\(\left(6\dfrac{1}{5}:21,7-1,2\times5^3\right)\times1\dfrac{3}{4}...">
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2 tháng 9 2017

(31/5:217/10-1,2*125)*7/4-1/2

=(2/7-150)*7/4-1/2

=-1048/7*7/4-1/2

=-262-1/2

=-525/2

\(\left(6\dfrac{1}{5}:21,7-1,2\cdot5^3\right)\cdot1\dfrac{3}{4}-\dfrac{1}{2}\)

\(=\left(\dfrac{31}{5}\cdot\dfrac{10}{217}-1,2\cdot125\right)\cdot1,75-0,5\)

\(=\left(\dfrac{62}{217}-150\right)\cdot1,75-0,5\)

\(=\dfrac{62}{217}\cdot1,75-150\cdot1,75-0,5\)

\(=0,5-262,5-0,5\)

\(=-262,5\)

13 tháng 7 2018

\(a)\dfrac{-5}{21}-\dfrac{1}{3}+3\dfrac{1}{2}.\left(\dfrac{-2}{3}\right)^3\)

\(=\dfrac{-5}{21}+\dfrac{-7}{21}+\dfrac{7}{2}.\dfrac{-8}{27}\)

\(=-\dfrac{4}{7}+\dfrac{-28}{27}\)

\(=\dfrac{-108}{189}+\dfrac{-196}{189}\)

\(=-\dfrac{304}{189}\)

14 tháng 7 2018

\(b)-2\dfrac{1}{3}+\left(\dfrac{3}{8}-\dfrac{3}{4}\right)^3:\dfrac{5}{9}-\dfrac{1}{2}\)

\(=-\dfrac{7}{3}+\left(\dfrac{3}{8}-\dfrac{6}{8}\right)^3.\dfrac{9}{5}-\dfrac{1}{2}\)

\(=-\dfrac{7}{3}+\left(-\dfrac{3}{8}\right)^3.\dfrac{9}{5}-\dfrac{1}{2}\)

\(=-\dfrac{7}{3}+\dfrac{-27}{512}.\dfrac{9}{5}-\dfrac{1}{2}\)

\(=-\dfrac{7}{3}+\dfrac{-243}{2560}-\dfrac{1}{2}\)

\(=\dfrac{-17920}{7680}+\dfrac{-729}{7680}+\dfrac{-3840}{7680}\)

\(=\dfrac{-22489}{7680}\)

11 tháng 2 2018

\(B=\frac{1-\frac{1}{\sqrt{49}}+\frac{1}{49}-\frac{1}{\left(7\sqrt{7}\right)^2}}{\frac{\sqrt{64}}{2}-\frac{4}{7}+\frac{2^2}{7^2}-\frac{4}{343}}\)

\(B=\frac{1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}}{\frac{8}{2}-\frac{4}{7}+\frac{4}{49}-\frac{4}{343}}\)

\(B=\frac{\frac{343}{343}-\frac{49}{343}+\frac{7}{343}-\frac{1}{343}}{4-\frac{4}{7}+\frac{28}{343}-\frac{4}{343}}\)

\(B=\frac{\frac{300}{343}}{\frac{28}{7}-\frac{4}{7}+\frac{24}{343}}\)

\(B=\frac{\frac{300}{343}}{\frac{24}{7}+\frac{24}{343}}\)

\(B=\frac{\frac{300}{343}}{\frac{1323}{343}+\frac{24}{343}}\)

\(B=\frac{300}{343}:\frac{1347}{343}\)

\(B=\frac{100}{449}\)

11 tháng 2 2018

\(A=\frac{2^{12}.3^5-4^6.9^2}{\left(2^2.3\right)^6+8^4.3^5}-\frac{5^{10}.7^3-25^5.49^2}{\left(125.7\right)^3+5^9.14^3}\)

\(A=\frac{2^{12}.3^5-2^{12}.3^6}{2^{12}.3^6+2^{12}.3^5}-\frac{5^{10}.7^3-5^{10}.7^6}{5^9.7^3+5^9.2^3.7^3}\)

\(A=\frac{2^{12}.3^5\left(1-3\right)}{2^{12}.3^5.\left(3+1\right)}-\frac{5^{10}.7^3.\left(1-7^3\right)}{5^9.7^3.\left(1+8\right)}\)

\(A=\frac{-2}{4}-\frac{5.\left(-342\right)}{9}\)

\(A=\frac{-1}{2}+\frac{1710}{9}\)

\(A=\frac{-1}{2}+190\)

\(A=\frac{-1}{2}+\frac{380}{2}\)

\(A=\frac{379}{2}\)

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

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

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

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

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

27 tháng 8 2017

a)\(\frac{5}{3}\)+ \(\left(\frac{-2}{7}\right)\)-(-1,2)

=\(\frac{5}{3}+\left(\frac{-2}{7}\right)+\frac{6}{5}\)                                          

=\(\frac{175+\left(-30\right)+126}{105}\)

=\(\frac{271}{105}\)

b) \(\frac{-4}{9}+\frac{-5}{6}-\frac{17}{4}\)

=\(\frac{-16+\left(-30\right)-153}{36}\)

=\(\frac{-199}{36}\)

14 tháng 12 2017

\(\frac{5}{3}+\left(\frac{-2}{7}\right)-\left(\frac{-6}{5}\right)\)

=\(\frac{-2}{7}-\left(\frac{5}{3}+\frac{-6}{5}\right)\)

=\(\frac{-79}{105}\)\(\frac{-2}{7}-\frac{7}{15}\)

26 tháng 7 2018

a. = \(\dfrac{-1}{24}-\left\{\dfrac{1}{4}-\dfrac{-3}{8}\right\}\)

= \(\dfrac{-1}{24}-\left\{\dfrac{1}{4}+\dfrac{3}{8}\right\}\)

= \(\dfrac{-1}{24}-\dfrac{5}{8}\)

= \(\dfrac{-2}{3}\)

b. = \(12\dfrac{7}{88}-3\dfrac{5}{11}\)

= \(8\dfrac{5}{8}\)

c. = \(\dfrac{-28}{9}+\dfrac{-413}{9}\)

= \(-49\)

d. = \(\dfrac{8}{35}:\dfrac{2}{11}+\dfrac{-8}{35}:\dfrac{2}{11}\)

= \(\dfrac{2}{11}:\left(\dfrac{8}{35}+\dfrac{-8}{35}\right)\)

= 0

19 tháng 11 2018

5) \(\left(-2\right)^2+\sqrt{36}-\sqrt{9}+\sqrt{25}\)

=\(4+6-3+5\)

=\(12\)

19 tháng 11 2018

2) \(\dfrac{11}{25}.\left(-24,8\right)-\dfrac{11}{25}.75,2\)

=\(\dfrac{11}{25}.\left(-24,8-75,2\right)\)

=\(\dfrac{11}{25}.\left(-100\right)\)

=\(-44\)

3 tháng 8 2017

a, \(\dfrac{20^5.5^{10}}{100^5}=\dfrac{20^5.5^{10}}{\left(20.5\right)^5}=\dfrac{20^5.5^{10}}{20^5.5^5}=5^5\)

3 tháng 8 2017

b,\(\dfrac{\left(0,9\right)^5}{\left(0,3\right)^6}=\dfrac{\left(0,3.3\right)^5}{\left(0,3\right)^6}=\dfrac{\left(0,3\right)^5.3^5}{\left(0,3\right)^6}=\dfrac{3^5}{\left(0,3\right)}\)

2 tháng 9 2017

B=8+3.1/4-1/4+(4:1/2).8

=8+3/4-1/4+8.8

=8+3/4-1/4+64

=35/4-275/4

=-60

23 tháng 9 2018

Ta có:

\(A=\frac{3}{1\cdot5}+\frac{3}{5\cdot10}+...+\frac{3}{100\cdot105}\)

\(=\frac{3}{5}\cdot\left(\frac{5}{1\cdot5}+\frac{5}{5\cdot10}+...+\frac{5}{100\cdot105}\right)\)

\(=\frac{3}{5}\cdot\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{10}+...+\frac{1}{100}-\frac{1}{105}\right)\)

\(=\frac{3}{5}\left(1-\frac{1}{105}\right)=\frac{3}{5}\cdot\frac{104}{105}=\frac{312}{525}\)