\(\frac{\left(\frac{3}{15}+\frac{1}{4}+\frac{7}{20}\right)\times\frac{17}{49}}{5\frac{1}{3}+\frac{2}{5}}\)
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23 tháng 9 2018
\(\frac{\left(\frac{3}{15}+\frac{1}{4}+\frac{7}{20}\right)\times\frac{17}{49}}{5\frac{1}{3}+\frac{2}{5}}\)
\(=\frac{\left(\frac{12}{60}+\frac{15}{60}+\frac{21}{60}\right)\times\frac{17}{49}}{\frac{16}{3}\times\frac{2}{5}}\)
\(=\frac{\frac{48}{60}\times\frac{17}{49}}{\frac{80}{15}+\frac{6}{15}}\)
\(=\frac{\frac{816}{2940}}{\frac{86}{15}}\)
\(=\frac{816}{2940}:\frac{86}{15}\)
\(=\frac{816}{2940}\times\frac{15}{86}\)
\(=\frac{68}{245}\times\frac{15}{86}\)
\(=\frac{102}{2107}\)
9 tháng 8 2017
a) \(\frac{\frac{2}{7}+\frac{2}{5}+\frac{2}{17}+\frac{2}{293}}{\frac{3}{7}+\frac{3}{5}+\frac{3}{17}+\frac{3}{293}}+\frac{\frac{7}{12}+\frac{5}{6}-1}{5-\frac{3}{4}+\frac{1}{3}}\) \(=\frac{2\left(\frac{1}{7}+\frac{1}{5}+\frac{1}{17}+\frac{1}{293}\right)}{3\left(\frac{1}{7}+\frac{1}{5}+\frac{1}{17}+\frac{1}{293}\right)}+\frac{\frac{5}{12}}{\frac{55}{12}}\)
\(=\frac{2}{3}+\frac{1}{11}=\frac{25}{33}\)
b) \(\left(1-\frac{1}{7}\right).\left(1-\frac{2}{7}\right)....\left(1-\frac{10}{7}\right)=\left(1-\frac{1}{7}\right).\left(1-\frac{2}{7}\right)...\left(1-\frac{7}{7}\right).\left(1-\frac{8}{7}\right).\left(1-\frac{9}{7}\right).\) \(\left(1-\frac{10}{7}\right)\) = 0
9 tháng 8 2017
a)\(\frac{\frac{2}{7}+\frac{2}{5}+\frac{2}{17}+\frac{2}{293}}{\frac{3}{7}+\frac{3}{5}+\frac{3}{17}+\frac{3}{293}}+\frac{\frac{7}{12}+\frac{5}{6}-1}{5-\frac{3}{4}+\frac{1}{3}}\)
\(=\frac{2\left(\frac{1}{7}+\frac{1}{5}+\frac{1}{17}+\frac{1}{293}\right)}{3\left(\frac{1}{7}+\frac{1}{5}+\frac{1}{17}+\frac{1}{293}\right)}+\frac{\frac{7}{12}+\frac{10}{12}-\frac{12}{12}}{\frac{60}{12}-\frac{9}{12}+\frac{4}{12}}\)
\(=\frac{2}{3}+\frac{\frac{5}{12}}{\frac{55}{12}}\)
\(=\frac{2}{3}+\frac{1}{11}\)
\(=\frac{25}{33}\)
b)\(\left(1-\frac{1}{7}\right)\cdot\left(1-\frac{2}{7}\right)\cdot...\cdot\left(1-\frac{10}{7}\right)\)
Ta nhận thấy trong tích này có 1 thừa số là\(\left(1-\frac{7}{7}\right)=0\)nên tích trên sẽ bằng 0.
10 tháng 8 2016
a) \(\frac{1}{3}.\frac{-6}{13}.\frac{-9}{10}.\frac{-13}{36}\)
\(=\left(\frac{1}{3}.\frac{-9}{10}\right)\left(\frac{-6}{13}.\frac{-13}{36}\right)\)
\(=\frac{-3}{10}.\frac{1}{6}\)
\(=\frac{-1}{20}\)
b) \(\frac{-1}{3}.\frac{-15}{17}.\frac{34}{45}\)
\(=\frac{-1}{3}.\frac{-2}{3}\)
\(=\frac{2}{9}\)
c) \(\left(1-\frac{1}{5}\right)\left(\frac{-3}{10}+\frac{1}{5}\right)\)
\(=\frac{4}{5}.\frac{-1}{10}\)
\(=\frac{-2}{25}\)
d) \(A=\frac{1}{3}.\frac{4}{5}+\frac{1}{3}.\frac{6}{5}+\frac{2}{3}\)
\(=\frac{1}{3}\left(\frac{4}{5}+\frac{6}{5}\right)+\frac{2}{3}\)
\(=\frac{1}{3}.2+\frac{2}{3}\)
\(=\frac{2}{3}+\frac{2}{3}\)
\(=\frac{4}{3}\)
e) \(11\frac{1}{4}-\left(2\frac{5}{7}+5\frac{1}{4}\right)\)
\(=\left(11\frac{1}{4}-5\frac{1}{4}\right)-2\frac{5}{7}\)
\(=6-2\frac{5}{7}\)
\(=5\frac{7}{7}-2\frac{5}{7}\)
\(=3\frac{2}{7}\)
\(\frac{\left(\frac{3}{15}+\frac{1}{4}+\frac{7}{20}\right)\times\frac{17}{49}}{5\frac{1}{3}+\frac{2}{5}}=\frac{\left(\frac{1}{5}+\frac{1}{4}+\frac{7}{20}\right)\times\frac{17}{49}}{\frac{16}{3}+\frac{2}{5}}\)
=\(\frac{\left(\frac{4}{20}+\frac{5}{20}+\frac{7}{20}\right)\times\frac{17}{49}}{\frac{80}{15}+\frac{6}{15}}=\frac{\frac{16}{20}\times\frac{17}{49}}{\frac{86}{15}}=\frac{\frac{4}{5}\times\frac{17}{49}}{\frac{86}{15}}\)
=\(\frac{68}{245}\times\frac{15}{86}=\frac{102}{2107}\)
=\(\frac{102}{2107}\)