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\(\frac{\left(\frac{3}{10}-\frac{4}{15}-\frac{7}{20}\right)\cdot\frac{5}{19}}{\left(\frac{1}{14}+\frac{1}{7}+\frac{3}{35}\right)\cdot\left(\frac{-4}{3}\right)}=\frac{\left(\frac{18}{60}-\frac{16}{60}-\frac{21}{60}\right)\cdot\frac{5}{19}}{\left(\frac{5}{70}+\frac{10}{70}+\frac{6}{70}\right)\cdot\left(\frac{-4}{3}\right)}=\frac{\left(\frac{-19}{60}\right)\cdot\frac{5}{19}}{\frac{3}{10}\cdot\left(\frac{-4}{3}\right)}\)
\(=\frac{\frac{-1}{12}}{\frac{-2}{5}}=\left(\frac{-1}{12}\right):\left(\frac{-2}{5}\right)=\frac{5}{24}\)
\(\dfrac{\left[\left(\dfrac{3}{10}-\dfrac{4}{15}-\dfrac{7}{20}\right)\cdot\dfrac{5}{19}\right]}{\left[\dfrac{1}{14}+\dfrac{1}{7}-\dfrac{\left(-3\right)}{35}\right]\cdot\left(-\dfrac{4}{3}\right)}\)
\(=\dfrac{\left(\dfrac{18}{60}-\dfrac{16}{60}-\dfrac{21}{60}\right)\cdot\dfrac{5}{19}}{\left(\dfrac{5}{70}+\dfrac{10}{70}+\dfrac{6}{70}\right)\cdot\dfrac{-4}{3}}=\dfrac{\dfrac{-19}{60}\cdot\dfrac{5}{19}}{\dfrac{21}{70}\cdot\dfrac{-4}{3}}\)
\(=-\dfrac{1}{12}:\dfrac{-2}{5}=\dfrac{1}{12}\cdot\dfrac{5}{2}=\dfrac{5}{24}\)
*\(\frac{\left(\frac{3}{10}-\frac{4}{15}-\frac{7}{20}\right).\frac{5}{19}}{\left[\frac{1}{14}+\frac{1}{7}-\left(-\frac{3}{35}\right)\right].\frac{4}{3}}=\frac{\left(\frac{18}{60}-\frac{16}{60}-\frac{21}{60}\right).\frac{5}{19}}{\left(\frac{5}{70}+\frac{10}{70}+\frac{6}{70}\right).\frac{4}{3}}=\frac{\frac{-19}{60}.\frac{5}{19}}{\frac{21}{70}.\frac{4}{3}}=\frac{\frac{-1}{12}}{\frac{14}{35}}=-\frac{1}{12}.\frac{35}{14}=\frac{-35}{168}\)
*\(\frac{\left(1+2+3+...+100\right).\left(\frac{1}{3}-\frac{1}{5}-\frac{1}{7}-\frac{1}{9}\right).\left(6,3.12-21.3,6\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}\)
=\(\frac{\left(1+2+3+...+100\right)\left(\frac{1}{3}-\frac{1}{5}-\frac{1}{7}-\frac{1}{9}\right).\left(\frac{63}{10}.12-21.\frac{18}{5}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}\)
=\(\frac{\left(1+2+3+...+100\right)\left(\frac{1}{3}-\frac{1}{5}-\frac{1}{7}-\frac{1}{9}\right).\left(\frac{378}{5}-\frac{378}{5}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}\)
=\(\frac{\left(1+2+3+...+100\right)\left(\frac{1}{3}-\frac{1}{5}-\frac{1}{7}-\frac{1}{9}\right).0}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}=0\)
Silver bullet,Võ Đông Anh Tuấn,soyeon_Tiểubàng giải,Hoàng Lê Bảo Ngọc,
Lê Nguyên Hạo giúp cái.
\(\frac{\left(\frac{3}{10}-\frac{4}{15}-\frac{7}{20}\right).\frac{5}{19}}{\left(\frac{1}{14}+\frac{1}{7}-\frac{-3}{35}\right).\frac{-4}{3}}=\frac{\left(\frac{18}{60}-\frac{16}{60}-\frac{21}{60}\right).\frac{5}{19}}{\left(\frac{5}{70}+\frac{7}{70}-\frac{-6}{70}\right).\frac{-4}{3}}\)
\(=\frac{-\frac{19}{60}.\frac{5}{19}}{\frac{3}{10}.\frac{-4}{3}}=\frac{-\frac{1}{12}}{-\frac{2}{5}}=\frac{5}{24}\)