Chứng minh rằng: \(\dfrac{1}{5}+\dfrac{1}{15}+\dfrac{1}{25}+.....+\dfrac{1}{1985}< \dfrac{9}{20}\)
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Chứng minh rằng : \(\dfrac{1}{5}+\dfrac{1}{15}+\dfrac{1}{25}+...+\dfrac{1}{1985}< \dfrac{9}{20}\)
Chứng minh rằng : \(\dfrac{1}{5}+\dfrac{1}{15} +\dfrac{1}{25}+...+\dfrac{1}{1985}<\dfrac{9}{20}\)
23 tháng 4 2023 lúc 22:20
\(\dfrac{1}{5}\)+\(\dfrac{1}{13}\)+\(\dfrac{1}{25}\)+...+\(\dfrac{1}{10^2}\)+\(\dfrac{1}{11^2}\)< \(\dfrac{9}{20}\)
Chứng tỏ rằng biểu thức trên bé hơn 9/20
Chứng minh rằng với mọi số tự nhiên n:
\(\dfrac{1}{5}+\dfrac{1}{13}+\dfrac{1}{25}...+\dfrac{1}{n^2+\left(n+1\right)^2}< \dfrac{9}{20}\)
\(a^2+\left(a+1\right)^2=a^2+a^2+2a+1\\ =2a^2+2a+1>2a\left(a+1\right)\\ \Rightarrow\dfrac{1}{a^2+\left(a+1\right)^2}< \dfrac{1}{2a\left(a+1\right)}\)
\(\dfrac{1}{5}+\dfrac{1}{13}+\dfrac{1}{25}+...+\dfrac{1}{n^2+\left(n+1\right)^{^2}}\\ =\dfrac{1}{1^2+2^2}+\dfrac{1}{2^2+3^2}+\dfrac{1}{3^2+4^2}+...+\dfrac{1}{n^2+\left(n+1\right)^2}\\ < \dfrac{1}{2.1.\left(1+2\right)}+\dfrac{1}{2.2\left(2+1\right)}+....+\dfrac{1}{2n\left(n+1\right)}\\ =\dfrac{1}{2}\left(\dfrac{1}{3}+\dfrac{1}{2.3}+...+\dfrac{1}{n\left(n+1\right)}\right)\\ =\dfrac{1}{2}\left(\dfrac{1}{3}+\dfrac{1}{2}-\dfrac{1}{n+1}\right)\\ =\dfrac{1}{2}\left(\dfrac{5}{6}-\dfrac{1}{n+1}\right)\\ =\dfrac{5}{12}-\dfrac{1}{2n+2}< \dfrac{5}{12}< \dfrac{9}{20}\)
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Cho A = \(\dfrac{\left(3\dfrac{2}{15}+\dfrac{1}{5}\right):2\dfrac{1}{2}}{\left(5\dfrac{3}{7}-2\dfrac{1}{4}\right):4\dfrac{43}{56}}\) ; B = \(\dfrac{1,2:\left(1\dfrac{1}{5}.1\dfrac{1}{4}\right)}{0,32+\dfrac{2}{25}}\)
Chứng minh rằng A= B
\(A=\dfrac{\left(3+\dfrac{2}{15}+\dfrac{1}{5}\right):\dfrac{5}{2}}{\left(5+\dfrac{3}{7}-2-\dfrac{1}{4}\right):\left(4+\dfrac{43}{56}\right)}\)
\(=\dfrac{\dfrac{10}{3}\cdot\dfrac{2}{5}}{\dfrac{89}{28}:\dfrac{267}{56}}=\dfrac{4}{3}:\dfrac{2}{3}=2\)
\(B=\dfrac{\dfrac{6}{5}:\left(\dfrac{6}{5}\cdot\dfrac{5}{4}\right)}{\dfrac{8}{25}+\dfrac{2}{25}}=\dfrac{\dfrac{6}{5}:\dfrac{3}{2}}{\dfrac{2}{5}}=2\)
Do đó: A=B
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Chứng minh rằng:
a) \(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{2}{63}>2\)
b) \(\dfrac{3}{9\cdot14}+\dfrac{3}{14\cdot19}+\dfrac{3}{19\cdot24}+...+\dfrac{3}{\left(5n-1\right)\left(5n+4\right)}< \dfrac{1}{15}\)
c) \(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{9^2}\). Chứng minh rằng : \(\dfrac{2}{5}< A< \dfrac{8}{9}\)
Câu a :
Chưa nghĩ ra! Sorry nhé!!
Câu b :
Câu hỏi của Trần Thùy Linh - Toán lớp 6 | Học trực tuyến
Câu c :
Câu hỏi của Trần Thùy Linh - Toán lớp 6 | Học trực tuyến
Vào link đó mà xem, t ngại chép lại
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Bài 1:
a, Tính \(\dfrac{5.4^{15}.9^{9} - 4.3^{20}.8^{9}}{5.2^{9}.6^{19}-7.2^{29}.27^{6}}\)
b,Tìm x biết:
\(1\dfrac{1}{30}:(24\dfrac{1}{6}-24\dfrac{1}{5}) -\dfrac{1\dfrac{1}{2}-\dfrac{3}{4}}{4x-\dfrac{1}{2}}=(-1\dfrac{1}{15}):(8\dfrac{1}{5}-8\dfrac{1}{3})\)
Bài 2: Chứng minh số:
222...22200333...333(2001c/s 2; 2003 c/s3)
4 tháng 8 2017 lúc 16:29
Bài 1:
\(a)5\dfrac{1}{20}+\dfrac{8}{9}-\dfrac{15}{25}+\dfrac{75}{-18}\)
\(b)\dfrac{1}{15}+\dfrac{1}{35}+\dfrac{1}{63}+\dfrac{1}{99}+\dfrac{1}{143}\)
\(c)\dfrac{6}{7}+\dfrac{5}{7}:5-\dfrac{8}{9}.\dfrac{6}{7}+\dfrac{5}{8}:5-\dfrac{3}{16}.\left(-2\right)^2\)
a)\(=\dfrac{211}{180}\)
b)\(=\dfrac{5}{39}\)
c)=\(=-\dfrac{65}{168}\)
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6 tháng 3 2021 lúc 12:52
Cho A = \(\dfrac{1}{5^2}+\dfrac{1}{6^2}+\dfrac{1}{7^2}+.....+\dfrac{1}{2019^2}\)
Chứng minh rằng \(\dfrac{20}{101}< A< \dfrac{1}{4}\)
\(A=\dfrac{1}{5^2}+\dfrac{1}{6^2}+\dfrac{1}{7^2}+...+\dfrac{1}{2019^2}>\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}+\dfrac{1}{7\cdot8}+...+\dfrac{1}{2019\cdot2020}=\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+...+\dfrac{1}{2019}-\dfrac{1}{2020}=\dfrac{1}{5}-\dfrac{1}{2020}=\dfrac{404-1}{2020}=\dfrac{403}{2020}>\dfrac{40}{2020}=\dfrac{20}{101}\left(1\right)\) \(A=\dfrac{1}{5^2}+\dfrac{1}{6^2}+\dfrac{1}{7^2}+...+\dfrac{1}{2019^2}< \dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}+...+\dfrac{1}{2018\cdot2019}=\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{2018}-\dfrac{1}{2019}=\dfrac{1}{4}-\dfrac{1}{2019}=\dfrac{2019-4}{4\cdot2019}=\dfrac{2015}{4\cdot2019}< \dfrac{2019}{4\cdot2019}=\dfrac{1}{4}\left(2\right)\) Từ (1) và (2) \(\Rightarrow\dfrac{20}{101}< A< \dfrac{1}{4}\)
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