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Ta có 180 = 22.32.5 và 234 = 2.32.13 ⇒ ƯCLN(180, 234) = 2.32 = 18.
ƯC(180 ; 234) = Ư(18) = {1; 2; 3; 6; 9; 18}.
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(180=2^2.3^2.5\)
\(234=2.3^2.13\)
\(UCLN\left(180;234\right)=2.3^2=18\)
\(UC\left(180;234\right)=\left\{1;2;3;6;9;18\right\}\) (các phần tử thuộc N)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) ƯCLN (16, 24) = 8, ƯC (16, 24) = {1; 2; 4; 8};
b) Ta có 180 = 22 . 32 . 5; 234 = 2 . 32 . 13;
ƯCLN (180, 234) = 2 . 32 = 18, ƯC (180, 234) = {1; 2; 3; 6; 9; 18};
c) Ta có 60 = 22 . 3 . 5; 90 = 2 . 32 . 5; 135 = 33 . 5. Do đó
ƯCLN (60, 90, 135) = 3 . 5 = 15; ƯC (60, 90, 135) = {1; 3; 5; 15}.
![](https://rs.olm.vn/images/avt/0.png?1311)
a)UCLN(180;234)=18
=> UC(180,234)={1;2;3;6;9;18}
b)UCLN(60;90;135)=15
=>UC(60;90;135)={1;3;5;15}
![](https://rs.olm.vn/images/avt/0.png?1311)
a) 30=2.3.5
135=32.5
=>UTLN(30,135)=3.5=15
b)30=2.3.5
234=2.32.13
=>UTLN(30,234)=2.3=6
c)134=2.67
60=22.3.5
=>UTLN(134,60)=2
d)176=22.4.11
234=2.32.13
=>UTLN(176,234)=2
e)180=22.32.5
176=24.11
=>UTLN(180,176)=22=4
![](https://rs.olm.vn/images/avt/0.png?1311)
a)16=24
24=23.3
=>UCLN(16,24)=23=8
b)180=22.32.5
234=2.32.13
=>UCLN(180,234)=2.32=18
c)60=22.3.5
90=2.32.5
135=33.5
=>UCLN(60,90,135)=3.5=15
a) ƯCLN (16, 24) = 8, ƯC (16, 24) = {1; 2; 4; 8};
b) Ta có 180 = 22 . 32 . 5; 234 = 2 . 32 . 13;
ƯCLN (180, 234) = 2 . 32 = 18, ƯC (180, 234) = {1; 2; 3; 6; 9; 18};
c) Ta có 60 = 22 . 3 . 5; 90 = 2 . 32 . 5; 135 = 33 . 5. Do đó
ƯCLN (60, 90, 135) = 3 . 5 = 15; ƯC (60, 90, 135) = {1; 3; 5; 15}.
![](https://rs.olm.vn/images/avt/0.png?1311)
a)16=24
24=23.3
=>UCLN(16,24)=23=8
b)180=22.32.5
234=2.32.13
=>UCLN(180,234)=2.32=18
c)60=22.3.5
90=2.32.5
135=33.5
=>UCLN(60,90,135)=3.5=15
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Ta có: 16=4^2=2^4
24=3.8=3.2^3
ƯCLN(16,24)=2^3=8
ƯC(16,24)=Ư(8)={1,2,4,8}
b) Ta có: 180=2^2.3^2.5
243=2.3^2.13
60=2^2.3.5
90=2.3^2.5
135=3^3.5
ƯCLN(234,180,60,90,135)=3
ƯC(234,180,60,90,135)=Ư(3)={1,3}
2,9,18
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