Tính:
A= \(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+......+\frac{1}{2^{2006}}\)
B= \(\frac{1+\frac{1}{3}+\frac{1}{5}+.....+\frac{1}{999}}{\frac{1}{1.999}+\frac{1}{3.997}+.....+\frac{1}{997.3}+\frac{1}{999.1}}\)
Những câu hỏi liên quan
\(\frac{1+\frac{1}{3}+\frac{1}{5}+.....+\frac{1}{999}}{\frac{1}{1.999}+\frac{1}{3.997}+....\frac{1}{997.3}+\frac{1}{991.1}}\)
\(=\frac{1000\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{999}\right)}{1000\left(\frac{1}{1.999}+\frac{1}{3.997}+...+\frac{1}{997.3}+\frac{1}{999.1}\right)}=\frac{1000\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{999}\right)}{\frac{1+999}{1.999}+\frac{3+997}{3.997}+...+\frac{997+3}{997.3}+\frac{999+1}{999.1}}\)
\(=\frac{1000\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{999}\right)}{1+\frac{1}{999}+\frac{1}{3}+\frac{1}{997}+...+\frac{1}{997}+\frac{1}{3}+\frac{1}{999}+1}=\frac{1000\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{999}\right)}{2\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{999}\right)}=500\)
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\(ChoM=1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+...+\frac{1}{999}\\ N=\frac{1}{1.999}+\frac{1}{3.997}+...+\frac{1}{997.3}+\frac{1}{999.1}\\ Tính\frac{M}{N}\)
\(N=\frac{1}{1.999}+\frac{1}{3.997}+...+\frac{1}{997.3}+\frac{1}{999.1}\)
\(1000N=1+\frac{1}{999}+\frac{1}{3}+\frac{1}{997}+...+\frac{1}{997}+\frac{1}{3}+1\)
\(1000N=2\left[1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{999}\right]\)
\(N=\frac{1}{50}\left[1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{999}\right]\)
\(\frac{M}{N}=\frac{1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{999}}{\frac{1}{50}\left[1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{999}\right]}=\frac{1}{\frac{1}{50}}=50\)
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Tính giá trị biểu thức sau: \(C=\frac{1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{999}}{\frac{1}{1.999}+\frac{1}{3.997}+...+\frac{1}{997.3}+\frac{1}{999.1}}\)
Tính nhanh \(C=\frac{1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{999}}{\frac{1}{1.999}+\frac{1}{3.997}+...+\frac{1}{997.3}+\frac{1}{999.1}}\)
Giải chi tiết cho mk nha, thanks!
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 ~
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Tính A=\(\frac{1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{999}}{\frac{1}{1.999}+\frac{1}{3.997}+...+\frac{1}{999.1}}\)
\(C=\frac{1}{1.999}+\frac{1}{3.997}+...+\frac{1}{997.3}+\frac{1}{999.1}\)
Afrac{1+frac{1}{3}+frac{1}{5}+frac{1}{7}+......+frac{1}{999}}{frac{1}{1.999}+frac{1}{3.997}+frac{1}{5.995}+......+frac{1}{999.1}}Bfrac{1+left(1+2right)+left(1+2+3right)+left(1+2+3+4right)+......+left(1+2+3+...+98right)}{1.2+2.3+3.4+4.5+......+98.99}Cfrac{frac{1}{1.300}+frac{1}{2.301}+frac{1}{3.302}+......+frac{1}{100.400}}{frac{1}{1.102}+frac{1}{2.103}+frac{1}{3.104}+......+frac{1}{299.400}}Dfrac{frac{1}{99}+frac{2}{98}+frac{3}{97}+......+frac{99}{1}}{frac{1}{2}+frac{1}{3}+frac{1}{4}+......+frac...
Đọc tiếp
\(A=\frac{1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+......+\frac{1}{999}}{\frac{1}{1.999}+\frac{1}{3.997}+\frac{1}{5.995}+......+\frac{1}{999.1}}\)
\(B=\frac{1+\left(1+2\right)+\left(1+2+3\right)+\left(1+2+3+4\right)+......+\left(1+2+3+...+98\right)}{1.2+2.3+3.4+4.5+......+98.99}\)
\(C=\frac{\frac{1}{1.300}+\frac{1}{2.301}+\frac{1}{3.302}+......+\frac{1}{100.400}}{\frac{1}{1.102}+\frac{1}{2.103}+\frac{1}{3.104}+......+\frac{1}{299.400}}\)
\(D=\frac{\frac{1}{99}+\frac{2}{98}+\frac{3}{97}+......+\frac{99}{1}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+......+\frac{1}{100}}:\frac{92-\frac{1}{9}-\frac{2}{10}-\frac{3}{97}-......-\frac{92}{100}}{\frac{1}{45}+\frac{1}{50}+\frac{1}{55}+......+\frac{1}{500}}\)
Giup tui voi !!!!!!!!!!!!!!!!!!!!!!!!!!! Mai phai nop roi !!!!!!!!!!!!!!!!!!!
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Tìm x biết: frac{3}{2x+1}+ frac{10}{4x+2}- frac{6}{6x+3}frac{12}{26}Chứng minh rằng: frac{1}{5^2}+ frac{1}{6^2}+...+ frac{1}{2007^2} frac{1}{5}rút gọn: M ( 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})Giúp mình nha!!!
Đọc tiếp
Tìm x biết: \(\frac{3}{2x+1}\)+ \(\frac{10}{4x+2}\)- \(\frac{6}{6x+3}\)=\(\frac{12}{26}\)
Chứng minh rằng: \(\frac{1}{5^2}\)+ \(\frac{1}{6^2}\)+...+ \(\frac{1}{2007^2}\)> \(\frac{1}{5}\)
rút gọn: M = ( 1+\(\frac{1}{3}\)+ \(\frac{1}{5}\)+ ... + \(\frac{1}{999}\)) : (\(\frac{1}{1.999}\)+ \(\frac{1}{3.997}\)+...+\(\frac{1}{997.3}\)+\(\frac{1}{999.1}\))
Giúp mình nha!!!
\(\frac{3}{2x+1}+\frac{10}{4x+2}-\frac{6}{6x+3}=\frac{6}{13}\)
\(\Rightarrow\frac{3}{2x+1}+\frac{5}{2x+1}-\frac{2}{2x+1}=\frac{6}{13}\)
\(\Rightarrow\frac{6}{2x+1}=\frac{6}{13}\Rightarrow2x+1=13\Rightarrow x=6\)
mình giải hơi gọn có gì ko hiểu thì hỏi nha !
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Bài 1: Cho các số tự nhiên n1;n2;...;n10 thỏa mãn \(N=n_1+n_2+...+n_{10}=2013\)
Đặt \(S=n_1^2+n_2^2+...+n_{10}^2\)
Chứng minh S-1 chia hết cho 2
Bài 2: Tính
\(\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{999}\right):\left(\frac{1}{999}+\frac{1}{3.997}+...+\frac{1}{997.3}+\frac{1}{999}\right)\)
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
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Ác Mộng lam đủng rui. **** thui
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