Giúp nhanh cho mình với !!!!!!!
F=\(\frac{1}{0,1x2}+\frac{1}{0,2x3}+\frac{1}{0,3x4}+...+\frac{1}{9,9x100}\)
Ai nhanh mình tick cho !!!!!!!!!
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
A
21 tháng 6 2018
\(A=\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}+\frac{1}{512}\)
\(=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+....+\frac{1}{256}-\frac{1}{512}\)
\(=\frac{1}{2}-\frac{1}{512}\)
\(=\frac{255}{512}\)
Vậy \(A=\frac{255}{512}\)
BD
21 tháng 6 2018
A=14 +18 +116 +132 +164 +1128 +1256 +1512
=12 −14 +14 −18 +....+1256 −1512
=12 −1512
=255512
Vậy A=255512
Phạm Long Khánh
8 tháng 2 2019
\(\frac{1}{18}+\frac{1}{54}+\frac{1}{108}+...+\frac{1}{990}\)
\(=\frac{1}{3\cdot6}+\frac{1}{6\cdot9}+\frac{1}{9\cdot12}+...+\frac{1}{30\cdot33}\)
\(=\frac{1}{3}\left(\frac{3}{3\cdot6}+\frac{3}{6\cdot9}+\frac{3}{9\cdot12}+...+\frac{3}{30\cdot33}\right)\)
\(=\frac{1}{3}\left(\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+\frac{1}{9}-\frac{1}{12}+...+\frac{1}{30}-\frac{1}{33}\right)\)
\(=\frac{1}{3}\left(\frac{1}{3}-\frac{1}{33}\right)\)
\(=\frac{1}{3}\cdot\frac{10}{33}=\frac{10}{99}\)
5 tháng 7 2019
\(\frac{1}{18}+\frac{1}{54}+\frac{1}{108}+...+\frac{1}{990}\)
\(=\frac{1}{3.6}+\frac{1}{6.9}+\frac{1}{9.12}+...+\frac{1}{30.33}\)
\(=\frac{1}{3}\left(\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+\frac{1}{9}-\frac{1}{12}+...+\frac{1}{30}-\frac{1}{33}\right)\)
\(=\frac{1}{3}\left(\frac{1}{3}-\frac{1}{33}\right)\)
\(=\frac{1}{3}.\frac{10}{33}\)
\(=\frac{10}{99}\)
9 tháng 7 2021
Đặt \(A=\frac{1}{5}+\frac{1}{5^3}+...+\frac{1}{5^{101}}\)
\(\Rightarrow25A=5+\frac{1}{5}+\frac{1}{5^3}+...+\frac{1}{5^{99}}\)
\(\Rightarrow25A-A=\left(5+\frac{1}{5}+\frac{1}{5^3}+...+\frac{1}{5^{99}}\right)-\left(\frac{1}{5}+\frac{1}{5^3}+\frac{1}{5^5}+...+\frac{1}{5^{101}}\right)\)
hay \(24A=5-\frac{1}{5^{101}}\)
\(\Rightarrow A=\frac{5-\frac{1}{5^{101}}}{24}\)
\(\Rightarrow A:\left(1-\frac{1}{5^{102}}\right)=\frac{5-\frac{1}{5^{101}}}{24}.\frac{1}{1-\frac{1}{5^{102}}}\)
\(=\frac{5\left(1-\frac{1}{5^{102}}\right)}{24}.\frac{1}{1-\frac{1}{5^{102}}}=\frac{5}{24}\)
XO
9 tháng 7 2021
Đặt S = \(\frac{1}{7^2}+\frac{1}{7^4}+\frac{1}{7^6}+...+\frac{1}{7^{100}}\)
=> 72S = 49S = \(1+\frac{1}{7^2}+\frac{1}{7^4}+...+\frac{1}{7^{98}}\)
=> 49S - S = \(\left(1+\frac{1}{7^2}+\frac{1}{7^4}+...+\frac{1}{7^{98}}\right)-\left(\frac{1}{7^2}+\frac{1}{7^4}+\frac{1}{7^6}+...+\frac{1}{7^{100}}\right)\)
=> 48S = \(1-\frac{1}{7^{100}}\)
=> \(S=\frac{1-\frac{1}{7^{100}}}{48}\)
Khi đó A = \(\left(\frac{1-\frac{1}{7^{100}}}{48}\right):\left(1-\frac{1}{7^{100}}\right)=\frac{1}{48}\)
15 tháng 4 2019
\(\frac{1}{\left(a+1\right)a}=\frac{1}{a}-\frac{1}{a+1}\)
Áp dụng đẳng thức trên ta tính ĐƯỢC:
A= 1/100-(1/99-1/100+1/98-1/99+...+1/2-1/3+1/1-1/2)
=1/100-(-1/100+1)
=1/50+1=51/50
15 tháng 4 2019
\(A=\frac{1}{100}-\frac{1}{100.99}-\frac{1}{99.98}-\frac{1}{98.97}-...-\frac{1}{3.2}-\frac{1}{2.1}\)
\(A=\frac{1}{100}-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{98.99}+\frac{1}{99.100}\right)\)
\(A=\frac{1}{100}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)\)
\(A=\frac{1}{100}-\left(1-\frac{1}{100}\right)\)
\(A=\frac{1}{100}-\frac{99}{100}\)
\(A=\frac{-98}{100}=-\frac{49}{50}\)
7 tháng 6 2018
\(=\frac{1}{3}x\left(\frac{3}{1x4}+\frac{3}{4x7}+...+\frac{3}{77x80}\right)\)
\(=\frac{1}{3}x\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{77}-\frac{1}{80}\right)\)
\(=\frac{1}{3}x\left(\frac{1}{1}-\frac{1}{80}\right)\)
\(=\frac{1}{3}\times\frac{79.}{80}\)
\(=\frac{79}{240}\)
Tk giúp mk nha cảm ơn !!
Theo đề =>\(0,1F=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99,100}\)
\(\Rightarrow0,1F=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(\Rightarrow0,1F=1-\frac{1}{100}=\frac{99}{100}\)
\(\Rightarrow F=\frac{99}{10}\)
Chúc bạn hc tốt :)
cảm ơn bn!