Tính:
I=(\(\dfrac{1}{2}\)-\(\dfrac{1}{3}\)).(\(\dfrac{1}{2}\)-\(\dfrac{1}{5}\)).(\(\dfrac{1}{2}\)-\(\dfrac{1}{7}\)). ... .(\(\dfrac{1}{2}\)-\(\dfrac{1}{99}\))
Những câu hỏi liên quan
Chứng minh \(\dfrac{1}{5}\)< \(\dfrac{1}{4^2}\)+\(\dfrac{1}{5^2}\)+\(\dfrac{1}{6^2}\)+\(\dfrac{1}{7^2}\)+......+\(\dfrac{1}{99^2}\)+\(\dfrac{1}{100^2}\)<\(\dfrac{1}{3}\)
\(\dfrac{1}{4^2}+\dfrac{1}{5^2}+...+\dfrac{1}{100^2}>\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+...+\dfrac{1}{100\cdot101}=\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{100}-\dfrac{1}{101}=\dfrac{1}{4}-\dfrac{1}{101}>\dfrac{1}{4}-\dfrac{1}{20}=\dfrac{1}{5}\left(1\right)\)
\(\dfrac{1}{4^2}+\dfrac{1}{5^2}+...+\dfrac{1}{100^2}< \dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+...+\dfrac{1}{99\cdot100}=\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{100}=\dfrac{1}{3}-\dfrac{1}{100}< \dfrac{1}{3}\left(2\right)\) Từ (1) và (2) \(\Rightarrow\dfrac{1}{5}< \dfrac{1}{4^2}+\dfrac{1}{5^2}+...+\dfrac{1}{99^2}+\dfrac{1}{100^2}< \dfrac{1}{3}\)
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9 tháng 5 2021 lúc 20:08
\(\dfrac{1}{99}+\dfrac{2}{98}+\dfrac{3}{97}+...+\dfrac{99}{1}\)
\(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{100}\)
94-\(\dfrac{1}{7}-\dfrac{2}{8}-\dfrac{3}{9}-...-\dfrac{94}{100}\)
\(\dfrac{1}{35}+\dfrac{1}{40}+\dfrac{1}{45}+...+\dfrac{1}{500}\)
giúp mik nha mik cần gấp
Mình làm được một câu thôi, bạn dựa vào làm nha!
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11 tháng 4 2021 lúc 15:06
text{Bài 4. Chứng tỏ rằng:}a) dfrac{1}{2^2}+dfrac{1}{3^2}+dfrac{1}{4^2}+...+dfrac{1}{30^2} 1b) dfrac{1}{10}+dfrac{1}{11}+dfrac{1}{12}+...+dfrac{1}{99}+dfrac{1}{100}1c) dfrac{1}{5}+dfrac{1}{6}+dfrac{1}{7}+...+dfrac{1}{17} 2d) dfrac{1}{1.2}+dfrac{1}{2.3}+dfrac{1}{3.4}+...+dfrac{1}{29.30} 1
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\(\text{Bài 4. Chứng tỏ rằng:}\)
\(a\)) \(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{30^2}< 1\)
\(b\)) \(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}+...+\dfrac{1}{99}+\dfrac{1}{100}>1\)
\(c\)) \(\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+...+\dfrac{1}{17}< 2\)
\(d\)) \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{29.30}< 1\)
a)
\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{30^2}\\ < \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{29.30}\\ =1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{29}-\dfrac{1}{30}\\ =1-\dfrac{1}{30}=\dfrac{29}{30}< 1\left(dpcm\right)\)
b)
\(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}+...+\dfrac{1}{99}+\dfrac{1}{100}=\dfrac{1}{10}+\left(\dfrac{1}{11}+\dfrac{1}{12}+...+\dfrac{1}{99}+\dfrac{1}{100}\right)\\ >\dfrac{1}{10}+\dfrac{1}{100}+\dfrac{1}{100}+...+\dfrac{1}{100}=\dfrac{1}{10}+\dfrac{90}{100}\\ =\dfrac{110}{100}>1\left(đpcm\right).\)
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c)
\(\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+...+\dfrac{1}{17}\\ =\left(\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{9}\right)+\left(\dfrac{1}{10}+\dfrac{1}{11}+...+\dfrac{1}{17}\right)\\ < \dfrac{1}{5}.5+\dfrac{1}{8}.8=1+1=2\left(đpcm\right)\)
d) tương tự câu 1
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\(\dfrac{\left(13\dfrac{1}{4}-2\dfrac{5}{27}-10\dfrac{5}{6}\right).230\dfrac{1}{25}+46\dfrac{3}{4}}{\left(1\dfrac{3}{10}+\dfrac{10}{3}\right):\left(12\dfrac{1}{3}-14\dfrac{2}{7}\right)}\)
\(\dfrac{\left(1+2+3+...+99+100\right)\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{9}\right)\left(63.1,2-21.3,6\right)}{1-2+3-4+.....+99-100}\)
Tính giá trị biểu thức :
\(\dfrac{1}{2}-\left(\dfrac{1}{3}+\dfrac{2}{3}\right)+\left(\dfrac{1}{4}+\dfrac{2}{4}+\dfrac{3}{4}\right)-\left(\dfrac{1}{5}+\dfrac{2}{5}+\dfrac{3}{5}+\dfrac{4}{5}\right)+\left(\dfrac{1}{6}+\dfrac{2}{6}+\dfrac{3}{6}+\dfrac{4}{6}+\dfrac{5}{6}\right)-\left(\dfrac{1}{7}+\dfrac{2}{7}+\dfrac{3}{7}+\dfrac{4}{7}+\dfrac{5}{7}+\dfrac{6}{7}\right)+...+\left(100+...+\dfrac{99}{100}\right)\)
11 tháng 5 2022 lúc 7:54
Chứng minh rằng:
\(\dfrac{1}{3^2}+\dfrac{1}{5^2}+\dfrac{1}{7^2}+...+\dfrac{1}{99^2}< \dfrac{5}{18}\)
20 tháng 3 2021 lúc 16:09
Giúp mk vớiCâu 1:Cho A dfrac{1}{dfrac{99}{dfrac{1}{2}+}}+dfrac{2}{dfrac{98}{dfrac{1}{3}+}}+dfrac{3}{dfrac{97}{dfrac{1}{4}+....}}+...+dfrac{99}{dfrac{1}{dfrac{1}{100}}}. B dfrac{92}{dfrac{1}{45}+}-dfrac{1}{dfrac{9}{dfrac{1}{50}+}}-dfrac{2}{dfrac{10}{dfrac{1}{55}+}}-dfrac{3}{dfrac{11}{dfrac{1}{60}+....}}-...dfrac{92}{dfrac{100}{dfrac{1}{500}}}. Tính dfrac{A}{B}
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Giúp mk với
Câu 1:
Cho A = \(\dfrac{1}{\dfrac{99}{\dfrac{1}{2}+}}+\dfrac{2}{\dfrac{98}{\dfrac{1}{3}+}}+\dfrac{3}{\dfrac{97}{\dfrac{1}{4}+....}}+...+\dfrac{99}{\dfrac{1}{\dfrac{1}{100}}}\).
B =\(\dfrac{92}{\dfrac{1}{45}+}-\dfrac{1}{\dfrac{9}{\dfrac{1}{50}+}}-\dfrac{2}{\dfrac{10}{\dfrac{1}{55}+}}-\dfrac{3}{\dfrac{11}{\dfrac{1}{60}+....}}-...\dfrac{92}{\dfrac{100}{\dfrac{1}{500}}}\). Tính \(\dfrac{A}{B}\)
CMR:a)\(\dfrac{1}{3}< \dfrac{1}{11}+\dfrac{1}{12}+\dfrac{1}{13}+....+\dfrac{1}{30}< \dfrac{5}{2}\)
b)\(\dfrac{1}{5}< \dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{4}-\dfrac{1}{5}+.....-\dfrac{1}{99}< \dfrac{2}{5}\)
c)\(\dfrac{1}{15}< \dfrac{1}{2}.\dfrac{3}{4}......\dfrac{99}{100}< \dfrac{1}{10}\)
T làm biếng lắm; làm C thôi
\(A=\dfrac{1}{2}.\dfrac{3}{4}.\dfrac{5}{6}...\dfrac{99}{100}\\ \Rightarrow A< \dfrac{2}{3}.\dfrac{4}{5}.\dfrac{6}{7}...\dfrac{100}{101}\\ \Rightarrow A^2< \left(\dfrac{1}{2}.\dfrac{3}{4}.\dfrac{5}{6}...\dfrac{99}{100}\right).\left(\dfrac{2}{3}.\dfrac{4}{5}.\dfrac{6}{7}...\dfrac{100}{101}\right)\\ =\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}.\dfrac{4}{5}...\dfrac{99}{100}.\dfrac{100}{101}\\ =\dfrac{1}{101}< \dfrac{1}{100}\\ \Rightarrow A< \dfrac{1}{10}\)
Làm tương tự ta được A > 1/15
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câu a
\(A=\dfrac{1}{11}+\dfrac{1}{12}+...+\dfrac{1}{30}>\dfrac{20}{30}=\dfrac{2}{3}>\dfrac{1}{3}\)
\(A=\left(\dfrac{1}{11}+..+\dfrac{1}{15}\right)+\left(\dfrac{1}{16}+...+\dfrac{1}{30}\right)< 5.\dfrac{1}{10}+25.\dfrac{1}{15}=\dfrac{1}{2}+\dfrac{5}{3}=\dfrac{8}{6}=\dfrac{4}{3}< \dfrac{5}{2}\)
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Xem thêm câu trả lời
K =\(\dfrac{1}{2}-\dfrac{1}{2^3}-\dfrac{1}{2^5}-\dfrac{1}{2^7}-.........-\dfrac{1}{2^{99}}\)
4K= \(2-\dfrac{1}{2}-\dfrac{1}{2^3}-\dfrac{1}{2^5}-.....-\dfrac{1}{2^{97}}\)
4K - K = \(2-\dfrac{1}{2^{99}}\)
3K= \(2-\dfrac{1}{2^{99}}\)
K=\(\left(2-\dfrac{1}{2^{99}}\right):3\)
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\(4K=2-\dfrac{1}{2}-\dfrac{1}{2^5}-...-\dfrac{1}{2^{97}}\)
\(4K-K=2-\dfrac{1}{2^{99}}\)
\(3K=2-\dfrac{1}{2^{90}}\)
\(K=\left(2-\dfrac{1}{2^{90}}\right):3\)
\(K=\dfrac{2}{3}\)
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