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\(44.179+20^2-79.44\\ =44\left(179-79\right)+400\\ =44.100+400\\ =100\left(44+4\right)=4800\\ 3.4^2:\left[500-\left(7.35+125\right)\right]\\ =3.16:\left[500-370\right]\\ =48:130=\dfrac{24}{65}\\ 187+\left[921-\left(921+887\right)\right]\\ =187+921-921-887\\ =187-887=-700\)
44. 179+ 202 - 79.44 3. 42 : [500- (7.35+125)]
= 44. 179+400 - 79.44 = 3. 16: [500- (245+125)]
= 44. (179-79) +400 = 48: [ 500-370 ]
= 44. 100+400 = 48:130
= 4400+400 = 0, 369...
= 4800
187+ [ 921-( 921+887) = 187+ [921- 1808]
= 187+ (-159)
= 28
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a ) 4 7 + 1 7 + 3 7 = 8 7 b ) 4 7 + − 3 7 + − 1 2 + 1 2 = 1 7
c ) − 27 10 + − 3 10 + 1 4 + 5 4 + 13 2 = − 3 + 3 2 + 13 2 = 5 d ) 8 5 + 7 5 + − 11 4 + − 9 4 + − 7 20 = 3 − 5 + − 7 20 = − 47 20 e ) 7 5 + − 1 15 + − 11 3 + 11 2 + 1 6 + ( − 2 ) = 4 3 + 2 + ( − 2 ) = 4 3
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a: 43/52>26/52=1/2=60/120
b: 17/68=1/4<1/3=35/105<35/103
c: \(\dfrac{2018\cdot2019-1}{2018\cdot2019}=1-\dfrac{1}{2018\cdot2019}\)
\(\dfrac{2019\cdot2020-1}{2019\cdot2020}=1-\dfrac{1}{2019\cdot2020}\)
2018*2019<2019*2020
=>-1/2018*2019<-1/2019*2020
=>\(\dfrac{2018\cdot2019-1}{2018\cdot2019}< \dfrac{2019\cdot2020-1}{2019\cdot2020}\)
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mik ko hiểu cách bạn ghi
a) 187+ [921-(921+887)] ??? ri hã
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\(\dfrac{19}{19}\) = 1 < \(\dfrac{2005}{2004}\) vậy \(\dfrac{19}{19}\) < \(\dfrac{2005}{2004}\)
\(\dfrac{72}{73}\) = 1 - \(\dfrac{1}{73}\)
\(\dfrac{98}{99}\) = 1 - \(\dfrac{1}{99}\)
Vì \(\dfrac{1}{73}\) > \(\dfrac{1}{99}\) nên \(\dfrac{72}{73}\) < \(\dfrac{98}{99}\)
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a) ta có: \(1-\frac{2012}{2013}=\frac{1}{2013}\)
\(1-\frac{2013}{2014}=\frac{1}{2014}\)
mà \(\frac{1}{2013}>\frac{1}{2014}\) nên \(\frac{2013}{2014}>\frac{2012}{2013}\)
\(9^{21}\)và \(729^7\)
Ta có : \(9^{21}=\left(9^3\right)^7=729^7\)
Vì \(729^7=729^7\)nên \(9^{21}=729^7\)
So sánh 921 và 7297
Ta có :
\(9^{21}=\left(9^3\right)^7=729^7\)
Vì \(729^7=729^7\)
Nên \(9^{21}=729^7\)