D

D

D

Ư(12) ={1; 2; 3; 4; 6; 12} Ư(18) {1; 2; 3; 6; 9; 18}

 ƯC(12, 18) = {1; 2; 3; 6 }

vậy  ƯCLN (12, 18) = 6

ƯC(12,18)={1;2;3;6}

ƯCLN(12,18)={6}

Ư(12)={ 1,2,3,4,6,12 }

Ư(18)={1,2,3,6,9,18 }

ƯC(12,18)= {1,2,3,6 }

ƯCLN(12, 18)={6}

24=2³.3 ; 84=2².3.7

=> ƯCLN(24;84)=2².3=12

UCLN(12;52)=4

UCLN(36;990)=18

UCLN(54;36)=18

UCLN(12;52)=4

UCLN(36;990)=18

UCLN(54;36)=18

UCLN(10;20;70)=10

UCLN(25;55;75)=5

UCLN(63;2970)=9

UCLN(24;36;60)=6

UCLN(65;125)=5

UCLN(9;18;72)=9

12 = 2² . 3

52 = 2² . 13

ƯCLN(12,52) = 2² = 4

ƯC(12,52) = Ư(4) = 1,2,4

b) 36  = 2² . 3²

990 = 2 . 5 . 3² . 11

ƯCLN(36,990) = 2 . 3² = 18

ƯC(36,990) = Ư(18) = 1,2,3,6,9,18

c) 10 = 2 . 5

20 = 2² . 5

70 = 7 . 2 . 5

ƯCLN(10,20,70) = 2 . 5 = 10

ƯC(10,20,70) = Ư(10) = 1,2,5,10

d) 54 = 2 . 3³

36 = 2² . 3²

ƯCLN(54,36) = 2 . 3² = 18

ƯC(54,36) = Ư(18) = 1,2,3,6,9,18

e) 25 = 5²

55 = 11.5

75 = 5² . 3

ƯCLN(25,55,75) = 5

ƯC(25,55,75) = Ư(5) = 1,5

24={1;2;3;4;6;8;12;24}

\(ƯCLN\left(24;180\right)=12\)

24 = 2³.3

180 = 2².3².5

ƯCLN (24;180)= 2².3= 12

Ta có: 24=2\(^3\).3

           12=2\(^2\).3

ƯCLN(24, 12)=2\(^2\).3= 4.3=12.

`24 = 2^{3}.3`

`12 = 2^{2}.3`

`⇒ƯCLN(24;12)=2^{2}.3=12`

a) 16=2 mũ4

24=3.2 mũ 2

Suy ra ƯCLN (16,24) =2 mũ 2 =4

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}.

b)180 =3 mũ 2 nhân 2 mũ 2 nhân 5

234=13 nhân 3 mũ 2 nhân 2

suy ra ƯCLN(180,234)=3 mũ 2 nhân 2=18