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31 tháng 5 2022
\(E=\left(\dfrac{1}{3}-1\right)\cdot\left(\dfrac{1}{6}-1\right)\cdot\left(\dfrac{1}{10}-1\right)\cdot...\cdot\left(\dfrac{1}{45}-1\right)\)
\(\text{E=}\left(\dfrac{1}{3+6+10+...+45}\cdot1\right)\)
\(\text{E = }\left(\dfrac{1}{\left(45-3\right)\div2+1}\right)\)
\(\text{E = }\dfrac{1}{22}\)
23 tháng 5 2022
\(E=\left(\dfrac{1}{3}-1\right)\cdot\left(\dfrac{1}{6}-1\right)\cdot\left(\dfrac{1}{10}-1\right)\cdot...\cdot\left(\dfrac{1}{45}-1\right)\)
\(E=1\cdot\left(\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+...+\dfrac{1}{45}\right)\)
\(E=1\cdot\left(\dfrac{1}{45}-\dfrac{1}{3}\right)\div2+1\)
\(E=\dfrac{1}{22}\)
28 tháng 6 2016
Nhân cả tử cả mẫu của các phân số trong A với 2 ta có:
\(A=\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+..........+\frac{2}{90}\)
\(=2\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+.........+\frac{1}{90}\right)\)
\(=2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+........+\frac{1}{9.10}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+......+\frac{1}{9}-\frac{1}{10}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{10}\right)\)
\(=2.\frac{2}{5}\)
\(=\frac{4}{5}\)
28 tháng 6 2016
\(A=\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+...+\frac{1}{45}\)
\(A=\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+\frac{2}{30}+...+\frac{2}{90}\)
\(A=2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{9.10}\right)\)
\(A=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(A=2\left(\frac{1}{2}-\frac{1}{10}\right)\)
\(A=\frac{2}{5}\)
4 tháng 5 2018
\(\dfrac{1}{2}\) + \(\dfrac{1}{3}\) + \(\dfrac{1}{6}\) + \(\dfrac{1}{10}\) + \(\dfrac{1}{15}\) + ... + \(\dfrac{1}{36}\) + \(\dfrac{1}{45}\)
= \(\dfrac{2}{4}\) + \(\dfrac{2}{6}\) + \(\dfrac{2}{12}\) + \(\dfrac{2}{20}\) + \(\dfrac{2}{30}\) + ... + \(\dfrac{2}{72}\) + \(\dfrac{2}{90}\)
= \(\dfrac{2}{2.2}\) + \(\dfrac{2}{2.3}\) + \(\dfrac{2}{3.4}\) + \(\dfrac{2}{4.5}\) + \(\dfrac{2}{5.6}\) + ... + \(\dfrac{2}{8.9}\) + \(\dfrac{2}{9.10}\)
= 2 (\(\dfrac{1}{2.2}\) + \(\dfrac{1}{2.3}\) + \(\dfrac{1}{3.4}\) + \(\dfrac{1}{4.5}\) + \(\dfrac{1}{5.6}\) + ... + \(\dfrac{1}{8.9}\) + \(\dfrac{1}{9.10}\))
= 2 (\(\dfrac{1}{2}\) - \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) - \(\dfrac{1}{3}\) + \(\dfrac{1}{3}\) - \(\dfrac{1}{4}\) + ... + \(\dfrac{1}{8}\) - \(\dfrac{1}{9}\) + \(\dfrac{1}{9}\) - \(\dfrac{1}{10}\)) = 2 (\(\dfrac{1}{2}\) - \(\dfrac{1}{10}\)) = 2 . \(\dfrac{2}{5}\) = \(\dfrac{4}{5}\)
Mặc dù có 1 thiếu!
NT
0
12 tháng 9 2015
Coi \(A=\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+\frac{1}{45}\)
\(\Rightarrow\frac{1}{2}A=\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+\frac{1}{45}\right).\frac{1}{2}\)
\(=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}\)
\(=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\)
\(=\frac{1}{2}-\frac{1}{10}=\frac{2}{5}\)
\(\Rightarrow A=\frac{2}{5}:\frac{1}{2}=\frac{4}{5}\)
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{45}\)
= \(2.\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}\right)\)
= \(2.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\right)\)
= \(2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\right)\)
= \(2.\left(\frac{1}{2}-\frac{1}{10}\right)\)
= \(2.\frac{2}{5}\)
= \(\frac{4}{5}\)
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{45}\\ =\frac{2}{6}+\frac{2}{12}+...+\frac{2}{90}\\ =2\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\right)\)
\(2\left(\frac{1}{2}-\frac{1}{10}\right)=\frac{4}{5}\)