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\(\left(\dfrac{1}{2}\right)^3\cdot\left(\dfrac{1}{4}\right)^2\)
\(=\left(\dfrac{1}{2}\right)^3\cdot\left[\left(\dfrac{1}{2}\right)^2\right]^2\)
\(=\left(\dfrac{1}{2}\right)^3\cdot\left(\dfrac{1}{2}\right)^{2\cdot2}\)
\(=\left(\dfrac{1}{2}\right)^3\cdot\left(\dfrac{1}{2}\right)^4\)
\(=\left(\dfrac{1}{2}\right)^{3+4}\)
\(=\left(\dfrac{1}{2}\right)^7\)
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(- 47 ) - 53 = (- 47 ) + ( - 53 ) = -100
(- 43 ) - ( -43 ) =0
Ta có : -(4+7) =-11
( -4 ) - 7 = -11
Vì - ( 4+7 ) = -11 , ( -4 ) - 7 =-11 nên -(4 +7) = ( - 4) -7
Ta có : -( 12 -25 ) = 13
( -12 + 25 )=13
Vì -(12 -25 ) = 13; (-12 +25) = 13 nên - (12 - 25 ) = ( -12 +25 )
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`x/2+x+x/3+x+x+x/4=5 3/4`
`=>3x+x/2+x/3+x/4=23/4`
`=>49/12x=23/4`
`=>x=69/49`
Vậy `x=69/49`
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33000 = ( 33 )1000 = 271000
42000 =( 42 )1000 = 161000
Vì 271000 > 161000 nên 33000 > 42000
Ta co : 33000 = (3.3)1000 = 91000
42000 = (4.2)1000 = 81000
Vì : 9 > 8 nên 91000 > 81000.
Vay 33000 > 42000
k mk nha bn
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*Chứng minh A chia hết cho 4
Ta có: \(A=\left(3^1+3^2\right)+\left(3^3+3^4\right)+...+\left(3^{2015}+3^{2016}\right)\)
\(=3^1.\left(1+3\right)+3^3\left(1+3\right)+...+3^{2015}\left(1+3\right)\)
\(=4\left(3^1+3^3+...+3^{2015}\right)⋮4^{\left(đpcm\right)}\)
*Chứng minh A chia hết cho 13
Ta có: \(A=\left(3^1+3^2+3^3\right)+...+\left(3^{2014}+3^{2015}+3^{2016}\right)\)
\(=3\left(1+3^1+3^2\right)+...+3^{2014}\left(1+3^1+3^2\right)\)
\(=13\left(3+...+3^{2014}\right)⋮13^{\left(đpcm\right)}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Câu 7:
Gọi số học sinh là x
Theo đề, ta có: \(x\in BC\left(3;4;6;8\right)\)
hay x=72
3,
A=1-1/2+1/2-1/3+1/3-1/4+...+1/99-1/100
A=1-1/100(1)
B=1/2^2+1/3^2+..+1/100^2
B<1/1.2 +1/2.3+...+1/99.100
B<1-1/2+1/2-1/3+...+1/99-1/100
B<1-1/100(2)
từ (1) (2) => A>B