a,2

b,6

a,\(42=2.3.7\)

\(58=2.29\)

\(\RightarrowƯCLN\left(42;58\right)=2\)

b, \(18=2.3^2\\ 30=2.3.5\\ 42=2.3.7\)

\(\RightarrowƯCLN\left(18,30,42\right)=2.3=6\)

a) 42 = 2.3.7

    58 = 2.29

ƯCLN ( 42,58 ) = 2

b) 18 = 2.3²

     30 = 2.3.5

     42 = 2.3.7

ƯCLN ( 18,30,42 ) = 2.3 = 6

a: UCLN(16;42)=2

UC(16;42)={1;2}

b: UCLN(16;42;86)=2

UC(16;42;86)={1;2}

c: UCLN(25;75)=25

UC(25;75)={1;5;25}

d: UCLN(25;55;75)=5

UC(25;55;75)={1;5}

a: UCLN(16;42)=2

UC(16;42)={1;2}

b: UCLN(16;42;86)=2

UC(16;42;86)={1;2}

c: UCLN(25;75)=25

UC(25;75)={1;5;25}

d: UCLN(25;55;75)=5

UC(25;55;75)={1;5}

a,8

b,30

c,12

d,4

e,1

g,18

h,6

k,1

a) ƯCLN ( 16, 24 )

16 = 2³              24 = 2².3

ƯCLN ( 16;24 ) = 2²= 4

b) ƯCLN ( 60, 90 )

60 = 2².3.5          90=2.3².5

ƯCLN ( 60;90 ) = 2.3.5 = 30

c)  ƯCLN ( 24, 84 )

24 = 2³.3                       84 = 2².3.7

ƯCLN ( 24;84 ) = 2².3 = 12

d) ƯCLN ( 16, 60 )

16 = 2⁴              60 = 2².3.5

ƯCLN ( 16;60 ) = 2²=4

e) ƯCLN ( 18, 77 )

18 = 2.3²             77=7.11

 ƯCLN ( 18; 77 ) = 1

g) ƯCLN ( 18, 90 )

18 = 2.3²           90=2.3².5

 ƯCLN ( 18;90 ) = 2.3² = 18

h) ƯCLN ( 18, 30, 42 )

18 = 2.3²             30 = 2.3.5         42 = 2.3.7

 ƯCLN ( 18;30;42 ) = 2.3=6

k) ƯCLN ( 26, 39, 48 )

26 = 2.13           39 = 3.13      48 = 2⁴.3

 ƯCLN ( 26;39;48 ) = 1

sai 30 :18 = ?

a) Ta có: 12 = 2 2 . 3  và 52 = 2 2 . 13 => ƯCLN(12,52) = 2 2  = 4

Vậy  ƯC(12,52) = Ư(4) = {1;2;4}

b) Ta có: 36 = 2 2 . 3 2  và 990 = 2 . 3 2 . 5 . 11  => ƯCLN(36,900) =  2 . 3 2 = 18

Vậy  ƯC(36,900) = Ư(18) = {1;2;3;6;9;18}

c) 25 =  5 2 ; 55 = 5.11 và 75 =  3 . 5 2  => ƯCLN(25,55,75) = 5

Vậy  ƯC(25,55,75) = Ư(5) = {1;5}

d, Ta có: 24 ⋮ 14; 112 ⋮ 14 => ƯCLN(14,42,112) = 14

Vậy ƯC(14,42,112) = Ư(14) = {1;2;7;14}

\(a,ƯCLN\left(540,168\right)=12\\ \RightarrowƯC\left(540,168\right)=Ư\left(12\right)=\left\{-12;-6;-4;-3;-2;-1;1;2;3;4;6;12\right\}\\ b,ƯCLN\left(735,350\right)=35\\ \RightarrowƯC\left(735,350\right)=Ư\left(35\right)=\left\{-35;-7;-5;-1;1;5;7;35\right\}\)

\(a.540=2^2.3^3.5\)

\(168=2^3.3.7\)

\(ƯCLN\left(540;168\right)=2^2.3=12\)

\(ƯC\left(540;168\right)=\left\{12;24;36;...\right\}\)

\(b.735=3.5.7^2\)

\(350=2.5^2.7\)

\(ƯCLN\left(735;350\right)=5.7=35\)

\(ƯC\left(735;350\right)=\left\{35;70;105;...\right\}\)

a) Ta có: 21 = 3.7;  98 = 2.7²

=> ƯCLN(21, 98) = 7 ; BCNN(21, 98) = 2.3.7² = 294

b) Ta có: 36 = 2².3², 54 = 2.3³

ƯCLN(36, 54) = 2.3² = 18; BCNN(36, 54) = 2².3³ = 108.

a) Ta có:

\(21=3\cdot7\\98=2\cdot7^2\)

Khi đó, ta được: 

\( UCLN(21,98)=7\)

\(BCNN(21,98)=2\cdot3\cdot7^2=294\)

b) Ta có:

\(36=2^2\cdot3^2\\54=2\cdot3^3\)

Khi đó, ta được:

\(UCLN\left(36,54\right)=2\cdot3^2=18\)

\(BCNN\left(36,54\right)=2^2\cdot3^3=108\)

Bài 3: 

1: 

a: UCLN(90;126)=18

UC(90;126)={1;2;3;6;9;18}