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\(N=\frac{1}{1.999}+\frac{1}{3.997}+...+\frac{1}{997.3}+\frac{1}{999.1}\)
\(1000N=1+\frac{1}{999}+\frac{1}{3}+\frac{1}{997}+...+\frac{1}{997}+\frac{1}{3}+1\)
\(1000N=2\left[1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{999}\right]\)
\(N=\frac{1}{50}\left[1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{999}\right]\)
\(\frac{M}{N}=\frac{1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{999}}{\frac{1}{50}\left[1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{999}\right]}=\frac{1}{\frac{1}{50}}=50\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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 ~
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt A=\(\frac{1}{1.999}+\frac{1}{3.997}+...+\frac{1}{3.997}+\frac{1}{1.999}\)
=>1000A=\(1+\frac{1}{999}+\frac{1}{3}+\frac{1}{997}+...+\frac{1}{997}+\frac{1}{3}+1=2\left(1+\frac{1}{3}+...+\frac{1}{997}+\frac{1}{999}\right)\)
=>A=\(\frac{1}{50}\left(1+\frac{1}{3}+...+\frac{1}{997}+\frac{1}{999}\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2006}}\)
\(2A=2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2005}}\)
\(2A-A=\left(2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2005}}\right)-\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2006}}\right)\)
\(A=2-\frac{1}{2^{2006}}\)
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