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20 tháng 11 2016
\(\frac{\left(-5\right)^2.20^4}{8^2.125}=\frac{25.20^4}{8^2.25.5}=\frac{20^4}{8^2.5}=500\)
SD
1
25 tháng 11 2018
\(\dfrac{\left(-5\right)^2.20^2}{8^2.125}=\dfrac{\left[\left(-5\right).20\right]^2}{2^{3^2}.5^3}=\dfrac{\left(-100\right)^2}{\text{4^3}.5^3}=\dfrac{10000}{\left(4.5\right)^3}=\dfrac{10000}{20^3}\dfrac{10000}{8000}=\dfrac{5}{4}\)
ST
6
12 tháng 3 2020
có
(-5)^2*20^4/862*125
=25*(4*5)^4/64*125
=25*4^4*5^4/64*125
=25*5*4^3*4*5^3/64*125
=125*64*4*5^3/64*125
=4*5^3
=500
PT
1
PK
3
17 tháng 9 2018
\(\dfrac{25^2.125^2}{625^5}=\dfrac{5^4.5^6}{5^{20}}=\dfrac{5^{10}}{5^{20}}=\dfrac{1}{5^{10}}.\)
TT
17 tháng 9 2018
\(\dfrac{25^2.125^2}{625^5}=\dfrac{\left(5^2\right)^2.\left(5^3\right)^2}{\left(5^4\right)^5}\) \(=\dfrac{5^4.5^6}{5^{20}}\)=\(\dfrac{5^{10}}{5^{20}}\) = \(5^{-5}\)=\(\dfrac{1}{9765626}\)
LN
0
S
19 tháng 3 2020
a) \(9,2\cdot2\frac{1}{2}-\left(2\cdot0,125-1\frac{5}{12}\right):\frac{1}{4}\)
\(=23-\left(0,25-\frac{17}{12}\right):\frac{1}{4}\)
\(=23-\left(-\frac{7}{6}\right):\frac{1}{4}\)
\(=23-\left(-\frac{14}{3}\right)\)
\(=\frac{83}{3}\)
b) \(\frac{\sqrt{3^2}-\sqrt{39^2}}{\sqrt{59-10-\sqrt{91^2}}}\)
\(=\frac{3-39}{\sqrt{59-10-91}}\)
\(=\frac{-36}{\sqrt{-42}}\)
Vì -42 < 0 \(\Leftrightarrow\)Biểu thức không tồn tại
c) \(\frac{5}{18}-1,456:\frac{7}{25}+4,5\cdot\frac{4}{5}\)
\(=\frac{5}{18}-\frac{26}{5}+\frac{18}{5}\)
\(=\frac{5}{18}-\frac{8}{5}\)
\(=\frac{-119}{90}\)
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
LH
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) \(\left(-5\right).12:\left[\left(-\dfrac{1}{4}\right)+\dfrac{1}{2}:\left(-2\right)\right]+1\dfrac{1}{3}\)
\(=-60:\left[\dfrac{1}{4}+\dfrac{1}{2}.\dfrac{-1}{2}\right]+1.\dfrac{1}{3}\)
\(=-60:\left[-\dfrac{1}{4}-\dfrac{1}{4}\right]+1\dfrac{1}{3}\)
\(=-60:\left(\dfrac{1}{2}\right)+1\dfrac{1}{3}\)
\(=121\dfrac{1}{3}\)
52.125.(2,5)-5.0,04
=52.53.\(\dfrac{32}{3125}\).\(\dfrac{4}{100}\)
=55.\(\dfrac{32}{5^5}\).\(\dfrac{1}{25}\)
= \(\dfrac{32}{25}\)
32/25