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19 tháng 7 2016
\(A=\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+...+\frac{1}{120}\)
\(A=2.\left(\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+...+\frac{1}{240}\right)\)
\(A=2.\left(\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{15.16}\right)\)
\(A=2.\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{15}-\frac{1}{16}\right)\)
\(A=2.\left(\frac{1}{4}-\frac{1}{16}\right)\)
\(A=2.\frac{3}{16}=\frac{3}{8}\)
S
19 tháng 7 2016
\(A=\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+...+\frac{1}{120}\)
\(=2.\left(\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+...+\frac{1}{240}\right)\)
\(=2.\left(\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{15.16}\right)\)
\(=2.\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{15}-\frac{1}{16}\right)\)
\(=2.\left(\frac{1}{4}-\frac{1}{16}\right)\)
\(=2.\frac{3}{16}=\frac{3}{8}\)
27 tháng 7 2017
\(A=\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+...+\frac{1}{120}\)
\(A=\frac{2}{20}+\frac{2}{30}+\frac{2}{42}+...+\frac{2}{240}\)
\(A=2.\left(\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+...+\frac{1}{240}\right)\)
\(A=2.\left(\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{15.16}\right)\)
\(A=2.\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{15}-\frac{1}{16}\right)\)
\(A=2.\left(\frac{1}{4}-\frac{1}{16}\right)\)
\(A=2.\frac{3}{16}\)
\(A=\frac{3}{8}\)
27 tháng 7 2017
\(B=\frac{2}{15}+\frac{2}{35}+\frac{2}{63}+...+\frac{2}{399}\)
\(B=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{19.21}\)
\(B=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{19}-\frac{1}{21}\)
\(B=\frac{1}{3}-\frac{1}{21}\)
\(B=\frac{2}{7}\)
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) \(B=\frac{10}{56}+\frac{10}{140}+\frac{10}{260}+...+\frac{10}{1400}\)
\(B=\frac{5}{28}+\frac{5}{70}+\frac{5}{130}+...+\frac{5}{700}\)
\(B=\frac{5}{3}\cdot\left(\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+\frac{3}{10\cdot13}+...+\frac{3}{25\cdot28}\right)\)
\(B=\frac{5}{3}\cdot\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{25}-\frac{1}{28}\right)\)
\(B=\frac{5}{3}\cdot\left(\frac{1}{4}-\frac{1}{28}\right)=\frac{5}{3}\cdot\frac{3}{14}=\frac{5}{14}\)
AK
10 tháng 4 2018
\(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{5.6}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{5}-\frac{1}{6}\)
\(A=1-\frac{1}{6}\)
\(A=\frac{6}{6}-\frac{1}{6}\)
\(A=\frac{5}{6}\)
\(B=\frac{1}{10}+\frac{1}{15}+...+\frac{1}{120}\)
\(B=2.\left(\frac{1}{20}+\frac{1}{30}+...+\frac{1}{240}\right)\)
\(B=2.\left(\frac{1}{4.5}+\frac{1}{5.6}+..+\frac{1}{15.16}\right)\)
\(B=2.\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{15}-\frac{1}{16}\right)\)
\(B=2.\left(\frac{1}{4}-\frac{1}{16}\right)\)
\(B=2.\left(\frac{4}{16}-\frac{1}{16}\right)\)
\(B=2.\frac{3}{16}\)
\(B=\frac{3}{8}\)
Chúc bạn học tốt !!!
10 tháng 4 2018
\(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{5.6}\)
\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{5}-\frac{1}{6}\)
\(A=\frac{1}{1}-\frac{1}{6}\)
\(A=\frac{5}{6}\)
\(B=\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+...+\frac{1}{120}\)
\(B=\frac{2}{20}+\frac{2}{30}+\frac{2}{42}+...+\frac{2}{240}\)
\(B=\frac{2}{4.5}+\frac{2}{5.6}+\frac{2}{6.7}+...+\frac{2}{15.16}\)
\(B=2\left(\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{15.16}\right)\)
\(B=2.\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{15}-\frac{1}{16}\right)\)
\(B=2.\left(\frac{1}{4}-\frac{1}{16}\right)\)
\(B=2.\frac{3}{16}\)
\(B=\frac{6}{16}=\frac{3}{13}\)
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: \(\frac{1}{4.5}=\frac{1}{4}-\frac{1}{5};\frac{1}{5.6}=\frac{1}{5}-\frac{1}{6};\frac{1}{6.7}=\frac{1}{6}-\frac{1}{7};.....\)
\(\frac{1}{2}B=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-.....-\frac{1}{15}+\frac{1}{15}-\frac{1}{16}\)
\(\frac{1}{2}B=\frac{1}{4}-\frac{1}{16}=\frac{3}{16}\)
\(B=\frac{3}{16}:\frac{1}{2}=\frac{3}{16}.2=\frac{3}{8}\)
TV
0
H
1 tháng 3 2020
\(\frac{1}{2}B=\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{15.16}\)
\(=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{15}-\frac{1}{16}=\frac{1}{4}-\frac{1}{16}=\frac{3}{16}\)
=>\(B=\frac{3}{16}:\frac{1}{2}=\frac{3}{8}\)
\(C=\left(\frac{3}{29}-\frac{1}{5}\right)\cdot\frac{29}{3}=1-\frac{1}{5}\cdot\frac{29}{3}=1-\frac{29}{15}=-\frac{14}{15}\)
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 \(B=\frac{3}{8}\)