Chứng minh:
\(\dfrac{1}{5}+\dfrac{1}{6}+....+\dfrac{1}{17}>1\)
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Chứng minh:
\(\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+...+\dfrac{1}{17}+\dfrac{1}{18}+\dfrac{1}{19}< 2\)
Chứng minh:
C = \(\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+...+\dfrac{1}{17}< 2\)
1/5<2;1/6<2;...;1/17<2
suy ra 1/5+1/6+1/7+...+1/17<2
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\(\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+...+\dfrac{1}{17}\)
\(=\left(\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{10}\right)+\left(\dfrac{1}{11}+...+\dfrac{1}{17}\right)< \dfrac{1}{5}.6+\dfrac{1}{11}.7=\dfrac{6}{5}+\dfrac{7}{11}\)
\(=1\dfrac{46}{55}< 2\)
\(\Rightarrow dpcm\)
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chứng minh rằng
A= \(\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+...+\dfrac{1}{17}< 2\)
B=\(\dfrac{5}{11}+\dfrac{5}{12}+\dfrac{5}{13}+\dfrac{5}{14},1< B< 2\)
Chứng minh:
a) \(\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+...+\dfrac{1}{17}< 2\)
b) \(\dfrac{1}{101}+\dfrac{1}{102}+...+\dfrac{1}{299}+\dfrac{1}{300}>\dfrac{2}{3}\)
a)
Ta thấy:
\(\dfrac{1}{6}< \dfrac{1}{5}\)
\(\dfrac{1}{7}< \dfrac{1}{5}\)
\(\dfrac{1}{8}< \dfrac{1}{5}\)
\(\dfrac{1}{9}< \dfrac{1}{5}\)
\(\dfrac{1}{11}< \dfrac{1}{10}\)
\(\dfrac{1}{12}< \dfrac{1}{10}\)
\(\dfrac{1}{13}< \dfrac{1}{10}\)
...
\(\dfrac{1}{17}< \dfrac{1}{10}\)
\(\Rightarrow\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+...+\dfrac{1}{17}< 5\cdot\dfrac{1}{5}+8\cdot\dfrac{1}{10}=1+\dfrac{4}{5}=\dfrac{9}{5}< 2\)
Vậy \(\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+...+\dfrac{1}{17}< 2\)
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b)
Ta thấy:
\(\dfrac{1}{101}>\dfrac{1}{300}\)
\(\dfrac{1}{102}>\dfrac{1}{300}\)
\(\dfrac{1}{103}>\dfrac{1}{300}\)
...
\(\dfrac{1}{299}>\dfrac{1}{300}\)
\(\Rightarrow\dfrac{1}{101}+\dfrac{1}{102}+\dfrac{1}{103}+...+\dfrac{1}{300}>200\cdot\dfrac{1}{300}=\dfrac{2}{3}\)
Vậy \(\dfrac{1}{101}+\dfrac{1}{102}+\dfrac{1}{103}+...+\dfrac{1}{300}>\dfrac{2}{3}\)
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Cho S = \(\dfrac{6}{15}+\dfrac{6}{16}+\dfrac{6}{17}+\dfrac{6}{18}+\dfrac{6}{19}\) Chứng minh rằng 1<S<2
Ta có \(\dfrac{6}{15}>\dfrac{6}{16}>...>\dfrac{6}{19}\) nên \(S< \dfrac{6}{15}.5=2\).
Lại có \(S>\dfrac{6}{19}.5>1\) nên \(1< S< 2\)
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a) Chứng minh rằng: \(\dfrac{1}{6}< \dfrac{1}{5^2}+\dfrac{1}{6^2}+\dfrac{1}{7^2}+...+\dfrac{1}{100^2}< \dfrac{1}{4}\)
b) Tìm số nguyên a để: \(\dfrac{2a+9}{a+3}+\dfrac{5a+17}{a+3}-\dfrac{3a}{a+3}\) là số nguyên.
25 tháng 7 2021 lúc 14:43
chứng minh rằng :
a) \(\dfrac{1}{5^2}+\dfrac{1}{6^2}+...+\dfrac{1}{100^2}< \dfrac{1}{4}\) b)\(\dfrac{1}{5^2}+\dfrac{1}{6^5}+...+\dfrac{1}{2013^2}+\dfrac{1}{2014}>\dfrac{1}{5}\)
7 tháng 7 2021 lúc 16:51
Bài 3:
c) Chứng tỏ rằng \(\dfrac{1}{15}+\dfrac{1}{16}+\dfrac{1}{17}+...+\dfrac{1}{43}+\dfrac{1}{44}>\dfrac{5}{6}\)
Giúp mik vs! Thanks nhiều nha!
25 tháng 7 2021 lúc 11:54
chứng minh :
a) \(\dfrac{1}{5^2}+\dfrac{1}{6^2}+...+\dfrac{1}{100^2}>\dfrac{1}{4}\) b) \(\dfrac{1}{5^2}+\dfrac{1}{6^2}+...+\dfrac{1}{2013^2}+\dfrac{1}{2014}>\dfrac{1}{5}\)
