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27 tháng 7 2015
\(\frac{1}{2}\times\left(\frac{2}{1\times3}+\frac{2}{3\times5}+\frac{2}{5\times7}+\frac{2}{7\times9}+\frac{2}{9\times11}\right)\times y=\frac{2}{3}\)
\(\frac{1}{2}\times\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right)\times y=\frac{2}{3}\)
\(\frac{1}{2}\times\left(\frac{1}{1}-\frac{1}{11}\right)\times y=\frac{2}{3}\)
\(\frac{1}{2}\times\frac{10}{11}\times y=\frac{2}{3}\)
\(\frac{5}{11}\times y=\frac{2}{3}\) => \(y=\frac{2}{3}:\frac{5}{11}=\frac{2}{3}\times\frac{11}{5}=\frac{22}{15}\)
DP
1 tháng 7 2017
2/
a) \(\frac{4}{1\cdot5}+\frac{4}{5\cdot9}+\frac{4}{9\cdot13}+\frac{4}{13\cdot17}+\frac{4}{17\cdot21}\)
\(=\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+....+\frac{1}{17}-\frac{1}{21}\right)\)
\(=1-\frac{1}{21}=\frac{20}{21}\)
b) \(\left(1-\frac{1}{2}\right)\cdot\left(1-\frac{1}{3}\right)\cdot\left(1-\frac{1}{4}\right)\cdot...\cdot\left(1-\frac{1}{2017}\right)\)
\(=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot..\cdot\frac{2016}{2017}\)
\(=\frac{1}{2017}\)
c) \(A=2000-5-5-5-..-5\)(có 200 số 5)
\(A=2000-\left(5\cdot200\right)\)
\(A=2000-1000\)
\(A=1000\)
2 tháng 8 2016
\(=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{11}\)
\(=\frac{1}{1}-\frac{1}{11}=\frac{10}{11}\)
29 tháng 8 2018
\(\frac{12}{35}:\frac{35}{25}=\frac{12}{35}.\frac{25}{35}=\frac{12.25}{35.35}=\frac{12.5.5}{7.5.7.5}=\frac{12}{49}\)
\(\frac{9}{22}.\frac{33}{18}=\frac{9.33}{22.18}=\frac{9.3.11}{11.2.9.2}=\frac{3}{4}\)
1 tháng 5 2015
\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{8.9.10}=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}\right)+\frac{1}{2}.\left(\frac{1}{2.3}-\frac{1}{3.4}\right)+...+\frac{1}{2}.\left(\frac{1}{8.9}-\frac{1}{9.10}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{8.9}-\frac{1}{9.10}\right)\)
\(\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{9.10}\right)=\frac{1}{2}.\frac{22}{45}=\frac{11}{45}\)
ML
26 tháng 6 2015
\(3A=\frac{6}{3\times\left(3+6\right)}+\frac{15}{9\times\left(9+15\right)}+...+\frac{39}{84\times\left(84+39\right)}\)
\(=\frac{1}{3}-\frac{1}{9}+\frac{1}{9}-\frac{1}{24}+...+\frac{1}{84}-\frac{1}{123}=\frac{1}{3}-\frac{1}{123}=\frac{40}{123}\)
\(\Rightarrow A=\frac{40}{3.123}=\frac{40}{369}\)
MD
2
27 tháng 2 2017
\(\frac{1}{1.5}+\frac{1}{5.9}+\frac{1}{9.13}+.....+\frac{1}{97.101}\)
\(=\frac{1}{4}\left(\frac{4}{1.5}+\frac{4}{5.9}+\frac{4}{9.13}+......+\frac{4}{97.101}\right)\)
\(=\frac{1}{4}\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+.....+\frac{1}{97}-\frac{1}{101}\right)\)
\(=\frac{1}{4}\left(1-\frac{1}{101}\right)\)
\(=\frac{1}{4}.\frac{100}{101}=\frac{25}{101}\)
\(45\cdot\left(-\frac{5}{7}\right)+9\cdot5+\frac{8}{9}+135+45\cdot\frac{52}{63}\)
\(=-\frac{225}{7}+45+\frac{8}{9}+135+\frac{260}{7}\)
\(=\left(-\frac{225}{7}+\frac{260}{7}\right)+\left(45+135\right)+\frac{8}{9}\)
\(=\frac{35}{7}+180+\frac{8}{9}\)
\(=5+180+\frac{8}{9}\)
\(=185+\frac{8}{9}=185\frac{8}{9}\)
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