\(\frac{1}{4.6}\)+\(\frac{1}{2.6.8}\)+\(\frac{1}{3.8.10}\)+\(\frac{1}{4.10.12}\)
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21 tháng 5 2018
Đặt \(A=\dfrac{1}{4.6}+\dfrac{1}{2.6.8}+\dfrac{1}{3.8.10}+\dfrac{1}{4.10.12}\)
\(\dfrac{1}{2}A=\dfrac{1}{2.4.6}+\dfrac{1}{4.6.8}+\dfrac{1}{6.8.10}+\dfrac{1}{8.10.12}\)
\(\left(\dfrac{1}{2}A\right).4=\dfrac{4}{2.4.6}+\dfrac{4}{4.6.8}+\dfrac{4}{6.8.10}+\dfrac{4}{8.10.12}\)
\(\dfrac{1}{2}.4.A=\dfrac{6-2}{2.4.6}+\dfrac{8-4}{4.6.8}+\dfrac{10-6}{6.8.10}+\dfrac{12-8}{8.10.12}\)
\(2A=\dfrac{6}{2.4.6}-\dfrac{2}{2.4.6}+\dfrac{8}{4.6.8}-\dfrac{4}{4.6.8}+\dfrac{10}{6.8.10}-\dfrac{6}{6.8.10}+\dfrac{12}{8.10.12}-\dfrac{8}{8.10.12}\)
\(2A=\dfrac{1}{2.4}-\dfrac{1}{4.6}+\dfrac{1}{4.6}-\dfrac{1}{6.8}+\dfrac{1}{6.8}-\dfrac{1}{8.10}+\dfrac{1}{8.10}-\dfrac{1}{10.12}\)
\(2A=\dfrac{1}{2.4}-\dfrac{1}{10.12}\)
\(2A=\dfrac{1}{8}-\dfrac{1}{120}\)
\(2A=\dfrac{7}{60}\)
\(A=\dfrac{7}{120}\)
YY
13 tháng 7 2018
\(A=\left(\frac{1}{2\cdot4}+\frac{1}{4\cdot6}+.........+\frac{1}{96\cdot98}\right)-\left(\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+.......+\frac{1}{97\cdot99}\right)\)
\(=\frac{1}{2}\left(\frac{2}{2\cdot4}+\frac{2}{4\cdot6}+....+\frac{2}{96\cdot98}\right)-\frac{1}{2}\left(\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+.....+\frac{2}{97\cdot99}\right)\)
\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+......+\frac{1}{96}-\frac{1}{98}\right)-\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\right)\)
\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{98}\right)-\frac{1}{2}\left(\frac{1}{3}-\frac{1}{99}\right)\)
\(=\frac{12}{49}-\frac{16}{99}=\frac{404}{4851}\)
8 tháng 9 2016
\(S=\frac{1}{1.3}-\frac{1}{2.4}+\frac{1}{3.5}-\frac{1}{4.6}+\frac{1}{5.7}-\frac{1}{6.8}+\frac{1}{7.9}-\frac{1}{8.10}\)
\(=\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}\right)-\left(\frac{1}{2.4}-\frac{1}{4.6}-\frac{1}{6.8}-\frac{1}{8.10}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-...+\frac{1}{7}-\frac{1}{9}\right)-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-...+\frac{1}{8}-\frac{1}{10}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{9}\right)-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{10}\right)\)
\(=\frac{1}{2}.\frac{8}{9}-\frac{1}{2}.\frac{2}{5}\)
\(=\frac{4}{9}-\frac{1}{5}\)
\(=\frac{11}{45}\)
MA
7 tháng 9 2016
\(A=\frac{1}{1.3}-\frac{1}{2.4}+\frac{1}{3.5}-\frac{1}{4.6}+\frac{1}{5.7}-\frac{1}{6.8}+\frac{1}{7.9}-\frac{1}{8.10}\)
\(A=\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}\right)-\left(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+\frac{1}{8.10}\right)\)
\(A=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}\right)-\frac{1}{2}\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+\frac{2}{8.10}\right)\)
\(A=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}\right)-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+\frac{1}{8}-\frac{1}{10}\right)\)
\(A=\frac{1}{2}\left(1-\frac{1}{9}\right)-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{10}\right)\)
\(A=\frac{4}{9}-\frac{1}{5}=\frac{11}{45}\)
HT
7 tháng 9 2016
\(S=\frac{1}{1.3}-\frac{1}{2.4}+\frac{1}{3.5}-\frac{1}{4.6}+\frac{1}{5.7}-\frac{1}{6.8}+\frac{1}{7.9}-\frac{1}{8.10}\)
\(S=\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}\right)-\left(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+\frac{1}{8.10}\right)\)
\(S=\frac{1}{2}\left(1-\frac{1}{3}+...+\frac{1}{7}-\frac{1}{9}\right)-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{10}\right)\)
\(S=\frac{1}{2}\left(1-\frac{1}{9}\right)-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{10}\right)\)
\(S=\frac{1}{2}.\frac{8}{9}-\frac{1}{2}.\frac{2}{5}\)
\(S=\frac{4}{9}-\frac{1}{5}\)
\(S=\frac{11}{45}\)
PT
4
W
19 tháng 7 2018
= \(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{200+202}\)
= \(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-...+\frac{1}{200}-\frac{1}{202}\)
= \(\frac{1}{2}-\frac{1}{202}\)
= \(\frac{404}{202}-\frac{1}{202}\)
= \(\frac{403}{202}\)
24 tháng 10 2016
=\(\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{2.4}+...+\frac{2}{8.10}\right)\)
= \(\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}+....+\frac{1}{8}-\frac{1}{10}\right)\)
= \(\frac{1}{2}.\left(1+\frac{1}{2}-\frac{1}{9}-\frac{1}{10}\right)\)
=\(\frac{29}{45}\)
PA
3 tháng 12 2016
\(A=\frac{1}{1\times3}+\frac{1}{2\times4}+\frac{1}{3\times5}+\frac{1}{4\times6}+\frac{1}{5\times7}+\frac{1}{6\times8}+\frac{1}{7\times9}+\frac{1}{8\times10}\)
\(2A=\frac{2}{1\times3}+\frac{2}{2\times4}+\frac{2}{3\times5}+\frac{2}{4\times6}+\frac{2}{5\times7}+\frac{2}{6\times8}+\frac{2}{7\times9}+\frac{2}{8\times10}\)
\(2A=1-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-\frac{1}{5}+\frac{1}{4}-\frac{1}{6}+\frac{1}{5}-\frac{1}{7}+\frac{1}{6}-\frac{1}{8}+\frac{1}{7}-\frac{1}{9}+\frac{1}{8}-\frac{1}{10}\)
\(2A=1+\frac{1}{2}-\frac{1}{9}-\frac{1}{10}\)
\(2A=\frac{58}{45}\)
\(A=\frac{58}{45}\div2\)
\(A=\frac{29}{45}\)
LT
3 tháng 12 2016
\(2A=\frac{2}{1.3}+\frac{2}{2.4}+...+\frac{2}{8.10}=1-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-....+\frac{1}{8}-\frac{1}{10}\)
\(=1+\frac{1}{2}-\frac{1}{9}-\frac{1}{10}=\frac{58}{45}\)
\(A=\frac{29}{45}\)
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