Tính:
A= \(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+......+\frac{1}{2^{2006}}\)
B= \(\frac{1+\frac{1}{3}+\frac{1}{5}+.....+\frac{1}{999}}{\frac{1}{1.999}+\frac{1}{3.997}+.....+\frac{1}{997.3}+\frac{1}{999.1}}\)
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DN
9 tháng 2 2018
\(=\frac{1000\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{999}\right)}{1000\left(\frac{1}{1.999}+\frac{1}{3.997}+...+\frac{1}{997.3}+\frac{1}{999.1}\right)}=\frac{1000\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{999}\right)}{\frac{1+999}{1.999}+\frac{3+997}{3.997}+...+\frac{997+3}{997.3}+\frac{999+1}{999.1}}\)
\(=\frac{1000\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{999}\right)}{1+\frac{1}{999}+\frac{1}{3}+\frac{1}{997}+...+\frac{1}{997}+\frac{1}{3}+\frac{1}{999}+1}=\frac{1000\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{999}\right)}{2\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{999}\right)}=500\)
23 tháng 5 2017
\(N=\frac{1}{1.999}+\frac{1}{3.997}+...+\frac{1}{997.3}+\frac{1}{999.1}\)
\(1000N=1+\frac{1}{999}+\frac{1}{3}+\frac{1}{997}+...+\frac{1}{997}+\frac{1}{3}+1\)
\(1000N=2\left[1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{999}\right]\)
\(N=\frac{1}{50}\left[1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{999}\right]\)
\(\frac{M}{N}=\frac{1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{999}}{\frac{1}{50}\left[1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{999}\right]}=\frac{1}{\frac{1}{50}}=50\)
1 tháng 5 2018
Làm thử thoi nhé :)
\(C=\frac{1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{999}}{\frac{1}{1.999}+\frac{1}{3.997}+...+\frac{1}{997.3}+\frac{1}{999.1}}\)
\(\frac{1}{1000}C=\frac{1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{999}}{\frac{1000}{1.999}+\frac{1000}{3.997}+...+\frac{1000}{997.3}+\frac{1000}{999.1}}\)
\(\frac{1}{1000}C=\frac{1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{999}}{\frac{1+999}{1.999}+\frac{3+997}{3.997}+...+\frac{997+3}{997.3}+\frac{999+1}{999.1}}\)
\(\frac{1}{1000}C=\frac{1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{999}}{\frac{1}{1.999}+\frac{999}{1.999}+\frac{3}{3.997}+\frac{997}{3.997}+...+\frac{997}{997.3}+\frac{3}{997.3}+\frac{999}{999.1}+\frac{1}{999.1}}\)
\(\frac{1}{1000}C=\frac{1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{999}}{\frac{1}{999}+\frac{1}{1}+\frac{1}{997}+\frac{1}{3}+...+\frac{1}{3}+\frac{1}{997}+\frac{1}{1}+\frac{1}{999}}\)
\(\frac{1}{1000}C=\frac{1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{999}}{2\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{999}\right)}\)
\(\frac{1}{1000}C=\frac{1}{2}\)
\(C=\frac{1}{2}.1000\)
\(C=500\)
Vậy \(C=500\)
Chúc bạn học tốt ~
MA
0
NH
14 tháng 3 2018
Giup tui voi !!!!!!!!!!!!!!!!!!!!!!!!!!! Mai phai nop roi !!!!!!!!!!!!!!!!!!!
22 tháng 7 2016
\(\frac{3}{2x+1}+\frac{10}{4x+2}-\frac{6}{6x+3}=\frac{6}{13}\)
\(\Rightarrow\frac{3}{2x+1}+\frac{5}{2x+1}-\frac{2}{2x+1}=\frac{6}{13}\)
\(\Rightarrow\frac{6}{2x+1}=\frac{6}{13}\Rightarrow2x+1=13\Rightarrow x=6\)
mình giải hơi gọn có gì ko hiểu thì hỏi nha !
AM
21 tháng 6 2015
1)Ta có:S=\(n_1^2+n_2^2+...+n_{10}^2\)=\(\left(n_1+n_2+...+n_{10}\right)^2-2.\left(n_1n_2+n_2n_3+.....+n_{10}.n_1\right)=2013^2-2.\left(n_1n_2+n_2n_3+.....+n_{10}.n_1\right)\)
Do 20132 chia 2 dư 1
\(2.\left(n_1n_2+n_2n_3+.....+n_{10}.n_1\right)\) chia hết cho 2
=>\(2013^2-2.\left(n_1n_2+n_2n_3+.....+n_{10}.n_1\right)-1\) chia hết cho 2
=>S-1 chia hết cho 2
\(A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2006}}\)
\(2A=2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2005}}\)
\(2A-A=\left(2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2005}}\right)-\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2006}}\right)\)
\(A=2-\frac{1}{2^{2006}}\)