\(CM:\frac{1}{51}+\frac{1}{52}+....+\frac{1}{100}< \frac{5}{6}\)
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\(CM:\) \(\frac{1}{51}+\frac{1}{52}+....+\frac{1}{100}<\frac{5}{6}\)
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 rằng \(\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)=\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\)
\(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}=\left(1+\frac{1}{2}+...+\frac{1}{100}\right)-\left(1+\frac{1}{2}+...+\frac{1}{50}\right)\)
\(=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}\right)+\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)
\(=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)
\(\Rightarrowđpcm\)
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Chứng minh rằng: B= \(\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+...+\frac{1}{100}<\frac{5}{6}\)
CMR :
\(\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)\)
\(=\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+\frac{1}{100}\)
\(\left(1+\frac{1}{3}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)
\(=1+\frac{1}{2}+...+\frac{1}{100}-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)
\(=1+\frac{1}{2}+...+\frac{1}{100}-\left(1+\frac{1}{2}+...+\frac{1}{50}\right)\)
\(=\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\)(ĐPCM)
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\(\frac{1}{51}+\frac{1}{52}+......\frac{1}{100}\)
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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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\(choA=\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+\frac{1}{5\cdot6}+...+\frac{1}{99\cdot100};B=\frac{1}{51\cdot100}+\frac{1}{52\cdot99}+...+\frac{1}{52\cdot99}+\frac{1}{100\cdot51}\)
\(CMR:\) \(1-\frac{1}{2}+\frac{1}{3}-...+\frac{1}{99}-\frac{1}{100}=\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\)\(=\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\)
Biến đổi vp của đẳng thức :
\(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\)
\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{99}+\frac{1}{100}-1-\frac{1}{2}-\frac{1}{3}-...-\frac{1}{50}\)
\(=1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}-2\left[\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right]\)
\(=1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}-\frac{1}{2}-\frac{1}{4}-...-\frac{1}{200}\)
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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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