\(\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{1100}\right)×\left(\frac{1}{3}+\frac{1}{2}-\frac{5}{6}\right)\)
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20 tháng 9 2017
đổi hết thành ps rồi giải
gttđ > 0
cái này bn tự lm đc mà
cố lên
HQ
Hà Quang Minh
Giáo viên
19 tháng 9 2023
a)
\(\begin{array}{l}1\frac{1}{2} + \frac{1}{5}.\left[ {\left( { - 2\frac{5}{6} + \frac{1}{3}} \right)} \right]\\ = \frac{3}{2} + \frac{1}{5}.\left[ {\left( { - \frac{{17}}{6} + \frac{2}{6}} \right)} \right]\\ = \frac{3}{2} + \frac{1}{5}.\frac{{ - 15}}{6}\\ = \frac{3}{2} + \frac{{ - 1}}{2}\\ = \frac{2}{2}\\=1\end{array}\)
b)
\(\begin{array}{l}\frac{1}{3}.\left( {\frac{2}{5} - \frac{1}{2}} \right):{\left( {\frac{1}{6} - \frac{1}{5}} \right)^2}\\ = \frac{1}{3}.\left( {\frac{4}{{10}} - \frac{5}{{10}}} \right):{\left( {\frac{5}{{30}} - \frac{6}{{30}}} \right)^2}\\ = \frac{1}{3}.\frac{{ - 1}}{{10}}:{\left( {\frac{{ - 1}}{{30}}} \right)^2}\\ = \frac{{ - 1}}{{30}}:\frac{1}{{{{30}^2}}}\\ = \frac{{ - 1}}{{30}}{.30^2}\\ = - 30\end{array}\)
18 tháng 8 2020
[\(\frac{-75}{59}\).\(\frac{-107}{93}\)]\(\frac{31}{50}\)=\(\frac{2675}{1829}\).\(\frac{31}{50}\)=\(\frac{107}{118}\)
18 tháng 8 2020
\(\left[\frac{1\frac{11}{31}\cdot4\frac{3}{7}-\left(15-6\frac{1}{3}\cdot\frac{2}{19}\right)}{4\frac{5}{6}+\frac{1}{6}\left(12-5\frac{1}{3}\right)}\cdot\left(-1\frac{14}{93}\right)\right]\cdot\frac{31}{50}\)
\(=\left[\frac{\frac{42}{31}\cdot\frac{31}{7}-\left(15-\frac{19}{3}\cdot\frac{2}{19}\right)}{4\frac{5}{6}+\frac{1}{6}\left(12-\frac{16}{3}\right)}\cdot\left(-\frac{107}{93}\right)\right]\cdot\frac{31}{50}\)
\(=\left[\frac{6-\left(15-\frac{2}{3}\right)}{\frac{29}{6}+\frac{1}{6}\cdot\frac{20}{3}}\cdot\left(-\frac{107}{93}\right)\right]\cdot\frac{31}{50}\)
\(=\left[\frac{6-15+\frac{2}{3}}{\frac{29}{6}+\frac{10}{9}}\cdot\left(-\frac{107}{93}\right)\right]\cdot\frac{31}{50}\)
\(=\left[\frac{-\frac{25}{3}}{\frac{107}{18}}\cdot\left(-\frac{107}{93}\right)\right]\cdot\frac{31}{50}\)
\(=\left[\left(-\frac{150}{107}\right)\cdot\left(-\frac{107}{93}\right)\right]\cdot\frac{31}{50}=\frac{50}{31}\cdot\frac{31}{50}=1\)
Ta có:
\(\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{1100}\right).\left(\frac{1}{3}+\frac{1}{2}-\frac{5}{6}\right)\)
\(=\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{1100}\right).\left(\frac{5}{6}-\frac{5}{6}\right)\)
\(=\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{1100}\right).0\)
\(=0\)
Ta có:
\(\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{1100}\right).\left(\frac{1}{3}+\frac{1}{2}-\frac{5}{6}\right)\)
\(\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{1100}\right).\left(\frac{5}{6}-\frac{5}{6}\right)\)
\(\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{1100}\right).0\)
\(=0\)