Tính thuận tiện A=\(\dfrac{3}{2}-\dfrac{5}{6}+\dfrac{7}{12}-\dfrac{9}{20}+\dfrac{11}{30}-\dfrac{13}{42}+\dfrac{15}{56}-\dfrac{17}{72}\)
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23 tháng 3 2022
= \(\dfrac{5}{2}(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2019}-\dfrac{1}{2021})\)
= \(\dfrac{5}{2}\left(1-\dfrac{1}{101}\right)\)
= \(\dfrac{5}{2}.\dfrac{100}{101}\)
= \(\dfrac{250}{101}\)
18 tháng 9 2021
a: Ta có: \(\dfrac{8}{9}-\left(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{72}\right)\)
\(=\dfrac{8}{9}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{3}-...+\dfrac{1}{8}-\dfrac{1}{9}\right)\)
=0
C
18 tháng 3 2018
B= \(\dfrac{3}{2}-\dfrac{5}{6}+\dfrac{7}{12}-\dfrac{9}{20}+\dfrac{11}{30}-\dfrac{13}{42}+\dfrac{15}{56}\)
= \(\left(1+\dfrac{1}{2}\right)-\left(\dfrac{1}{2}+\dfrac{1}{3}\right)+\left(\dfrac{1}{3}+\dfrac{1}{4}\right)-\left(\dfrac{1}{4}+\dfrac{1}{5}\right)+\left(\dfrac{1}{5}+\dfrac{1}{6}\right)-\left(\dfrac{1}{6}+\dfrac{1}{7}\right)+\left(\dfrac{1}{7}+\dfrac{1}{8}\right)\)
= 1+\(\dfrac{1}{8}\)=\(\dfrac{9}{8}\)
28 tháng 6 2022
\(=\left(\dfrac{3}{2}-\dfrac{5}{6}+\dfrac{7}{12}\right)+\left(-\dfrac{9}{20}+\dfrac{11}{30}\right)+\left(\dfrac{-13}{42}+\dfrac{15}{56}\right)\)
\(=\dfrac{18-10+7}{12}+\dfrac{-27+22}{60}+\dfrac{-1}{24}\)
\(=\dfrac{15}{12}+\dfrac{-5}{60}+\dfrac{-1}{24}\)
\(=\dfrac{30-1+\left(-2\right)}{24}=\dfrac{27}{24}=\dfrac{9}{8}\)
MD
18 tháng 8 2017
1,Ta có:\(\dfrac{9}{10}-\dfrac{1}{90}-\dfrac{1}{72}-\dfrac{1}{57}-\dfrac{1}{42}-\dfrac{1}{30}-\dfrac{1}{20}-\dfrac{1}{12}-\dfrac{1}{6}-\dfrac{1}{2}\) =\(\dfrac{9}{10}-\left(\dfrac{1}{90}+\dfrac{1}{72}+...+\dfrac{1}{2}\right)\)
= \(\dfrac{9}{10}-\left\{\dfrac{1}{\left(9.10\right)}+\dfrac{1}{\left(9.8\right)}+...+\dfrac{1}{\left(2.1\right)}\right\}\)
= \(\dfrac{9}{10}-\left(\dfrac{1}{9}-\dfrac{1}{10}+\dfrac{1}{8}-\dfrac{1}{9}+...+\dfrac{1}{1}-\dfrac{1}{2}\right).\left(\dfrac{1}{90}=\dfrac{1}{9.10}=\dfrac{1}{9}-\dfrac{1}{10}\right)\)=\(\dfrac{9}{10}-\left(1-\dfrac{1}{10}\right)\)
=\(\dfrac{9}{10}-\dfrac{9}{10}\)
= 0
Ý 2 dễ rồi bạn tự tính
BN
18 tháng 8 2017
1, \(\dfrac{9}{10}-\dfrac{1}{90}-\dfrac{1}{72}-\dfrac{1}{56}-\dfrac{1}{42}-\dfrac{1}{30}-\dfrac{1}{20}-\dfrac{1}{12}-\dfrac{1}{6}-\dfrac{1}{2}\)
\(=\dfrac{9}{10}-\left(\dfrac{1}{90}+\dfrac{1}{72}+\dfrac{1}{56}+\dfrac{1}{42}+\dfrac{1}{30}+\dfrac{1}{20}+\dfrac{1}{6}+\dfrac{1}{2}\right)\)
\(=\dfrac{9}{10}-\left(\dfrac{1}{9.10}+\dfrac{1}{8.9}+...+\dfrac{1}{1.2}\right)\)
\(=\dfrac{9}{10}-\left(\dfrac{1}{9}-\dfrac{1}{10}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{7}-\dfrac{1}{8}+...+1-\dfrac{1}{2}\right)\)
\(=\dfrac{9}{10}-\left(\dfrac{-1}{10}+1\right)=\dfrac{9}{10}-\dfrac{9}{10}=0\)
2, \(\dfrac{-5}{11}\cdot\dfrac{13}{17}-\dfrac{5}{11}.\dfrac{4}{17}\)
\(=\dfrac{-5}{11}\cdot\dfrac{13}{17}+\dfrac{-5}{11}.\dfrac{4}{17}\)
\(=\dfrac{-5}{11}\left(\dfrac{13}{17}+\dfrac{4}{17}\right)=\dfrac{-5}{11}.1=\dfrac{-5}{11}\)
16 tháng 4 2021
a) \(A=\dfrac{3}{5}+6\dfrac{5}{6}+\left(11\dfrac{5}{20}-9\dfrac{1}{4}\right):8\dfrac{1}{3}\)
\(=\dfrac{3}{5}+\dfrac{41}{6}\left(11\dfrac{1}{4}-9\dfrac{1}{4}\right):8\dfrac{1}{3}\)
\(=\dfrac{3}{5}+\dfrac{41}{6}.2.\dfrac{3}{25}\)
\(=\dfrac{3}{5}+\dfrac{41}{25}\)
\(=\dfrac{15}{25}+\dfrac{41}{25}\)
\(=\dfrac{56}{25}\)
16 tháng 4 2021
a) A = \(\dfrac{3}{5}+6\dfrac{5}{6}\left(11\dfrac{5}{20}-9\dfrac{1}{4}\right):8\dfrac{1}{3}\)
A = \(\dfrac{3}{5}+\dfrac{41}{6}\) \(\left(\dfrac{45}{4}-\dfrac{37}{4}\right)\) : \(\dfrac{25}{3}\)
A = \(\dfrac{3}{5}+\dfrac{41}{6}\) . 2 : \(\dfrac{25}{3}\)
A = \(\dfrac{3}{5}\) + \(\dfrac{41}{3}\) : \(\dfrac{25}{3}\)
A = \(\dfrac{3}{5}\) + \(\dfrac{41}{25}\)
A = \(\dfrac{56}{25}\)
A = \(\dfrac{3}{2}\) - \(\dfrac{5}{6}\) + \(\dfrac{7}{12}\) - \(\dfrac{9}{20}\) + \(\dfrac{11}{30}\) - \(\dfrac{13}{42}\) + \(\dfrac{15}{56}\) - \(\dfrac{17}{72}\)
A = (1 + \(\dfrac{1}{2}\)) - (\(\dfrac{1}{2}\) + \(\dfrac{1}{3}\)) + (\(\dfrac{1}{3}\) + \(\dfrac{1}{4}\)) - (\(\dfrac{1}{4}\) + \(\dfrac{1}{5}\)) + (\(\dfrac{1}{5}\) + \(\dfrac{1}{6}\)) - (\(\dfrac{1}{6}\) + \(\dfrac{1}{7}\)) + (\(\dfrac{1}{7}\) + \(\dfrac{1}{8}\)) - (\(\dfrac{1}{8}\) + \(\dfrac{1}{9}\))
A = 1 + \(\dfrac{1}{2}\) - \(\dfrac{1}{2}\) - \(\dfrac{1}{3}\) + \(\dfrac{1}{3}\) + \(\dfrac{1}{4}\) - \(\dfrac{1}{4}\) - \(\dfrac{1}{5}\) + \(\dfrac{1}{5}\) + \(\dfrac{1}{6}\) - \(\dfrac{1}{6}\) - \(\dfrac{1}{7}\) + \(\dfrac{1}{7}\) + \(\dfrac{1}{8}\) - \(\dfrac{1}{8}\) - \(\dfrac{1}{9}\)
A = 1 - \(\dfrac{1}{9}\)
A = \(\dfrac{8}{9}\)\(A=\left(1+\dfrac{1}{2}\right)-\left(\dfrac{1}{2}+\dfrac{1}{3}\right)+\left(\dfrac{1}{3}+\dfrac{1}{4}\right)-\left(\dfrac{1}{4}+\dfrac{1}{5}\right)+\left(\dfrac{1}{5}+\dfrac{1}{6}\right)-\left(\dfrac{1}{6}+\dfrac{1}{7}\right)+\left(\dfrac{1}{7}+\dfrac{1}{8}\right)-\left(\dfrac{1}{8}+\dfrac{1}{9}\right)\)
\(A=1+\dfrac{1}{2}-\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}+\dfrac{1}{4}-\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}+\dfrac{1}{6}-\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}+\dfrac{1}{8}-\dfrac{1}{8}-\dfrac{1}{9}\)
\(A=1+\dfrac{1}{9}=\dfrac{10}{9}\)