a) \(49\dfrac{8}{23}-\left(5\dfrac{7}{32}+14\dfrac{8}{23}\right)\)
\(=\dfrac{1135}{23}-\left(\left(5+14\right)+\left(\dfrac{7}{32}+\dfrac{8}{23}\right)\right)\)
\(=\dfrac{1135}{23}-\left(19+\dfrac{417}{736}\right)\)
\(=\dfrac{1135}{23}-19\dfrac{417}{736}\)
\(=\dfrac{1135}{23}-\dfrac{14401}{736}\)
\(=\dfrac{953}{32}\)
b) \(-\dfrac{3}{7}\cdot\dfrac{5}{9}+\dfrac{4}{9}\cdot\dfrac{-3}{7}+2\dfrac{3}{7}\)
\(=-\dfrac{1}{7}\cdot\dfrac{5}{3}-\dfrac{4}{3}\cdot\dfrac{1}{7}+\dfrac{17}{7}\)
\(=-\dfrac{5}{21}-\dfrac{4}{21}+\dfrac{17}{7}\)
\(=2\)
c) \(0,7\cdot2\dfrac{2}{3}\cdot20\cdot0,375\cdot\dfrac{5}{28}\)
\(=\dfrac{7}{10}\cdot\dfrac{8}{3}\cdot20\cdot\dfrac{3}{8}\cdot\dfrac{5}{28}\)
\(=2\cdot\dfrac{5}{4}\)
\(=\dfrac{5}{2}\)
d) \(\left(9\dfrac{30303}{80808}+7\dfrac{303030}{484848}\right)+4,03\)
\(=\left(9\cdot\dfrac{3}{8}+\dfrac{303030}{69264}\right)+\dfrac{403}{100}\)
\(=\left(\dfrac{27}{8}+\dfrac{35}{8}\right)+\dfrac{403}{100}\)
\(=\dfrac{31}{4}+\dfrac{403}{100}\)
\(=\dfrac{589}{50}\)
P/s: Đánh dấu phẩy, dấu chấm (dấu nhân) cần rõ ràng (vì dấu chấm người ta sẽ hiểu là dấu nhân thay vì hiểu là dấu phẩy)
a) \(49\dfrac{8}{23}\)- \(\left(5\dfrac{7}{32}+14\dfrac{8}{23}\right)\)
= \(\left(49\dfrac{8}{23}-14\dfrac{8}{23}\right)+5\dfrac{7}{32}\)
=35+\(5\dfrac{7}{32}\)
=\(\dfrac{1287}{32}\)
b)\(-\dfrac{3}{7}.\dfrac{5}{9}+\dfrac{4}{9}.\dfrac{-3}{7}+2\dfrac{3}{7}\)
=\(\left[\left(\dfrac{-3}{7}\right).\left(\dfrac{5}{9}+\dfrac{4}{9}\right)\right]+2\dfrac{3}{7}\)
=\(\left[\left(\dfrac{-3}{7}\right).\dfrac{9}{9}\right]+2\dfrac{3}{7}\)
=\(\left[\left(\dfrac{-3}{7}\right).1\right]+2\dfrac{3}{7}\)
=\(\left(\dfrac{-3}{7}\right)+2\dfrac{3}{7}\)
=2
c) 0,7.\(2\dfrac{2}{3}\).20.0.375.\(\dfrac{5}{28}\)
=0 (Vì có một thừ số là 0 nên nguyên cả tích là 0)
d)\(\left(9\dfrac{30303}{80808}+7\dfrac{303030}{484848}\right)+4,03\)
=17+4,03
=21,03
