tinh
A= \(\frac{1}{100}-\frac{1}{100.98}-\frac{1}{98.96}-...-\frac{1}{6.4}-\frac{1}{4.2}\)
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5 tháng 4 2019
\(\frac{7}{4}-\left(\frac{1}{2.2}+\frac{1}{4.3}+\frac{1}{6.4}+\frac{1}{8.5}+\frac{1}{10.6}+\frac{1}{12.7}+\frac{1}{14.8}\right)\div x=0\)
\((\frac{1}{2.2}+\frac{1}{4.3}+\frac{1}{6.4}+\frac{1}{8.5}+\frac{1}{10.6}+\frac{1}{12.7}+\frac{1}{14.8})\div x=\frac{7}{4}\)
\((\frac{1}{4}+\frac{1}{12}+\frac{1}{24}+\frac{1}{40}+\frac{1}{60}+\frac{1}{84}+\frac{1}{112})\div x=\frac{7}{4}\)
\(\left[\frac{1}{2}\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}\right)\right]\div x=\frac{7}{4}\)
\(\left[\frac{1}{2}\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}\right)\right]\div x=\frac{7}{4}\)
\(\left[\frac{1}{2}\left(1-\frac{1}{8}\right)\right]\div x=\frac{7}{4}\)
\(\left(\frac{1}{2}.\frac{7}{8}\right)\div x=\frac{7}{4}\)
\(\frac{7}{16}\div x=\frac{7}{4}\)
\(x=\frac{7}{16}\div\frac{7}{4}\)
\(x=\frac{7}{16}\times\frac{4}{7}\)
\(x=\frac{1}{4}\)
5 tháng 4 2019
\(\frac{7}{4}-\left(\frac{1}{2\cdot2}+\frac{1}{4\cdot3}+\frac{1}{6\cdot4}+\frac{1}{8\cdot5}+\frac{1}{10\cdot6}+\frac{1}{12\cdot7}+\frac{1}{14\cdot8}\right)\)
\(=\frac{7}{4}-\left(\frac{1}{4}+\frac{1}{12}+\frac{1}{24}+\frac{1}{40}+\frac{1}{60}+\frac{1}{84}+\frac{1}{112}\right)\)
\(=\frac{7}{4}-\frac{1}{2}\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}\right)\)
\(=\frac{7}{4}-\frac{1}{2}\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}\right)\)
\(=\frac{7}{4}-\frac{1}{2}\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}\right)\)
\(=\frac{7}{4}-\frac{1}{2}\left(1-\frac{1}{8}\right)\)
\(=\frac{7}{4}-\frac{1}{2}\cdot\frac{7}{8}\)
\(=\frac{7}{4}-\frac{7}{16}=\frac{28}{16}-\frac{7}{16}=\frac{21}{16}\)
HL
1
NV
Nguyễn Việt Lâm
Giáo viên
19 tháng 6 2019
\(\frac{4k}{4k^4+1}=\frac{4k}{4k^4+4k^2+1-4k^2}=\frac{4k}{\left(2k^2+1\right)^2-\left(2k\right)^2}=\frac{4k}{\left(2k^2+2k+1\right)\left(2k^2-2k+1\right)}=\frac{1}{2k^2-2k+1}-\frac{1}{2k^2+2k+1}\)
\(=\frac{1}{2k\left(k-1\right)+1}-\frac{1}{2k\left(k+1\right)+1}\)
\(A=\frac{1}{1}-\frac{1}{5}+\frac{1}{5}-\frac{1}{13}+...+\frac{1}{2k\left(k-1\right)+1}-\frac{1}{2k\left(k+1\right)+1}\)
\(=1-\frac{1}{2k\left(k+1\right)+1}=...\)
11 tháng 2 2018
Ta có: \(4n^4+1=\left(4n^4+4n^2+1\right)-4n^2=\left(2n^2+2n+1\right)\left(2n^2-2n+1\right)\)
\(\frac{4n}{4n^4+1}=\frac{\left(2n^2+2n+1\right)-\left(2n^2-2n+1\right)}{\left(2n^2-2n+1\right)\left(2n^2+2n+1\right)}=\frac{1}{2n^2-2n+1}-\frac{1}{2n^2+2n+1}\)
Thay vào ta có:
\(\frac{4.1}{4.1^4+1}+\frac{4.2}{4.2^2+1}+...+\frac{4n}{4n^4+1}=\frac{220}{221}\)
\(\Leftrightarrow1-\frac{1}{5}+\frac{1}{5}-\frac{1}{13}+...+\frac{1}{2n^2-2n+1}-\frac{1}{2n^2+2n+1}=\frac{220}{221}\)
\(\Leftrightarrow1-\frac{1}{2n^2+2n+1}=\frac{220}{221}\)
\(\Leftrightarrow\frac{2n^2+2n}{2n^2+2n+1}=\frac{220}{221}\Rightarrow n=10\)
TT
1
14 tháng 4 2018
Đặt \(A=\frac{1}{50.48}-\frac{1}{48.46}-...-\frac{1}{4.2}\) ta có :
\(A=\frac{1}{48.50}-\left(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{46.48}\right)\) ( xắp sếp lại cho đẹp đội hình thôi :)
Đặt \(B=\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{46.48}\) ta có :
\(2B=\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{46.48}\)
\(2B=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{46}-\frac{1}{48}\)
\(2B=\frac{1}{2}-\frac{1}{48}\)
\(2B=\frac{23}{48}\)
\(B=\frac{23}{48}:2\)
\(B=\frac{23}{48}.\frac{1}{2}\)
\(B=\frac{23}{96}\)
\(\Rightarrow\)\(A=\frac{1}{48.50}-B=\frac{1}{48.50}-\frac{23}{96}=\frac{1}{2400}-\frac{23}{96}=\frac{-287}{1200}\)
Vậy \(A=\frac{-287}{1200}\)
Chúc bạn học tốt ~
\(A=\frac{1}{100}-\frac{1}{100.98}-\frac{1}{98.96}-....-\frac{1}{6.4}-\frac{1}{4.2}\)
\(\Rightarrow A=\frac{1}{100}-\left(\frac{1}{100.98}+\frac{1}{98.96}+....+\frac{1}{6.4}+\frac{1}{4.2}\right)\)
\(\Rightarrow A=\frac{1}{100}-\left(\frac{1}{100}-\frac{1}{98}+\frac{1}{98}-\frac{1}{96}+.....+\frac{1}{6}-\frac{1}{4}+\frac{1}{4}-\frac{1}{2}\right)\)
\(\Rightarrow A=\frac{1}{100}-\left(\frac{1}{100}-\frac{1}{2}\right)\Rightarrow A=\frac{1}{100}-\frac{1}{100}+\frac{1}{2}\Rightarrow A=\frac{1}{2}\)
\(A=\frac{1}{100}-\frac{1}{100.98}-\frac{1}{98.96}-...-\frac{1}{6.4}-\frac{1}{4.2}\)
\(A=\frac{1}{100}-\left(\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{96.98}+\frac{1}{98.100}\right)\)
\(A=\frac{1}{100}-\frac{1}{2.2}.\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{48.49}+\frac{1}{49.50}\right)\)
\(A=\frac{1}{100}-\frac{1}{4}.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{48}-\frac{1}{49}+\frac{1}{49}-\frac{1}{50}\right)\)
\(A=\frac{1}{100}-\frac{1}{4}.\left(1-\frac{1}{50}\right)\)
\(A=\frac{1}{100}-\frac{1}{4}.\frac{49}{50}\)
\(A=\frac{2}{200}-\frac{49}{200}=-\frac{47}{200}\)