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banana
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Nguyễn Châu Mỹ Linh
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trunghocgiaovien123
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_Shadow_
14 tháng 4 2019 lúc 11:25

\(T=\frac{4}{2.4}+\frac{4}{4.6}+\frac{4}{6.8}+...+\frac{4}{2008.2010}\)

\(T=2.\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{2008.2010}\right)\)

\(T=2.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{2008}-\frac{1}{2010}\right)\)

\(T=2.\left(\frac{1}{2}-\frac{1}{2010}\right)\)

\(T=2.\frac{502}{1005}=\frac{1004}{1005}\)

\(\Rightarrow T=\frac{1004}{1005}\)

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_Shadow_
14 tháng 4 2019 lúc 11:29

\(A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2007.2009}+\frac{1}{2009+2011}\)

\(A=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{2009+2011}\right)\)

\(A=\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2009}-\frac{1}{2011}\right)\)

\(A=\frac{1}{2}.\left(1-\frac{1}{2011}\right)\)

\(A=\frac{1}{2}.\frac{2010}{2011}\)

\(\Rightarrow A=\frac{1005}{2011}\)

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_Shadow_
14 tháng 4 2019 lúc 11:34

\(C=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{100}\right)\)

\(C=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{99}{100}\)

\(C=\frac{1.2.3...99}{2.3.4...100}\)

\(\Rightarrow C=\frac{1}{100}\)

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Duong Thi Nhuong
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Võ Đông Anh Tuấn
12 tháng 11 2016 lúc 10:07

\(A=1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+\frac{1}{4}\left(1+2+3+4\right)+...+\frac{1}{16}\left(1+2+3+...+2016\right)\)

\(A=1+\frac{1}{2}.\frac{\left(1+2\right).2}{2}+\frac{1}{3}.\frac{\left(1+3\right).3}{2}+\frac{1}{4}.\frac{\left(1+4\right).4}{2}+...+\frac{1}{16}.\frac{\left(1+16\right).16}{2}\)

\(A=1+\frac{1}{2}.\frac{3.2}{2}+\frac{1}{3}.\frac{4.3}{2}+\frac{1}{4}.\frac{5.4}{2}+...+\frac{1}{16}.\frac{17.16}{2}\)

\(A=1+\frac{3}{2}+\frac{4}{2}+\frac{5}{2}+...+\frac{17}{2}\)

\(A=\frac{1}{2}.\left(2+3+4+5+...+17\right)\)

\(A=\frac{1}{2}.\frac{\left(2+17\right).16}{2}=19.4=76\)

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Trần Hà Mi
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Đỗ Lê Mỹ Hạnh
4 tháng 1 2017 lúc 22:07

a) \(A=\left(1:\frac{1}{4}\right).4+25\left(1:\frac{16}{9}:\frac{125}{64}\right):\left(-\frac{27}{8}\right)\)

\(=4.4+25.\frac{36}{125}:\frac{-27}{8}\)

\(=16-\frac{32}{15}=\frac{240}{15}-\frac{32}{15}=\frac{208}{15}\)

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Nguyễn Châu Mỹ Linh
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Hatsune Miku
19 tháng 6 2018 lúc 15:39
​29/152323/125/65/4-31/1231/6-13/31087/1801/61/62-67/24
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Phí Nhật Minh
11 tháng 4 2022 lúc 16:57
Ôi mẹ ơi dài khiếp
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 Khách vãng lai đã xóa
Giang Đặng
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Trần Quốc Lộc
29 tháng 10 2017 lúc 10:43

\(A=\dfrac{1}{2}-\left(\dfrac{1}{2}\right)^2-\left(\dfrac{1}{2}\right)^3-...-\left(\dfrac{1}{2}\right)^{10}\\ \\2A=1-\dfrac{1}{2}-\left(\dfrac{1}{2}\right)^2-...-\left(\dfrac{1}{2}\right)^9\\ 2A-A=\left[1-\dfrac{1}{2}-\left(\dfrac{1}{2}\right)^2-...-\left(\dfrac{1}{2}\right)^9\right]-\left[\dfrac{1}{2}-\left(\dfrac{1}{2}\right)^2-\left(\dfrac{1}{2}\right)^3-...-\left(\dfrac{1}{2}\right)^{10}\right]\\ A=1-\dfrac{1}{4}+\left(\dfrac{1}{2}\right)^{10}\)

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Trần Minh Hoàng
29 tháng 10 2017 lúc 14:09

Ta có:

\(A=\dfrac{1}{2}-\left(\dfrac{1}{2}\right)^2-\left(\dfrac{1}{2}\right)^3-...-\left(\dfrac{1}{2}\right)^{10}\)

\(\Rightarrow\dfrac{1}{2}-A=\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3+...+\left(\dfrac{1}{2}\right)^{10}\)

\(\Rightarrow\dfrac{1}{2}\left(\dfrac{1}{2}-A\right)=\left(\dfrac{1}{2}\right)^3+\left(\dfrac{1}{2}\right)^4+...+\left(\dfrac{1}{2}\right)^{11}\)

\(\Rightarrow\left(\dfrac{1}{2}-A\right)-\dfrac{1}{2}\left(\dfrac{1}{2}-A\right)=\left[\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3+...+\left(\dfrac{1}{2}\right)^{10}\right]-\left[\left(\dfrac{1}{2}\right)^3+\left(\dfrac{1}{2}\right)^4+...+\left(\dfrac{1}{2}\right)^{11}\right]\)

\(\Rightarrow\dfrac{1}{2}\left(\dfrac{1}{2}-A\right)=\left(\dfrac{1}{2}\right)^2-\left(\dfrac{1}{2}\right)^{11}\)

\(\Rightarrow\dfrac{1}{2}-A=\dfrac{1}{2}-\left(\dfrac{1}{2}\right)^{10}\)

\(\Rightarrow A=\left(\dfrac{1}{2}\right)^{10}\)

\(\left(\dfrac{1}{2}\right)^{10}\) > 0 nên A > 0

Vậy, A > 0

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Luong Dinh Sy
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Linh Linh
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Đinh Thùy Linh
5 tháng 7 2016 lúc 10:12

\(D=-\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\left(1-\frac{1}{4^2}\right)\cdot...\cdot\left(1-\frac{1}{100^2}\right).\)

\(D=-\frac{2^2-1}{2^2}\cdot\frac{3^2-1}{3^2}\cdot\frac{4^2-1}{4^2}\cdot...\cdot\frac{100^2-1}{100^2}.\)

\(D=-\frac{1\cdot3}{2^2}\cdot\frac{2\cdot4}{3^2}\cdot\frac{3\cdot5}{4^2}\cdot\frac{4\cdot6}{5^2}\cdot...\cdot\frac{98\cdot100}{99^2}\cdot\frac{99\cdot101}{100^2}=-\frac{1}{2}\cdot\frac{101}{100}=-\frac{101}{200}\)

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