TL :

ƯCLN( 18 ; 45 ; 36 )

Ta có :

18 = 2 . 3²

45 = 3² . 5

36 = 2² . 3²

=> ƯCLN( 18 ; 45 ; 36 ) = 3²  = 9

ƯCLN( 25 ; 125 ; 75 ) 

Ta có :

25 = 5²

125 = 5³

75 = 3 . 5²

=> ƯCLN( 25 ; 125 ; 75 ) = 5² = 25

UCLL(18,45,36)

\(18=2\times3^2\)

\(45=3^2\times5\)

\(36=2^2\times3^2\)

UCLL(18,45,36)=\(3^2=9\)

UCLL(25,125,75)

\(25=5^2\)

\(125=5^3\)

\(75=3\times5^2\)

UCLL(25,125,75)=\(5^2=25\)

UCLN(10 ;12 ;15)=1

Vi 10 12 15 la cac so nguyen to cung nhau
 

1. UCLN ( 56 , 140 ) ==>28

2.UCLN (15 , 19 )==>1

3.UCLN (60 , 180 )==>60

4. UCLN ( 24 , 84 , 180 )==>12

5. UCLN ( 16 , 80 , 176 )==>16

6. UCLN ( 18 , 30 , 77 )==>1

a)UCLN(56,140)

56=2³.7

140=2.5.7

=>UCLN(56,140)=2.7=14

b)UCLN(15,19)

15=3.5

19=1.19

=>UCLN(15,19)=1

c)UCLN(60,180)

60=2².3.5

180=2².3².5

=>UCLN(60,180)=2².3.5=60

d)UCLN(24,84,180)

24=2³.3

84=2².3.7

180=2².3³.5

=>UCLN(24,84,180)=2².3=12

e)UCLN(16,80,176)

16=2⁴

80=2⁴.5

176=2⁴.11

=>UCLN(16,80,176)=2⁴=16

f)UCLN(18,30,77)

18=3².2

30=2.3.5

77=7.11

=>UCLN(18,30,77)=1

1.28

2.1

3.60

4.12

5.16

6.1

a: \(12=2^2\cdot3;18=3^2\cdot2\)

=>\(ƯCLN\left(12;18\right)=2\cdot3=6\)

b: \(24=2^3\cdot3;48=2^4\cdot3\)

=>\(ƯCLN\left(24;48\right)=2^3\cdot3=24\)

c: \(44=2^2\cdot11;121=11^2\)

=>\(ƯCLN\left(44;121\right)=11\)

d: \(36=3^2\cdot2^2;108=3^3\cdot2^2;224=2^5\cdot7\)

=>\(ƯCLN\left(36;108;224\right)=2^2=4\)

a, Ta có :

\(12=2\cdot2\cdot3=2^2\cdot3\)

\(18=2\cdot3\cdot3=3^2\cdot2\)

Thừa số nguyên tố chung là : \(2;3\)

\(ƯCLN\) \(\left(12;18\right)=2\cdot3=6\)

⇒ \(ƯCLN\) \(\left(12;18\right)\) là \(6\)

b, Ta có:

\(24=2\cdot2\cdot2\cdot3=2^3\cdot3\)

\(48=2\cdot2\cdot2\cdot2\cdot3=2^4\cdot3\)

Thừa số nguyên tố chung là : \(2;3\)

\(ƯCLN\) \(\left(24;48\right)\) \(=2^3\cdot3=8\cdot3=24\)

⇒ \(ƯCLN\) \(\left(24;48\right)\) là \(24\)

c, Ta có

\(44=2\cdot2\cdot11=2^2\cdot11\)

\(121=11\cdot11=11^2\)

Thừa số nguyên tố chung là : \(11\)

\(ƯCLN\) \(\left(44;121\right)\) \(=11\)

⇒ \(ƯCLN\) là \(11\)

d, Ta có :

\(36=2\cdot2\cdot3\cdot3=2^2\cdot3^3\)

\(108=2\cdot2\cdot3\cdot3\cdot3=2^2\cdot3^3\)

\(224=2\cdot2\cdot2\cdot2\cdot2\cdot7=2^5\cdot7\)

Thừa số nguyên tố chung là : \(2\)

\(ƯCLN\) \(\left(36;108;224\right)\) \(=2\cdot2=2^2=4\)

⇒ \(ƯCLN\) \(\left(36;108;224\right)\) là \(4\)

\(#thaolinh\)

e) \(24=2^3.3\)

\(84=2^2.3.7\)

\(180=2^2.3^2.5\)

\(\RightarrowƯCLN\left(24;84;180\right)=2^2.3=12\)

b) \(24=2^2.3\)

\(36=2^2.3^2\)

\(\RightarrowƯCLN\left(24;36\right)=2^2.3=12\)

g) \(56=2^3.7\)

\(140=2^2.5.7\)

\(\RightarrowƯCLN\left(56;140\right)=2^2.7=28\)

h) \(12=2^2.3\)

\(14=2.7\)

\(8=2^3\)

\(20=2^2.5\)

\(\RightarrowƯCLN\left(12;14;8;20\right)=2\)

d) \(6=2.3\)

\(8=2^3\)

\(18=2.3^2\)

\(\RightarrowƯCLN\left(6;8;18\right)=2\)

k) \(7=7\)

\(9=3^2\)

\(12=2^2.3\)

\(21=3.7\)

\(\RightarrowƯCLN\left(7;9;12;21\right)=1\)

giup bai 1 gium di ma , thu 2 nop bai roi 

giup voi, tui ko biet lam bai 1

\(10=2.5\\ 12=2^2.3\\ \Rightarrow\text{Ư}CLN\left(10;12\right)=2\)

10=2.5
12=2.2.3
UCLN(10,12)=2