\(4\frac{3}{4}+\left(-0,37\right)+\left(-1.28\right)+\left(-2,5\right)+3\frac{1}{2}\)
(các bn giải giúp mình với)
K
Khách
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KN
21 tháng 8 2020
\(8\frac{2}{7}-\left(1\frac{1}{6}+25\%\right)=\frac{58}{7}-\left(\frac{7}{6}+\frac{1}{4}\right)=\frac{58}{7}-\frac{17}{12}=\frac{577}{84}\)
\(4\frac{3}{4}+\left(-0,37\right)+\left(-1,28\right)+\left(-2,5\right)+3\frac{1}{12}\)
\(=\frac{19}{4}+\left(-\frac{83}{20}\right)+\frac{37}{12}=\frac{3}{5}+\frac{37}{12}=\frac{221}{60}\)
21 tháng 8 2020
\(8\frac{2}{7}-\left(1\frac{1}{6}+25\%\right)=\frac{58}{7}-\left(\frac{7}{6}+\frac{1}{4}\right)=\frac{58}{7}-\frac{17}{12}=\frac{577}{84}\)
31 tháng 7 2016
\(\frac{19}{4}-\frac{37}{100}+\frac{1}{8}-\frac{32}{25}-\frac{5}{2}+\frac{7}{2}=\left(\frac{19}{4}-\frac{37}{100}-\frac{32}{25}\right)+\left(\frac{7}{2}-\frac{5}{2}\right)+\frac{1}{8}=\frac{31}{10}+1+\frac{1}{8}=\frac{169}{40}\)
24 tháng 5 2017
(1-1/3).(1-1/5).(1-1/7).(1-1/9).(1-1/11).(1-1/13).(1-1/2).(1-1/4).(1-1/6).(1-1/8).(1-1/10)
=2/3.4/5.6/7.8/9.10/11.12/13.1/2.3/4.5/6.7/8.9/10
=8/15.48/63.120/143.3/8.35/48.9/10
=384/945.360/1144.315/480
=138240/1081080.315/480
=43545600/518918400=84/1001
31 tháng 1 2022
a: \(=\dfrac{17}{4}-\dfrac{37}{100}+\dfrac{1}{8}-\dfrac{32}{25}-\dfrac{5}{2}+\dfrac{7}{2}\)
\(=\dfrac{35}{8}+\dfrac{8}{8}-\dfrac{37}{100}-\dfrac{128}{100}\)
\(=\dfrac{43}{8}-\dfrac{165}{100}=\dfrac{149}{40}\)
b: \(=\left(\dfrac{22\cdot26+3\cdot10-65}{130}\right):\left(\dfrac{4\cdot22-2\cdot26+3\cdot143}{286}\right)\)
\(=\dfrac{537}{130}\cdot\dfrac{286}{465}=\dfrac{1969}{775}\)
M
5 tháng 2 2016
Câu một \(=3^2.\frac{1}{3^5}.\left(3^4\right)^2.\frac{1}{3^3}=3^{10}.\frac{1}{3^8}=3^2=9\)
Câu hai \(=\left(2^2.2^5\right):\left(2^3.\frac{1}{2^4}\right)=\frac{2^7}{\frac{2^3}{2^4}}=2^8=256\)
Chờ chút nhá :D
27 tháng 3 2020
a) \(\frac{x+5}{4}-\frac{2x-3}{3}=\frac{6x-1}{8}+\frac{2x-1}{12}\)
<=> \(\frac{x}{4}+\frac{5}{4}-\frac{2x}{3}+1=\frac{6x}{8}-\frac{1}{8}+\frac{2x}{12}-\frac{1}{12}\)
<=> \(-\frac{4}{3}x=-\frac{59}{24}\)
<=> \(x=\frac{59}{32}\)
Vậy S = { 59/32}
b) \(\frac{\left(x+10\right)\left(x+4\right)}{12}-\frac{\left(x+4\right)\left(2-x\right)}{4}=\frac{\left(x+10\right)\left(x-2\right)}{3}\)
<=> \(\frac{x^2+14x+40}{12}-\frac{-x^2-2x+8}{4}=\frac{x^2+8x-20}{3}\)
<=> \(\left(\frac{x^2}{12}+\frac{x^2}{4}-\frac{x^2}{3}\right)+\left(\frac{14}{12}x+\frac{2}{4}x-\frac{8}{3}x\right)=-\frac{20}{8}+\frac{8}{4}-\frac{40}{12}\)
<=> \(-x=-8\)
<=> x = 8
Vậy S = { 8 }
\(=\frac{19}{4}+\left(-\frac{37}{100}\right)+\left(-\frac{32}{25}\right)+\left(-\frac{5}{2}\right)+\frac{7}{2}\)
\(=\frac{219}{50}+\left(-\frac{32}{25}\right)+\left(-\frac{5}{2}\right)+\frac{7}{2}\)
\(=\frac{31}{10}+\left(-\frac{5}{2}\right)+\frac{7}{2}\)
\(=\frac{31}{10}+\left(-\frac{25}{10}\right)+\frac{35}{10}\)
\(=\frac{41}{10}\)
\(4\frac{3}{4}+\left(-0,37\right)+\left(-1,28\right)+\left(-2,5\right)+3\frac{1}{2}\)
\(=\frac{19}{4}-\frac{37}{100}-\frac{128}{100}-\frac{25}{10}+\frac{7}{2}\)
\(=\frac{475}{100}-\frac{37}{100}-\frac{128}{100}-\frac{250}{100}+\frac{350}{100}\)
\(=\frac{410}{100}=4,1\)