bữa sau đăng ít thôi ._.
1,
\(a,\dfrac{3}{5}+\dfrac{7}{5}=\dfrac{10}{5}=2\)
\(b,\dfrac{1}{6}+\dfrac{-5}{3}=\dfrac{1}{6}+\dfrac{-10}{6}=\dfrac{-9}{6}=\dfrac{-3}{2}\)
\(c,\dfrac{1}{8}-\dfrac{1}{2}=\dfrac{1}{8}-\dfrac{4}{8}=\dfrac{-3}{8}\)
\(d,-5\cdot\dfrac{1}{3}=\dfrac{-5}{3}\)
\(e,\dfrac{-1}{3}\cdot\dfrac{5}{7}=\dfrac{-5}{21}\)
\(f,\dfrac{2}{7}:\dfrac{3}{4}=\dfrac{2}{7}\cdot\dfrac{4}{3}=\dfrac{8}{21}\)
2,
\(a,0,75+\dfrac{-1}{3}-\dfrac{5}{18}=\dfrac{3}{4}+\dfrac{-1}{3}-\dfrac{5}{18}\)
\(=\dfrac{5}{12}-\dfrac{5}{18}=\dfrac{5}{36}\)
\(b,7\dfrac{3}{5}-\left(2\dfrac{5}{7}+5\dfrac{3}{5}\right)\)
\(=\dfrac{38}{5}-\left(\dfrac{19}{7}+\dfrac{28}{5}\right)\)
\(=\dfrac{38}{5}-\dfrac{291}{35}\)
\(=-\dfrac{5}{7}\)
\(c,\dfrac{4}{15}\cdot\dfrac{1}{3}\cdot\dfrac{15}{20}=\dfrac{1}{15}\)
\(d,\dfrac{-1}{9}\cdot\dfrac{15}{22}:\dfrac{-25}{9}\)
\(=\dfrac{-1}{9}\cdot\dfrac{15}{22}\cdot\dfrac{9}{-25}\)
\(=\dfrac{3}{110}\)
3,
\(a,\dfrac{2}{3}+20\%\cdot\dfrac{10}{7}\)
\(=\dfrac{2}{3}+\dfrac{1}{5}\cdot\dfrac{10}{7}\)
\(=\dfrac{2}{3}+\dfrac{2}{7}\)
\(=\dfrac{20}{21}\)
\(b,\dfrac{3}{4}+1\dfrac{4}{5}:\dfrac{3}{2}-1\)
\(=\dfrac{3}{4}+\dfrac{9}{5}\cdot\dfrac{2}{3}-1\)
\(=\dfrac{3}{4}+\dfrac{6}{5}-1\)
\(=\dfrac{19}{20}\)
\(c,\left(2-\dfrac{1}{2}\right)\left(\dfrac{-3}{4}+\dfrac{1}{2}\right)\)
\(=\dfrac{3}{2}\cdot\dfrac{-1}{4}=\dfrac{-3}{8}\)
\(d,-1,5\cdot\left(\dfrac{7}{3}-\dfrac{5}{3}\cdot4\right)\)
\(=-\dfrac{3}{2}\cdot\dfrac{-13}{3}\)
\(=\dfrac{13}{2}\)
4,
\(a,\dfrac{-3}{4}+\dfrac{2}{7}+\dfrac{-1}{4}+\dfrac{3}{5}+\dfrac{5}{7}\)
\(=\left(\dfrac{-3}{4}+\dfrac{-1}{4}\right)+\left(\dfrac{2}{7}+\dfrac{5}{7}\right)+\dfrac{3}{5}\)
\(=-1+1+\dfrac{3}{5}=\dfrac{3}{5}\)
\(b,\dfrac{6}{21}-\dfrac{-12}{44}+\dfrac{10}{14}-\dfrac{1}{-4}-\dfrac{18}{33}\)
\(=\dfrac{43}{44}\)
\(c,-\dfrac{5}{8}\cdot\dfrac{-12}{29}\cdot\dfrac{8}{-10}\cdot5,8\)
\(=-\dfrac{6}{5}\)
\(d,\left(\dfrac{67}{111}+\dfrac{2}{33}-\dfrac{15}{117}\right)\left(\dfrac{1}{3}-25\%-\dfrac{1}{12}\right)\)
\(=\left(\dfrac{67}{111}+\dfrac{2}{33}-\dfrac{15}{117}\right)0\)
\(=0\)
5,
\(a,\dfrac{9}{17}\cdot\dfrac{3}{7}+\dfrac{9}{17}:\dfrac{7}{4}\)
\(=\dfrac{9}{17}\cdot\left(\dfrac{3}{7}+\dfrac{4}{7}\right)\)
\(=\dfrac{9}{17}\)
\(b,\left(\dfrac{-9}{21}\right)\cdot17\dfrac{2}{3}-\left(\dfrac{-3}{5}\right)^2\cdot\dfrac{22}{3}\)
\(=\left(\dfrac{-9}{21}\right)\cdot\dfrac{53}{3}-\dfrac{9}{25}\cdot\dfrac{22}{3}\)
\(=-\dfrac{53}{7}-\dfrac{66}{25}\)
\(=-\dfrac{1787}{175}\)
\(c,\dfrac{6}{7}\cdot\dfrac{8}{13}+\dfrac{6}{13}\cdot\dfrac{9}{7}-\dfrac{4}{13}\cdot\dfrac{6}{7}\)
\(=\dfrac{6}{7}\cdot\left(\dfrac{8}{13}+\dfrac{9}{13}-\dfrac{4}{13}\right)\)
\(=\dfrac{6}{7}\)
câu d bài 5 làm lộn ở bài 4 :")
\(4d,\dfrac{3}{7}\cdot\left(-\dfrac{2}{5}\right)\cdot2\dfrac{1}{3}\cdot20\cdot\dfrac{19}{72}\)
\(=-\dfrac{6}{35}\cdot\dfrac{7}{3}\cdot\dfrac{380}{72}\)
\(=-\dfrac{19}{9}\)
câu 6 ko lm .-.
