a, 56 =  2 3 . 7 ; 28 =  2 2 . 5

UCLN(56,28) =  2 2 = 4

b, Ta có: 156 =  2 2 . 3 . 13 ; 13 = 13 nên UCLN(156,13) = 13

c, Có 215 = 5.43; 216 = 2 3 . 3 3  nên UCLN(215,216) = 1

d, Ta có: 1111 = 11.101

Thấy 11111 không chia hết cho cả hai số 11 và 101 nên UCLN(11111;1111) = 1

a)

31 = 31

22=2.11

34=2.17

+) ƯCLN(31,22) = 1

+) ƯCLN(31,34) = 1

+) ƯCLN (22,34) = 2

b)

\(105 = 3.5.7;128 = {2^7};135 = 3.3.3.5 = {3^3}.5\)

+) ƯCLN (105,128) = 1

+) ƯCLN (128,135) = 1

+) ƯCLN (105,135) = 15.

a: \(31=31;22=2\cdot11;34=2\cdot17\)

=>\(ƯCLN\left(31;22\right)=1;ƯCLN\left(31;34\right)=1;ƯCLN\left(22;34\right)=2\)

b: \(105=3\cdot5\cdot7;128=2^7;135=3^3\cdot5\)

=>\(ƯCLN\left(105;135\right)=5;ƯCLN\left(128;135\right)=1;ƯCLN\left(105;128\right)=1\)

Có 16 =  2 4 ; 24 =  2 3 . 3 nên UCLN(16,24) =  2 3 = 8

Vậy UC(16,24) = {1;2;4;8}

b, Có 180 =  2 3 . 3 2 . 5 ; 234 =  2 . 3 2 . 13 nên UCLN(180,234) = 2.3 = 6

Vậy UC(180,234) = {1;2;3;6}

c, Có 60 =  2 2 . 3 . 5 ; 90 =  2 . 3 2 . 5 ; 135 =  3 3 . 5 nên UCLN(60,90,135) = 3.5 = 15

Vậy UC(60,90,135) = {1;3;5;15}

1)

a) 18 = 2.3²

30 = 2.3.5

ƯCLN(18; 30) = 2.3 = 6

b) 24 = 2³.3

48 = 2⁴.3

ƯCLN(24; 48) = 2³.3 = 24

c) 18 = 2.3²

30 = 2.3.5

15 = 3.5

ƯCLN(18; 30; 15) = 3

d) 24 = 2³.3

48 = 2⁴.3

36 = 2².3²

ƯCLN(24; 48; 36) = 2².3 = 12

2) a) 174 = 18 . 9 + 12

18 = 12 . 1 + 6

12 = 6 . 2

Vậy ƯCLN(174; 18) = 6

b) 124 = 16 . 7 + 12

16 = 12 . 1 + 4

12 = 4 . 3

⇒ ƯCLN(124; 16) = 4

⇒ BCNN(124; 16) = 124 . 16 : 4 = 496

\(b,ƯCLN\left(70,540,1225\right)=35\\ \RightarrowƯC\left(70,540,1225\right)=Ư\left(35\right)=\left\{1;5;7;35\right\}\\ d,ƯCLN\left(48,24,60\right)=12\\ \RightarrowƯC\left(48,24,60\right)=Ư\left(12\right)=\left\{1;2;3;4;6;12\right\}\\ g,ƯCLN\left(3780,4320,5184\right)=108\\ \RightarrowƯC\left(3780,4320,5184\right)=Ư\left(108\right)=\left\{1;2;3;4;6;9;12;18;27;36;54;108\right\}\)

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