chứng minh \(\frac{1}{3^3}+\frac{1}{5^3}+\frac{1}{7^3}+.........+\frac{1}{2019^3}< \frac{1}{12}\)
Những câu hỏi liên quan
chứng minh \(\frac{1}{3^2}+\frac{1}{5^2}+\frac{1}{7^2}+.....+\frac{1}{2019^2}< \frac{1}{12}\)
\(\frac{21}{36}\cdot\frac{5}{2}-\frac{7}{12}\cdot\frac{2}{7}+\left(2018-2019\right)^0\)
\(-12\frac{1}{3}-\frac{5}{7}+7\frac{1}{3}+1\frac{5}{7}+1^{2019}\)
GIUP MINH VOI!!!!!!
\(a,\frac{21}{36}.\frac{5}{2}-\frac{7}{12}.\frac{2}{7}+\left(2018-2019\right)^0\)
=\(\frac{7}{12}.\frac{5}{2}-\frac{7}{12}.\frac{2}{7}+\left(-1\right)\)
= \(\frac{7}{12}.\left(\frac{5}{2}+\frac{2}{7}\right)+\left(-1\right)\)
=\(\frac{7}{12}.\frac{39}{14}+\left(-1\right)\)
=\(\frac{13}{8}+\left(-1\right)\)
= \(\frac{5}{8}\)
\(b,-12\frac{1}{3}-\frac{5}{7}+7\frac{1}{3}+1\frac{5}{7}+1^{2019}\)
=\(-\frac{37}{3}+\frac{-5}{7}+\frac{22}{3}+\frac{12}{7}+1\)
=\(\left(\frac{-37+22}{3}\right)+\left(\frac{-5+12}{7}\right)+=1\)
= \(-5+1+1\)
=\(-3\)
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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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bài 1: tính A:=\(\frac{1}{2}-\frac{2}{3}+\frac{3}{4}-\frac{4}{5}+\frac{5}{6}-\frac{6}{7}-\frac{5}{6}+\frac{4}{5}-\frac{3}{4}+\frac{2}{3}-\frac{2}{3}-\frac{1}{2}\)
Bài 2: Cho B=\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+.....+\frac{1}{49}-\frac{1}{50}\)
Chứng minh rằng: \(\frac{7}{12}< A< \frac{5}{6}\)
Cho A=\(\frac{1}{3^2}+\frac{1}{5^5}+\frac{1}{7^2}+...+\frac{1}{2019^2}\)
Chứng minh A<\(\frac{1}{2}\)
Ta có :
\(A=\frac{1}{3^2}+\frac{1}{5^2}+...+\frac{1}{2019^2}< \frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{2018.2020}\)
Cho \(S=\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{2018.2020}\)
\(\Rightarrow S=\frac{1}{2}\left(\frac{2}{2.4}+\frac{2}{4.6}+...+\frac{2}{2018.2020}\right)\)
\(\Leftrightarrow S=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{2018}-\frac{1}{2020}\right)\)
\(\Leftrightarrow S=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{2020}\right)=\frac{1009}{4040}< \frac{1}{2}\)
Mà A < S ⇒ đpcm
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1.chứng minh rằng Afrac{1}{16} biết Afrac{1}{5^2}+frac{2}{5^3}+frac{3}{5^4}+.....+frac{99}{5^{100}}
2.tính (M-N)^3 biết:
M1-frac{1}{2}+frac{1}{3}-frac{1}{4}+.....+frac{1}{2017}-frac{1}{2018}+frac{1}{2019}
Nfrac{1}{1010}+frac{1}{1011}+.....+frac{1}{2019}
Đọc tiếp
1.chứng minh rằng A<\(\frac{1}{16}\) biết A=\(\frac{1}{5^2}+\frac{2}{5^3}+\frac{3}{5^4}+.....+\frac{99}{5^{100}}\)
2.tính (M-N)\(^3\) biết:
M=1-\(\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{2017}-\frac{1}{2018}+\frac{1}{2019}\)
N=\(\frac{1}{1010}+\frac{1}{1011}+.....+\frac{1}{2019}\)
\(5A=\frac{1}{5}+\frac{2}{5^2}+\frac{3}{5^3}+...+\frac{99}{5^{99}}\)
\(A=\frac{1}{5^2}+\frac{2}{5^3}+\frac{3}{5^4}+...+\frac{99}{5^{100}}\)
\(\Rightarrow4A=5A-A=\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{99}}-\frac{99}{5^{100}}\)
Đặt \(B=\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{99}}\)
Khi đó \(4A=B-\frac{99}{5^{100}}< B\)
\(5B=1+\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{98}}\)
\(B=\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{98}}+\frac{1}{5^{99}}\)
\(\Rightarrow4B=5B-B=1-\frac{1}{5^{99}}\)
\(\Rightarrow B=\frac{1}{4}-\frac{1}{4\cdot5^{99}}< \frac{1}{4}\)
\(\Rightarrow4A < B\Rightarrow4A< \frac{1}{4}\)
\(\Rightarrow A< \frac{1}{16}\) ( đpcm )
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2. \(M=\left(1+\frac{1}{3}+...+\frac{1}{2019}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2018}\right)\)
\(M=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}+\frac{1}{2019}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2018}\right)\)
\(M=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2019}\right)-\left(1+\frac{1}{2}+...+\frac{1}{1009}\right)\)
\(M=\frac{1}{1010}+\frac{1}{1011}+...+\frac{1}{2019}\)
\(\Rightarrow\left(M-N\right)^3=0\)
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16 tháng 3 2018 lúc 21:01
ai giúp mình bài này :
Chứng minh rằng :
\(\frac{1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{4035}}{1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}}>\frac{2019}{4036}\)
Nguyễn Văn Anh Kiệt
CTV thì vẫn đc hỏi!! Chỉ những thằng não ngắn mới nghĩ như vậy~~
ko lm đc thì ra chỗ khác cho ng` giỏi làm =))
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Chứng minh rằng: \(B=\frac{1}{2^3}+\frac{1}{3^3}+...+\frac{1}{2019^3}< \frac{1}{2^2}\)
+ \(n^3=n\cdot n^2>n\left(n^2-1\right)\)
\(\Rightarrow n^3>n\left(n^2+n-n-1\right)\)
\(\Rightarrow n^3>n\left[n\left(n+1\right)-\left(n-1\right)\right]\)
\(\Rightarrow n^3>n\left(n-1\right)\left(n+1\right)\)\(\Rightarrow\frac{1}{n^3}< \frac{1}{\left(n-1\right)n\left(n+1\right)}\)
\(\Rightarrow\frac{1}{n^3}< \frac{1}{2}\left[\frac{\left(n+1\right)-\left(n-1\right)}{\left(n-1\right)n\left(n+1\right)}\right]=\frac{1}{2}\left(\frac{1}{\left(n-1\right)n}-\frac{1}{n\left(n+1\right)}\right)\)
Do đó : \(B< \frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+...+\frac{1}{2018\cdot2019\cdot2020}\)
\(\Rightarrow B< \frac{1}{2}\left(\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+...+\frac{1}{2018\cdot2019}-\frac{1}{2019\cdot2020}\right)\)
\(\Rightarrow B< \frac{1}{2}\left(\frac{1}{2}-\frac{1}{2019\cdot2020}\right)< \frac{1}{4}\)
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CHO A = \(\frac{1}{3}+\frac{1}{^{3^2}}+\frac{1}{3^3}+...+\frac{1}{^{3^{2018}}}+\frac{1}{3^{2019}}\)
Chứng minh A<\(\frac{1}{2}\)
