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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)=\(5^2=25\)
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
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) \(7=7\)
\(9=3^2\)
\(12=2^2.3\)
\(21=3.7\)
\(\RightarrowƯCLN\left(7;9;12;21\right)=1\)
\(10=2.5\\ 12=2^2.3\\ \Rightarrow\text{Ư}CLN\left(10;12\right)=2\)
UCLL(10,12,15)
\(10=2\times5\)
\(12=2^2\times3\)
\(15=3\times5\)
UCLL(10,12,15)=1
UCLL(20,45,18)
\(20=2^2\times5\)
\(45=3^2\times5\)
\(18=2\times3^2\)
UCLL(20,45,18)=1