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K
Khách
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PT
1
18 tháng 12 2015
B = \(\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+....+\frac{1}{120}\)
\(B=\frac{1}{4.5:2}+\frac{1}{5.6:2}+\frac{1}{6.7:2}+.....+\frac{1}{15.16:2}\)
\(\frac{1}{2}B=\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+....+\frac{1}{15.16}\)
Ta thấy: \(\frac{1}{4.5}=\frac{1}{4}-\frac{1}{5};\frac{1}{5.6}=\frac{1}{5}-\frac{1}{6};\frac{1}{6.7}=\frac{1}{6}-\frac{1}{7};.....\)
\(\frac{1}{2}B=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-.....-\frac{1}{15}+\frac{1}{15}-\frac{1}{16}\)
\(\frac{1}{2}B=\frac{1}{4}-\frac{1}{16}=\frac{3}{16}\)
\(B=\frac{3}{16}:\frac{1}{2}=\frac{3}{16}.2=\frac{3}{8}\)
NH
0
2 tháng 8 2015
a) \(A=\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+...+\frac{1}{120}\)
\(A=\frac{2}{20}+\frac{2}{30}+\frac{2}{42}+...+\frac{2}{240}\)
\(A=2\cdot\left(\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+...+\frac{1}{15\cdot16}\right)\)
\(A=2\cdot\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{15}-\frac{1}{16}\right)\)
\(A=2\cdot\left(\frac{1}{4}-\frac{1}{16}\right)=2\cdot\frac{3}{16}=\frac{3}{8}\)
b) \(B=\frac{10}{56}+\frac{10}{140}+\frac{10}{260}+...+\frac{10}{1400}\)
\(B=\frac{5}{28}+\frac{5}{70}+\frac{5}{130}+...+\frac{5}{700}\)
\(B=\frac{5}{3}\cdot\left(\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+\frac{3}{10\cdot13}+...+\frac{3}{25\cdot28}\right)\)
\(B=\frac{5}{3}\cdot\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{25}-\frac{1}{28}\right)\)
\(B=\frac{5}{3}\cdot\left(\frac{1}{4}-\frac{1}{28}\right)=\frac{5}{3}\cdot\frac{3}{14}=\frac{5}{14}\)
RK
2
Q
16 tháng 3 2018
Có: \(N=\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+....+\frac{1}{120}\)
\(=>N=\frac{2}{20}+\frac{2}{30}+\frac{2}{42}+...+\frac{2}{240}\)
\(=>N=\frac{2}{4\cdot5}+\frac{2}{5\cdot6}+\frac{2}{6\cdot7}+...+\frac{2}{15\cdot16}\)
\(=>N=\left(\frac{2}{4}-\frac{2}{5}+\frac{2}{5}-\frac{2}{6}+...+\frac{2}{15}-\frac{2}{16}\right)\)
\(=>N=\frac{2}{4}-\frac{2}{16}\)
\(=>N=\frac{1}{2}-\frac{1}{8}\)
\(=>N=\frac{8-2}{16}=\frac{6}{16}=\frac{3}{8}\)
Vậy \(N=\frac{3}{8}\)
16 tháng 3 2018
Ta có :
\(N=\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+...+\frac{1}{120}\)
\(N=2\left(\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+...+\frac{1}{240}\right)\)
\(N=2\left(\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{15.16}\right)\)
\(N=2\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{15}-\frac{1}{16}\right)\)
\(N=2\left(\frac{1}{4}-\frac{1}{16}\right)\)
\(N=\frac{1}{2}-\frac{1}{8}\)
\(N=\frac{3}{8}\)
Vậy \(N=\frac{3}{8}\)
Chúc bạn học tốt ~
TH
1
TN
12 tháng 5 2016
\(=\frac{2}{20}+\frac{2}{30}+...+\frac{2}{420}\)
\(=\frac{2}{4.5}+\frac{2}{5.6}+...+\frac{2}{20.21}\)
\(=2\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{20}-\frac{1}{21}\right)\)
\(=2\left(\frac{1}{4}-\frac{1}{21}\right)\)
\(=2\times\frac{17}{84}\)
\(=\frac{17}{72}\)
T
2
6 tháng 4 2019
\(=\frac{2}{20}+\frac{2}{30}+\frac{2}{42}+...+\frac{2}{240}\)
\(=\frac{2}{4.5}+\frac{2}{5.6}+...+\frac{2}{15.16}\)
\(=2.\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{15}-\frac{1}{16}\right)\)
\(=2.\left(\frac{1}{4}-\frac{1}{16}\right)\)
=\(\frac{1}{8}\)
Tích cho mình nhé cảm ơn
KN
7 tháng 4 2019
\(\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+...+\frac{1}{120}\)
\(=\frac{2}{20}+\frac{2}{30}+\frac{2}{42}+...+\frac{2}{240}\)
\(=\frac{2}{4.5}+\frac{2}{5.6}+\frac{2}{6.7}+...+\frac{2}{15.16}\)
\(=2\left(\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{15.16}\right)\)
\(=2\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{15}-\frac{1}{16}\right)\)
\(=2\left(\frac{1}{4}-\frac{1}{16}\right)\)
\(=2\left(\frac{4}{16}-\frac{1}{16}\right)\)
\(=2\times\frac{3}{16}\)
\(=\frac{6}{16}=\frac{3}{8}\)
25 tháng 6 2016
\(\frac{x}{2008}-\frac{1}{10}-\frac{1}{15}-\frac{1}{21}-...-\frac{1}{120}=\frac{5}{8}\)
\(\frac{x}{2008}-\left(\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+...+\frac{1}{120}\right)=\frac{5}{8}\)
\(\frac{x}{2008}-\left(\frac{2}{20}+\frac{2}{30}+\frac{2}{42}+...+\frac{2}{240}\right)=\frac{5}{8}\)
\(\frac{x}{2008}-2.\left(\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+...+\frac{1}{240}\right)=\frac{5}{8}\)
\(\frac{x}{2008}-2.\left(\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{15.16}\right)=\frac{5}{8}\)
\(\frac{x}{2008}-2.\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{15}-\frac{1}{16}\right)=\frac{5}{8}\)
\(\frac{x}{2008}-2.\left(\frac{1}{4}-\frac{1}{16}\right)=\frac{5}{8}\)
\(\frac{x}{2008}-2.\frac{3}{16}=\frac{5}{8}\)
\(\frac{x}{2008}-\frac{3}{8}=\frac{5}{8}\)
\(\frac{x}{2008}=\frac{5}{8}+\frac{3}{8}\)
\(\frac{x}{2008}=1=\frac{2008}{2008}\)
=> x = 2008
Vậy x = 2008
Ta có: \(B=\frac{1}{10}+\frac{1}{15}+...+\frac{1}{120}\)
\(\Rightarrow B=\frac{2}{20}+\frac{2}{30}+...+\frac{2}{240}\)
\(\Rightarrow B=2.\left(\frac{1}{20}+\frac{1}{30}+...+\frac{1}{240}\right)\)
\(\Rightarrow B=2.\left(\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{15.16}\right)\)
\(\Rightarrow B=2.\left(\frac{1}{4}-\frac{1}{16}\right)=2.\frac{3}{16}=\frac{3}{8}\)
Vậy \(B=\frac{3}{8}\)
nha m.n
\(B=\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+.....+\frac{1}{120}\)
\(B=2.\left(\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+.....+\frac{1}{240}\right)\)
\(B=2.\left(\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+....+\frac{1}{15.16}\right)\)
\(B=2.\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+......+\frac{1}{15}-\frac{1}{16}\right)\)
\(B=2.\left(\frac{1}{4}-\frac{1}{16}\right)\)
\(B=2.\frac{3}{16}\)
\(B=\frac{3}{8}\)
Vậy \(B=\frac{3}{8}\)