Chứng minh rằng:
(1+\(\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}\))-(\(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\))=\(\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+...+\frac{1}{100}\)
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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 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 :\(\frac{1}{1\times2}+\frac{1}{3\times4}+\frac{1}{5\times6}+...+\frac{1}{99\times100}=\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+...+\frac{1}{100}\)
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 \(\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)=\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\)
\(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}=\left(1+\frac{1}{2}+...+\frac{1}{100}\right)-\left(1+\frac{1}{2}+...+\frac{1}{50}\right)\)
\(=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}\right)+\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)
\(=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)
\(\Rightarrowđpcm\)
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chứng minh rằng:\(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
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\(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+...+\frac{1}{99\cdot100}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(\Leftrightarrow\left(1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+...+\frac{1}{100}\right)\)
\(\Leftrightarrow\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{99}+\frac{1}{100}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)\)
\(\Leftrightarrow\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{99}+\frac{1}{100}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)\)
\(\Leftrightarrow\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+...+\frac{1}{100}\)
Ta có đpcm
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Cho \(A=\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+\frac{1}{5\cdot6}+...+\frac{1}{99\cdot100}\)
Chứng minh rằng:
a) \(A=\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+...+\frac{1}{100}\)
b) \(\frac{7}{12}< A< \frac{5}{6}\)
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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Chứng tỏ rằng: \(\frac{1}{1}\times\frac{1}{3}\times\frac{1}{5}\times.....\times\frac{1}{99}=\frac{2}{51}\times\frac{2}{52}\times\frac{2}{53}\times.....\times\frac{2}{100}\)
VÌ 1/1.1/3.......1/99=2/51.2/52.........2/100
VÀ 2/51.2/52.....2/100=1/1.1/3.......1/99
SUY RA BẰNG NHAU
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Chứng minh rằng: B= \(\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+...+\frac{1}{100}<\frac{5}{6}\)
