Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(F=\left(9,75.21\frac{3}{7}+\frac{39}{4}.18\frac{4}{7}\right).\frac{15}{78}=\left(\frac{39}{4}.21\frac{3}{7}+\frac{39}{4}.18\frac{4}{7}\right).\frac{15}{78}=\left(\frac{39}{4}\left(21\frac{3}{7}+18\frac{4}{7}\right)\right).\frac{15}{78}=\left(\frac{39}{4}.40\right).\frac{15}{78}=390.\frac{15}{78}=75\)\(H=\left(9\frac{30303}{80808}+7\frac{303030}{484848}\right)+4,03=\left(9\frac{3}{8}+7\frac{30}{48}\right)+4,03=17+4,03=21,03\)
I=10101(5/111111+5/222222+4/3.7.11.13.37)
I=10101(5/111111+5/222222+4/111111)
I=10101(10/222222+5/222222+8/222222)
I=10101.23/222222
I=232323/222222
\(H=\left(9\frac{3}{8}+7\frac{3}{8}\right)+4,03=16\frac{3}{8}+4,03=16,375+4,03=20,405\)
\(I=10101.\left(\frac{5}{111111}+\frac{2,5}{111111}-\frac{4}{111111}\right)=10101.\frac{3,5}{111111}=\frac{7}{22}\)
\(H=\left(9\frac{30303}{80808}+7\frac{303030}{484848}\right)+4.03\)
\(H=\left(\frac{75}{8}+\frac{61}{8}\right)+\frac{403}{100}\)
\(H=17+\frac{403}{100}\)
\(H=\frac{2103}{100}\)
\(I=10101.\left(\frac{5}{111111}+\frac{5}{222222}-\frac{4}{3.7.11.13.37}\right)\)
\(I=10101.\left(\frac{5}{111111}+\frac{5}{222222}-\frac{4}{111111}\right)\)
\(I=10101.\left(\frac{5}{111111}-\frac{4}{111111}+\frac{5}{222222}\right)\)
\(I=10101.\left(\frac{1}{111111}+\frac{5}{222222}\right)\)
\(I=10101.\frac{7}{222222}\)
\(I=\frac{7}{22}\)
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
\(A=\left(9+\frac{30303}{80808}+7+\frac{303030}{484848}\right)+4,03\)
\(A=\left(9+\frac{3\times10101}{8\times10101}+7+\frac{30\times10101}{48\times10101}\right)+4,03\)
\(A=\left(9+\frac{3}{8}+7+\frac{5}{8}\right)+4,03\)
\(A=\left(9+7+\frac{3}{8}+\frac{5}{8}\right)+4,03\)
\(A=\left(16+1\right)+4,03\)
\(A=17+4,03=21,03\)