\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+;;;;+\frac{1}{110}\)
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HQ
Hà Quang Minh
Giáo viên
2 tháng 11 2023
a) $\frac{1}{3} + \frac{1}{3} + \frac{1}{6} = \frac{2}{3} + \frac{1}{6} = \frac{4}{6} + \frac{1}{6} = \frac{5}{6}$
b) $\frac{1}{{12}} + \frac{3}{4} + \frac{2}{{12}} = \left( {\frac{1}{{12}} + \frac{2}{{12}}} \right) + \frac{3}{4} = \frac{1}{4} + \frac{3}{4} = \frac{4}{4} = 1$
c) $\frac{{19}}{{15}} + 0 + \frac{{11}}{{15}} = \frac{{19 + 11}}{{15}} = \frac{{30}}{{15}} = 2$
23 tháng 8 2017
Ta có:
\(\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{1100}\right).\left(\frac{1}{3}+\frac{1}{2}-\frac{5}{6}\right)\)
\(=\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{1100}\right).\left(\frac{5}{6}-\frac{5}{6}\right)\)
\(=\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{1100}\right).0\)
\(=0\)
24 tháng 8 2017
Ta có:
\(\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{1100}\right).\left(\frac{1}{3}+\frac{1}{2}-\frac{5}{6}\right)\)
\(\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{1100}\right).\left(\frac{5}{6}-\frac{5}{6}\right)\)
\(\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{1100}\right).0\)
\(=0\)
29 tháng 6 2022
a: \(=\dfrac{3\left(\dfrac{1}{41}-\dfrac{4}{47}+\dfrac{9}{53}\right)}{4\left(\dfrac{1}{41}-\dfrac{4}{47}+\dfrac{9}{53}\right)}+\dfrac{-\dfrac{1}{4}\cdot\dfrac{-2}{3}-\dfrac{3}{4}:\dfrac{1}{6}}{\dfrac{3}{2}\cdot\left(\dfrac{-2}{3}-\dfrac{3}{4}\cdot\dfrac{-2}{3}\right)}\)
\(=\dfrac{3}{4}+\dfrac{\dfrac{2}{12}-\dfrac{9}{2}}{\dfrac{3}{2}\cdot\dfrac{-1}{6}}=\dfrac{3}{4}+\dfrac{-13}{3}:\dfrac{-3}{12}\)
\(=\dfrac{3}{4}+\dfrac{13}{3}\cdot\dfrac{12}{3}=\dfrac{3}{4}+\dfrac{156}{9}=\dfrac{217}{12}\)
b: \(A=158\left(\dfrac{12\left(1-\dfrac{1}{7}-\dfrac{1}{289}-\dfrac{1}{85}\right)}{4\left(1-\dfrac{1}{7}-\dfrac{1}{289}-\dfrac{1}{85}\right)}:\dfrac{5\left(1+\dfrac{1}{13}+\dfrac{1}{169}+\dfrac{1}{91}\right)}{6\left(1+\dfrac{1}{13}+\dfrac{1}{169}+\dfrac{1}{91}\right)}\right)\cdot\dfrac{50550505}{711711711}\)
\(=158\cdot\left(3\cdot\dfrac{6}{5}\right)\cdot\dfrac{50550505}{711711711}\)
\(\simeq40.39\)
\(A=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{110}\)
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{10.11}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{10}-\frac{1}{11}\)
\(A=1-\frac{1}{11}\)
\(A=\frac{10}{11}\)