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\(=\left(\dfrac{1}{2}+\dfrac{4}{5}-\dfrac{4}{3}\right)\left(2.3+4.28-1.28\right)\)

\(=\dfrac{15+24-40}{30}\cdot\left(2.3+3\right)\)

\(=\dfrac{-1}{30}\cdot\dfrac{53}{10}=\dfrac{-53}{300}\)

19 tháng 4 2017

Thực hiện các phép tính:

a) 9,6.212−(2.125−1512):149,6.212−(2.125−1512):14

b) 518−1,456:725+4,5.45518−1,456:725+4,5.45;

c) (12+0,8−113).(2,3+4725−1,28)(12+0,8−113).(2,3+4725−1,28)

d) (−5).12:[(−14)+12:(−2)]+113(−5).12:[(−14)+12:(−2)]+113.

Hướng dẫn làm bài:

a) 9,6.212−(2.125−1512):149,6.212−(2.125−1512):14

=9,6.52−(250−1712)×4=9,6.52−(250−1712)×4

=4,8.5−(1000−173)=4,8.5−(1000−173)

=24−1000+173=24−1000+173

=−976+173=−976+173

=−97013=−97013

b) 518−1,456:725+4,5.45518−1,456:725+4,5.45;

=518−1,456×257+92.45=518−1,456×257+92.45

=518−0,208×25+185=518−0,208×25+185

=518−5,2+185=518−5,2+185

=25−468+32490=25−468+32490

=−11990=−11990

c) (12+0,8−113).(2,3+4725−1,28)(12+0,8−113).(2,3+4725−1,28)

=(12+45−43).(2310+10725−3225)=(12+45−43).(2310+10725−3225)

=(15+24−4030).(2310+10725−3225)=(15+24−4030).(2310+10725−3225)

=(15+24−4030).(115+214−6450)=(15+24−4030).(115+214−6450)

=−130.26550=−130.26550

=−53300=−53300

d) (−5).12:[(−14)+12:(−2)]+113(−5).12:[(−14)+12:(−2)]+113

=−60:[14+12×(−12)]+1.13=−60:[14+12×(−12)]+1.13

=−60:[−14−14]+113=−60:[−14−14]+113

=−60:(12)+113=−60:(12)+113

=120+113=120+113

=12113

19 tháng 4 2017

a) \(9,6.2\dfrac{1}{2}-\left(2.125-1\dfrac{5}{12}\right):\dfrac{1}{4}\)

\(=9,6.\dfrac{5}{2}-\left(250-\dfrac{17}{12}\right).4\)

\(=4,8.5-\left(1000-\dfrac{17}{3}\right)\)

\(=24-1000+\dfrac{17}{3}\)

\(=-976+\dfrac{17}{3}=-970\dfrac{1}{3}\)

b) \(\dfrac{5}{18}-1,456:\dfrac{7}{25}+4,5.\dfrac{4}{5}\)

\(=\dfrac{5}{18}-1,456.\dfrac{25}{7}+\dfrac{9}{2}.\dfrac{4}{5}\)

\(=\dfrac{5}{18}-0,208.25+\dfrac{18}{5}\)

\(=\dfrac{5}{18}-5,2+\dfrac{18}{5}\)

\(=-\dfrac{119}{90}\)

c) \(\left(\dfrac{1}{2}+0,8-1\dfrac{1}{3}\right).\left(2,3+4\dfrac{7}{25}-1,28\right)\)

\(=\left(\dfrac{1}{2}+\dfrac{4}{5}-\dfrac{4}{3}\right).\left(\dfrac{23}{10}+\dfrac{107}{25}-\dfrac{32}{25}\right)\)

\(=-\dfrac{1}{30}.\dfrac{265}{50}=-\dfrac{53}{300}\)

d) \(\left(-5\right).12:\left[\left(-\dfrac{1}{4}\right)+\dfrac{1}{2}:\left(-2\right)\right]+1\dfrac{1}{3}\)

\(=-60:\left[\dfrac{1}{4}+\dfrac{1}{2}.\dfrac{-1}{2}\right]+1.\dfrac{1}{3}\)

\(=-60:\left[-\dfrac{1}{4}-\dfrac{1}{4}\right]+1\dfrac{1}{3}\)

\(=-60:\left(\dfrac{1}{2}\right)+1\dfrac{1}{3}\)

\(=121\dfrac{1}{3}\)

18 tháng 10 2021

\(\dfrac{\dfrac{4}{5}:\left(\dfrac{4}{5}\cdot\dfrac{5}{4}\right)}{\dfrac{16}{25}-\dfrac{1}{25}}+\dfrac{\left(\dfrac{27}{25}-\dfrac{2}{25}\right):\dfrac{4}{7}}{\left(\dfrac{59}{9}-\dfrac{13}{4}\right)\cdot\dfrac{36}{17}}+\left(\dfrac{6}{5}\cdot\dfrac{1}{2}\right):\dfrac{4}{5}\)

\(=\dfrac{4}{5}:\dfrac{3}{5}+\dfrac{7}{4}:7+\dfrac{3}{5}:\dfrac{4}{5}\)

\(=\dfrac{4}{3}+\dfrac{1}{4}+\dfrac{3}{4}\)

\(=\dfrac{7}{3}\)

19 tháng 10 2021

thanks bn nhìuvuiyeu

9 tháng 7 2018

\(=\dfrac{0,8:1}{0,6}+\dfrac{1:\dfrac{4}{7}}{3\dfrac{11}{36}.2\dfrac{2}{17}}+0,6:\dfrac{4}{5}\)

\(=\dfrac{0,4}{0,3}+\dfrac{\dfrac{7}{4}}{7}+0,75\)

\(=1\dfrac{0,4}{0,3}\)

15 tháng 3 2019

đặt M=\(\dfrac{0,8:\left(\dfrac{4}{5}.1.25\right)}{0,64-\dfrac{1}{5}}=\dfrac{4}{3}\)

N=\(\dfrac{\left(1,08-\dfrac{2}{25}\right):\dfrac{4}{7}}{\left(6\dfrac{5}{9}-3\dfrac{1}{4}\right).2\dfrac{2}{17}=\dfrac{1}{4}}\)

M+N=\(\dfrac{4}{3}+\dfrac{1}{4}=1\dfrac{1}{17}\)

5 tháng 11 2017

a. \(10,\left(3\right)+0,\left(4\right)-8,\left(6\right)\)

\(\Leftrightarrow\dfrac{31}{3}+\dfrac{4}{9}-\dfrac{26}{3}\)

\(=\dfrac{19}{9}\)

b. \(\dfrac{0,8:\left(\dfrac{4}{5}.1,25\right)}{0,64-\dfrac{1}{25}}+\dfrac{\left(1,08-\dfrac{2}{25}\right):\dfrac{4}{7}}{\left(6\dfrac{5}{9}-3\dfrac{1}{4}\right).2\dfrac{2}{17}}+\left(1,2.0,5\right):\dfrac{4}{5}\)

\(=\dfrac{0,8}{0,6}+\dfrac{1,75}{7}+0,6:\dfrac{4}{5}\)

\(=\dfrac{0,8}{0,6}+\dfrac{1,75}{7}+\dfrac{3}{4}\)

\(=\dfrac{7}{3}\)

5 tháng 11 2017

a, \(\dfrac{19}{9}\)

b, \(\dfrac{7}{3}\)

26 tháng 6 2017

\(\dfrac{0,8\left(\dfrac{4}{5}.1,25\right)}{0,64-\dfrac{1}{25}}+\dfrac{\left(10,8-\dfrac{2}{25}\right):\dfrac{4}{7}}{\left(6\dfrac{5}{9}-3\dfrac{1}{4}.2\dfrac{2}{17}\right)}\)

\(=\dfrac{0,8}{0,6}+\dfrac{\dfrac{268}{25}.\dfrac{7}{4}}{\left(\dfrac{59}{9}-\dfrac{13}{4}.\dfrac{36}{17}\right)}\)

\(=\dfrac{4}{3}+\dfrac{\dfrac{469}{25}}{\dfrac{-50}{153}}\)

\(=\dfrac{4}{3}-57,4056\)

\(=-56,07\)

26 tháng 6 2017

Đấy là tính nhanh?

1: Ta có: \(23\dfrac{1}{4}\cdot\dfrac{7}{5}-13\dfrac{1}{4}:\dfrac{5}{7}\)

\(=\dfrac{93}{4}\cdot\dfrac{7}{5}-\dfrac{53}{4}\cdot\dfrac{7}{5}\)

\(=\dfrac{7}{5}\cdot10=14\)

2: Ta có: \(\left(1+\dfrac{2}{3}-\dfrac{1}{4}\right)\left(\dfrac{4}{5}-\dfrac{3}{4}\right)^2\)

\(=\dfrac{12+8-3}{12}\cdot\dfrac{1}{400}\)

\(=\dfrac{17}{12}\cdot\dfrac{1}{400}=\dfrac{17}{4800}\)

a: \(=\left(1+\dfrac{4}{23}-\dfrac{4}{23}\right)+\left(\dfrac{5}{21}+\dfrac{16}{21}\right)+\dfrac{1}{2}\)

\(=1+1+\dfrac{1}{2}=2+\dfrac{1}{2}=\dfrac{5}{2}\)

b: \(=\left(\dfrac{1}{25}+\dfrac{5}{25}+\dfrac{25}{25}\right):\left(\dfrac{1}{25}-\dfrac{5}{25}-\dfrac{25}{25}\right)\)

\(=\dfrac{31}{25}:\dfrac{-29}{25}=\dfrac{-31}{29}\)

c: \(=\dfrac{\dfrac{1}{9}-\dfrac{1}{7}-\dfrac{1}{11}}{\dfrac{4}{9}-\dfrac{4}{7}-\dfrac{4}{11}}+\dfrac{\dfrac{3}{5}-\dfrac{3}{25}-\dfrac{3}{125}-\dfrac{3}{625}}{\dfrac{4}{5}-\dfrac{4}{25}-\dfrac{4}{125}-\dfrac{4}{625}}\)

=1/4+3/4

=1

13 tháng 9 2017

\(\dfrac{\dfrac{1}{9}-\dfrac{1}{7}-\dfrac{1}{11}}{\dfrac{4}{9}-\dfrac{4}{7}-\dfrac{4}{11}}+\dfrac{3\left(\dfrac{1}{5}-\dfrac{1}{25}-\dfrac{1}{125}-\dfrac{1}{625}\right)}{4\left(\dfrac{1}{5}-\dfrac{1}{25}-\dfrac{1}{125}-\dfrac{1}{625}\right)}\)

\(=\dfrac{\dfrac{1}{9}-\dfrac{1}{7}-\dfrac{1}{11}}{4\left(\dfrac{1}{9}-\dfrac{1}{7}-\dfrac{1}{11}\right)}+\dfrac{3}{4}\)

\(=\dfrac{1}{4}+\dfrac{3}{4}\)

\(=1\)

13 tháng 9 2017

\(\dfrac{\dfrac{1}{9}-\dfrac{1}{7}-\dfrac{1}{11}}{\dfrac{4}{9}-\dfrac{4}{7}-\dfrac{4}{11}}+\dfrac{3\left(\dfrac{1}{5}-\dfrac{1}{25}-\dfrac{1}{125}-\dfrac{1}{625}\right)}{4\left(\dfrac{1}{5}-\dfrac{1}{25}-\dfrac{1}{125}-\dfrac{1}{625}\right)}\)

\(=\dfrac{\dfrac{1}{9}-\dfrac{1}{7}-\dfrac{1}{11}}{4\left(\dfrac{1}{9}-\dfrac{1}{7}-\dfrac{1}{11}\right)}+\dfrac{3\left(\dfrac{1}{5}-\dfrac{1}{25}-\dfrac{1}{125}-\dfrac{1}{625}\right)}{4\left(-\dfrac{1}{25}-\dfrac{1}{125}-\dfrac{1}{625}\right)}=\dfrac{1}{4}+\dfrac{3}{4}=1\)