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![](https://rs.olm.vn/images/avt/0.png?1311)
\(60\%.x+\frac{2}{3}.x+3=680\)
\(\frac{3}{5}.x+\frac{2}{3}.x+x=680\)
\(\left(\frac{3}{5}+\frac{2}{3}+1\right).x=680\)
\(\frac{34}{15}\cdot x=680\)
\(x=680\div\frac{34}{15}\)
\(x=300\)
Vậy ...
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
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\(a,2\times31\times12+4\times6\times42+8\times27\times3\)
\(=24\times31+24\times42+24\times27\)
\(=24\times\left(31+42+27\right)\)
\(=24\times100\)
\(=2400\)
\(b,583.931+69.\left(170+583\right)+31.170\)
\(=583.931+69.170+69.583+31.170\)
\(=583.\left(931+69\right)+170.\left(69+31\right)\)
\(=583.1000+170.100\)
\(=583000+17000\)
\(=600000\)
\(c,732.976+24.\left(680+732\right)+76.680\)
\(=732.976+24.680+24.732+76.680\)
\(=732.\left(976+24\right)+680.\left(24+76\right)\)
\(=732.1000+680.100\)
\(=732000+68000\)
\(=800000\)
Học tốt
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: 612=22.32.17 ; 680= 5.17.23
Ta có: ƯCLN(612;680)= 22.17=68
Vậy: \(x\inƯ\left(68\right),x\ge35\Rightarrow x=68\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Vì 612 chia hết cho a và 680 chia hết cho a nên a ∈ ƯC(612,680)
Ta có : 612 = 2 2 . 3 2 . 17 ; 680 = 2 3 . 5 . 17 => ƯCLN(612,680) = 2 2 . 17 = 68
Mà Ư(68) = {1;2;4;17;34;68}
=> ƯC(612,680) = {1;2;4;17;34;68}
=> a ∈ {1;2;4;17;34;68}
Vì a lớn hơn 30 nên a ∈ {34;68}
![](https://rs.olm.vn/images/avt/0.png?1311)
10 . [(56 - 23) + x] = 680
(56 - 23) + x = 680 : 10
(56 - 23) + x = 68
33 + x = 68
x = 68 - 33
x = 35
![](https://rs.olm.vn/images/avt/0.png?1311)
60%x + 2/3x +x=680
x(3/5+2/3+1)=680
Đến bước này bạn làm được chưa?
60%x+2/3x+x=680
=> 0,6x+2/3x+x=680
=> x.(0,6+2/3+1)=680
=> x.34/15 =680
=> x =680:34/15
=> x =300
k mik nhé
Vuthimailoan