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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
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.
\(\left(\frac{-2}{3}+\frac{3}{7}\right):\frac{4}{5}+\left(\frac{-1}{3}+\frac{4}{7}\right):\frac{4}{5}\)
\(=\left(\frac{-2}{3}+\frac{3}{7}+\frac{-1}{3}+\frac{4}{7}\right):\frac{4}{5}\)
\(=\left[\left(\frac{-2}{3}+\frac{-1}{3}\right)+\left(\frac{3}{7}+\frac{4}{7}\right)\right]:\frac{4}{5}\)
\(=\left[\left(-1+1\right)\right]:\frac{4}{5}\)
\(=0:\frac{4}{5}=0\)
~ Hok tốt ~
\(\left(\frac{3}{5}-\frac{1}{2}+\frac{7}{3}\right)-\left(\frac{1}{3}-\frac{5}{2}+\frac{1}{5}\right)-\left(-\frac{3}{5}+3-\frac{5}{3}\right)\)
\(=\frac{3}{5}-\frac{1}{2}+\frac{7}{3}-\frac{1}{3}+\frac{5}{2}-\frac{1}{5}+\frac{3}{5}-3+\frac{5}{3}\)
\(=\frac{3}{5}+\frac{-1}{2}+\frac{7}{3}+\frac{-1}{3}+\frac{5}{2}+\frac{-1}{5}+\frac{3}{5}+\left(-3\right)+\frac{5}{3}\)
\(=\left(\frac{3}{5}+\frac{-1}{5}+\frac{3}{5}\right)+\left(\frac{-1}{2}+\frac{5}{2}\right)+\left(\frac{7}{3}+\frac{-1}{3}+\frac{5}{3}\right)+\left(-3\right)\)
\(=\frac{5}{2}+\frac{4}{2}+\frac{11}{3}+\left(-3\right)\)
\(=1+2+\frac{11}{3}+\left(-3\right)\)
\(=0+\frac{11}{3}\)
\(=\frac{11}{3}\)
\(\left(\frac{3}{5}-\frac{1}{2}+\frac{7}{3}\right)-\left(\frac{1}{3}-\frac{5}{2}+\frac{1}{5}\right)-\left(-\frac{3}{5}+3-\frac{5}{3}\right)\)
\(=\frac{3}{5}-\frac{1}{2}+\frac{7}{3}-\frac{1}{3}+\frac{5}{2}-\frac{1}{5}+\frac{3}{5}-3+\frac{5}{3}\)
\(=\left(\frac{3}{5}-\frac{1}{5}+\frac{3}{5}\right)+\left(-\frac{1}{2}+\frac{5}{2}\right)+\left(\frac{7}{3}-\frac{1}{3}+\frac{5}{3}\right)-3\)
\(=1+2+\frac{11}{3}-3\)
\(=\left(1+2-3\right)+\frac{11}{3}\)
\(=0+\frac{11}{3}\)
=\(\frac{11}{3}\)
(Bài làm có gì ko hiểu bn cứ hỏi mk nhé
^...^ ^_^)