1}{51}+\frac{1}{52}+\frac{1}{53}+\frac{1}{54}+....+\frac{1}{100}< 1\)
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Chứng minh rằng :
\(\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)=\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+\frac{1}{54}+...+\frac{1}{100}\)
Ta có: \(\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)\)
\(=\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+...+\frac{1}{100}\)(đpcm)
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Chứng minh \(\frac{1}{2}< \frac{1}{51}+\frac{1}{52}+\frac{1}{53}+...+\frac{1}{100}< 1\)
Ta có:\(\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+............+\frac{1}{100}\)
\(=\left(\frac{1}{51}+\frac{1}{52}+.........+\frac{1}{75}\right)+\left(\frac{1}{76}+\frac{1}{77}+........+\frac{1}{100}\right)\)
\(>\frac{1}{75}.25+\frac{1}{100}.25=\frac{1}{3}+\frac{1}{4}=\frac{7}{12}>\frac{1}{2}\)
\(\left(\frac{1}{51}+\frac{1}{52}+..........+\frac{1}{75}\right)+\left(\frac{1}{76}+........+\frac{1}{100}\right)\)
\(< \frac{1}{50}.25+\frac{1}{75}.25=\frac{1}{2}+\frac{1}{3}=\frac{5}{6}< 1\)
\(\Rightarrowđpcm\)
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Chứng minh
H = \(\frac{1}{2}< \frac{1}{51}+\frac{1}{52}+\frac{1}{53}+...+\frac{1}{100}< 1\\\)
Ta có :
\(H=\frac{1}{51}+\frac{1}{52}+\frac{1}{52}+....+\frac{1}{100}\)
\(\Rightarrow H>\frac{1}{100}+\frac{1}{100}+\frac{1}{100}+....+\frac{1}{100}\)
\(\Rightarrow H>\frac{1}{100}.50\)
\(\Rightarrow H>\frac{1}{2}\)
Lại có :
\(H=\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+.....+\frac{1}{100}\)
\(\Rightarrow H< \frac{1}{51}+\frac{1}{51}+\frac{1}{51}+........+\frac{1}{51}\)
\(\Rightarrow H< \frac{1}{51}.50\)
\(\Rightarrow H< \frac{50}{51}\)
\(\Rightarrow H< 1\)
Vậy \(\frac{1}{2}< H< 1\left(ĐPCM\right)\)
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Chứng minh rằng :\(\frac{1}{1\times2}+\frac{1}{3\times4}+\frac{1}{5\times6}+...+\frac{1}{99\times100}=\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+...+\frac{1}{100}\)
Ta có :
\(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{99.100}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\)
\(=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{99}+\frac{1}{100}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{99}+\frac{1}{100}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)\)
\(=\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\)
cảm ơn bạn nha
Chứng minh rằng
\(\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)
\(=\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+...+\frac{1}{100}\)
ai giúp mk ik
mk đg cần gấp,còn nhìu đề chx lm
Chứng minh rằng: B= \(\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+...+\frac{1}{100}<\frac{5}{6}\)
Tính:\(\left(\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+\frac{1}{54}+...+\frac{1}{100}\right):\left(\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+\frac{1}{5\cdot6}+...+\frac{1}{99\cdot100}\right)\)
Số chia rút gọn thành 1/51+1/52+...+1/99+1/100
=> biểu thức bằng 1
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A = \(\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+...+\frac{1}{100}\) . Chứng minh : \(\frac{7}{12}< A< \frac{5}{6}\)
\(A=\left(\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+...+\frac{1}{75}\right)+\left(\frac{1}{76}+\frac{1}{77}+...+\frac{1}{100}\right)\)
Chia A làm 2 phần,mỗi phân 25 số hạng.
\(A>\frac{25.1}{75}+\frac{25.1}{100}\)
\(A>\frac{1}{3}+\frac{1}{4}=\frac{7}{12}\)
Bé hơn em làm tương tự có điều để nguyên cả 50 p/số.
Chúc em học tốt^^
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bạn có thể giải cụ thể hơn cho mình được ko ?
mình chả hiểu gì cả
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Chứng minh rằng:
(1+\(\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}\))-(\(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\))=\(\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+...+\frac{1}{100}\)
ta có:\(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}=\)\(\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)\)
=\(\left(1+\frac{1}{3}+...+\frac{1}{99}\right)+\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\) \(-\left(1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{50}\right)\)
=\(\left(1+\frac{1}{3}+...+\frac{1}{99}\right)+\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)-\) \(2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)
=\(\left(1+\frac{1}{3}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)
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