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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

Ác Mộng lam đủng rui. **** thui

23 tháng 5 2017

\(N=\frac{1}{1.999}+\frac{1}{3.997}+...+\frac{1}{997.3}+\frac{1}{999.1}\)

\(1000N=1+\frac{1}{999}+\frac{1}{3}+\frac{1}{997}+...+\frac{1}{997}+\frac{1}{3}+1\)

\(1000N=2\left[1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{999}\right]\)

\(N=\frac{1}{50}\left[1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{999}\right]\)

\(\frac{M}{N}=\frac{1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{999}}{\frac{1}{50}\left[1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{999}\right]}=\frac{1}{\frac{1}{50}}=50\)

9 tháng 2 2018

giup minh nhe

9 tháng 2 2018

\(=\frac{1000\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{999}\right)}{1000\left(\frac{1}{1.999}+\frac{1}{3.997}+...+\frac{1}{997.3}+\frac{1}{999.1}\right)}=\frac{1000\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{999}\right)}{\frac{1+999}{1.999}+\frac{3+997}{3.997}+...+\frac{997+3}{997.3}+\frac{999+1}{999.1}}\)
\(=\frac{1000\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{999}\right)}{1+\frac{1}{999}+\frac{1}{3}+\frac{1}{997}+...+\frac{1}{997}+\frac{1}{3}+\frac{1}{999}+1}=\frac{1000\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{999}\right)}{2\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{999}\right)}=500\)

28 tháng 7 2019

\(A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2006}}\)

\(2A=2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2005}}\)

\(2A-A=\left(2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2005}}\right)-\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2006}}\right)\)

\(A=2-\frac{1}{2^{2006}}\)

10 tháng 4 2018

\(Q=\left(\frac{1}{99}+\frac{12}{999}+\frac{123}{999}\right)\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{6}\right)\text{ }\)

\(Q=\left(\frac{1}{99}+\frac{12}{999}+\frac{123}{999}\right)\left(\frac{3}{6}-\frac{2}{6}-\frac{1}{6}\right)\)

\(Q=\left(\frac{1}{99}+\frac{12}{999}+\frac{123}{999}\right).0\)

\(Q=0\)

13 tháng 4 2018

Q=(1/99+12/999+123/999).(1/2-1/3-1/6) =(1/99+12/999+123/999).0 Q=0

14 tháng 3 2018

Giup tui voi !!!!!!!!!!!!!!!!!!!!!!!!!!! Mai phai nop roi !!!!!!!!!!!!!!!!!!!

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 ~ 

4 tháng 4 2018

HELP ME ! MK cần gấp