Chứng minh:\(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+...+\frac{1}{99}+\frac{1}{100}>1\)
Những câu hỏi liên quan
Chứng minh :
a) frac{1}{3}-frac{2}{3^2}+frac{3}{3^3}-frac{4}{3^4}+...+frac{99}{3^{99}}-frac{100}{3^{100}} frac{3}{16} frac{1}{3}-frac{2}{3^2}+frac{3}{3^3}+frac{4}{4^4}+...+frac{99}{3^{99}}-frac{100}{3^{100}} frac{3}{16}
b)frac{1}{41}+frac{1}{42}+frac{1}{43}+...+frac{1}{79}+frac{1}{80} frac{7}{12}
c) Cho Sfrac{3}{10}+frac{3}{11}+frac{3}{12}+frac{3}{13}+frac{3}{14}
Chứng minh 1 S 2
Đọc tiếp
Chứng minh :
a) \(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+...+\frac{99}{3^{99}}-\frac{100}{3^{100}}< \frac{3}{16}\) \(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}+\frac{4}{4^4}+...+\frac{99}{3^{99}}-\frac{100}{3^{100}}< \frac{3}{16}\)
b)\(\frac{1}{41}+\frac{1}{42}+\frac{1}{43}+...+\frac{1}{79}+\frac{1}{80}< \frac{7}{12}\)
c) Cho \(S=\frac{3}{10}+\frac{3}{11}+\frac{3}{12}+\frac{3}{13}+\frac{3}{14}\)
Chứng minh \(1< S< 2\)
Chứng tỏ:
\(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+...+\frac{1}{99}+\frac{1}{100}>1\)
cho A=\(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+...+\frac{1}{99}+\frac{1}{100}\) chứng tỏ A>1
Ta có:
\(\frac{1}{10}+\frac{1}{11}+...+\frac{1}{49}>\frac{1}{50}+\frac{1}{50}+...+\frac{1}{50}=\frac{40}{50}=\frac{4}{5}\)
\(\frac{1}{50}+\frac{1}{51}+...+\frac{1}{99}>\frac{1}{100}+\frac{1}{100}+...+\frac{1}{100}=\frac{50}{100}=\frac{1}{2}\)
Từ đây ta suy ra
A > \(\frac{4}{5}+\frac{1}{2}+\frac{1}{100}=1,31>1\)
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Cho tổng A =\(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+...+\frac{1}{99}+\frac{1}{100}\)
Chứng tỏ A > 1
Cho tổng A =\(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+...+\frac{1}{99}+\frac{1}{100}\)
Chứng tỏ A > 1
30 số hạng đầu lớn hơn 1
\(\frac{1}{10}+\frac{1}{11}+..+\frac{1}{19}>\frac{1}{20}+\frac{1}{20}+..+\frac{1}{20}=\frac{1}{2}\)\(\frac{1}{2}\)
\(\frac{1}{20}+\frac{1}{21}+..+\frac{1}{29}>\frac{1}{30}+\frac{1}{30}+..+\frac{1}{30}=\frac{1}{3}\)
\(\frac{1}{30}+\frac{1}{31}+...+\frac{1}{39}>\frac{1}{40}+\frac{1}{40}+...+\frac{1}{40}=\frac{1}{4}\)
=> \(\frac{1}{10}+\frac{1}{11}+...+\frac{1}{39}>\frac{1}{2}+\frac{1}{3}+\frac{1}{4}=\frac{13}{12}>1\)
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Cho \(A=\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+...+\frac{1}{99}+\frac{1}{100}\)
Chứng tỏ rằng A > 1
\(A=\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+...+\frac{1}{99}+\frac{1}{100}\)
\(A=\frac{1}{10}+\left(\frac{1}{11}+\frac{1}{12}+...+\frac{1}{99}+\frac{1}{100}\right)>\frac{1}{10}+\left(\frac{1}{100}+\frac{1}{100}+...+\frac{1}{100}\right)\)
\(A=\frac{1}{10}+\frac{99}{100}=1\)
=> A > 1
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\(A=\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+...+\frac{1}{99}+\frac{1}{100}\)
\(A=\frac{1}{10}+\frac{1}{11}+...+\frac{1}{19}>\frac{1}{20}+\frac{1}{20}+...+\frac{1}{20}=\frac{10}{20}=\frac{1}{2}\)
\(A=\frac{1}{20}+\frac{1}{21}+...+\frac{1}{29}>\frac{1}{30}+\frac{1}{30}+...+\frac{1}{30}=\frac{10}{30}=\frac{1}{3}\)
\(A=\frac{1}{30}+\frac{1}{31}+...+\frac{1}{39}>\frac{1}{40}+\frac{1}{40}+... +\frac{1}{40}=\frac{10}{40}=\frac{1}{4}\)
\(\Rightarrow A>1\)
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Ta thấy:1/10;1/11;1/12;1/13;...;1/99>1/100
=)1/10+1/11+1/12+1/13+...+1/100>1/100+1/100+1/100+1/100..+1/100=1/100.100=1
Vậy A>1
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22 tháng 6 2017 lúc 15:18
Cho tổng A=\(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+...\frac{1}{99}+\frac{1}{100}\)
Chứng tỏ rằng A > 1
\(A=\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+...+\frac{1}{100}\)
\(=\frac{1}{10}+\left(\frac{1}{11}+\frac{1}{12}+...+\frac{1}{99}+\frac{1}{100}\right)>\frac{1}{10}+\left(\frac{1}{100}+\frac{1}{100}+...+\frac{1}{100}\right)\)
\(=\frac{1}{10}+\frac{90}{100}>1\)
\(A>1\left(đpcm\right)\)
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a) Cho Sfrac{1}{31}+frac{1}{32}+frac{1}{33}+frac{1}{60}
Chứng minh frac{3}{5} S frac{4}{5}
b) Chứng minh frac{1}{41}+frac{1}{42}+frac{1}{43}+......+frac{1}{100}frac{7}{10}
c) Chứng minh frac{3}{10}+frac{3}{11}+frac{3}{12}+frac{3}{13}+frac{3}{14} không là số tự nhiên
d) Chứng minh frac{1}{15} D frac{1}{10}với Dfrac{1}{2}.frac{3}{4}.frac{5}{6}.....frac{99}{100}
Đọc tiếp
a) Cho \(S=\frac{1}{31}+\frac{1}{32}+\frac{1}{33}+\frac{1}{60}\)
Chứng minh \(\frac{3}{5}< S< \frac{4}{5}\)
b) Chứng minh \(\frac{1}{41}+\frac{1}{42}+\frac{1}{43}+......+\frac{1}{100}>\frac{7}{10}\)
c) Chứng minh \(\frac{3}{10}+\frac{3}{11}+\frac{3}{12}+\frac{3}{13}+\frac{3}{14}\) không là số tự nhiên d) Chứng minh \(\frac{1}{15}< D< \frac{1}{10}với\) \(D=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}.....\frac{99}{100}\)
Bạn tham khảo ở link này nhé :
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Chứng tỏ tổng sau lớn hơn 1
\(C=\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+...+\frac{1}{99}+\frac{1}{100}\)
Ta có : \(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+....+\frac{1}{19}>\frac{1}{20}+\frac{1}{20}+\frac{1}{20}+...+\frac{1}{20}=\frac{1}{20}.10=\frac{1}{2}\) ( 10 số hạng 1/20)
\(\frac{1}{21}+\frac{1}{22}+\frac{1}{23}+....+\frac{1}{29}>\frac{1}{30}+\frac{1}{30}+\frac{1}{30}+...+\frac{1}{30}=\frac{1}{30}.10=\frac{1}{3}\) ( 10 số hạng 1/30 )
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\(\frac{1}{90}+\frac{1}{91}+...+\frac{1}{99}>\frac{1}{100}+\frac{1}{100}+...+\frac{1}{100}=\frac{1}{100}.10=\frac{1}{10}\). Và: \(\frac{1}{100}=\frac{1}{100}\)
Nên: \(C=\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+...+\frac{1}{100}>\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{10}+\frac{1}{100}>1\) (đpcm)
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Ta có:
\(\frac{1}{10}+\frac{1}{11}+...+\frac{1}{19}>\frac{1}{20}+\frac{1}{20}+...+\frac{1}{20}=\frac{10}{20}=\frac{1}{2}\)
\(\frac{1}{20}+\frac{1}{21}+...+\frac{1}{19}>\frac{1}{30}+\frac{1}{30}+...+\frac{1}{30}=\frac{10}{30}=\frac{1}{3}\)
\(\frac{1}{30}+\frac{1}{31}+...+\frac{1}{39}>\frac{1}{40}+\frac{1}{40}+...+\frac{1}{40}+\frac{10}{40}+\frac{1}{4}\)
\(=>\frac{1}{10}+\frac{1}{11}+...+\frac{1}{39}>\frac{1}{2}+\frac{1}{3}+\frac{1}{4}=\frac{13}{12}>1\)
Vậy \(C>1\)
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