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a) 1 = 1
10 = 2 . 5
ƯCLN(1, 10) = 1
b) 11 = 11
15 = 3 . 5
ƯCLN(11, 15) = 1
c) 18 = 2 . 32
42 = 2 . 3 . 7
ƯCLN(18, 42) = 2 . 3 = 6
d) 24 = 23 . 3
16 = 24
ƯCLN(24, 16) = 23 = 8
a)ƯCLN (1,10)={1}
b)ƯCLN (11,15)={1}
c)Ta có :18=2.32
42=2.3.7
\(\Rightarrow\)ƯCLN(18,42)=2.3=6
d)Ta thấy: 24 ;16;8 \(⋮\)8 \(\Rightarrow\)ƯCLN (24;16;8)=8
a) 90 = 2.32.5
126 = 2.32.7
UCLN(90,126) = 2.32 = 18
ƯC(90,126) = Ư(18) = {1;2;3;6;9;18}
b) 180 = 22.32.5
234 = 2.32.13
UCLN(180,234) = 2.32 = 18
ƯC(180,234) = Ư(18) = {1;2;3;6;9;18}
a ) UCLN ( 90 , 126 ) = 18
UC ( 90 , 126 ) \(\in\left\{1;2;3;6;9;18\right\}\)
b ) UCLN ( 180 , 234 ) = 18
\(UC\left(180,234\right)\in\left\{1;2;3;6;9;18\right\}\)
Nhớ k cho mk nha 
nguyen huyen tram
a) UCLN ( 12;10 ) = 2
b) UCLN ( 9;81 ) = 9
c) UCLN ( 1;10 ) = 1
d) UCLN ( 32,192 ) = 32
e) UCLN ( 12,15,10 ) = 1
f) UCLN ( 28;48 ) = 4
g) UCLN ( 25;55;75 ) = 5
Ta có : \(65=5\cdot13\)
\(125=5^3\)
=> UCLN(65 125 ) = 5
Vậy UC( 65 , 125 ) = 1 , 5
65 = 5 x 13
125 = 53
=> UCLN(65,125) = 5
=> ƯC(65,125) = U(5) = {1;5}
a, \(40=2^3\cdot5\)
\(24=2^3\cdot3\)
\(ƯCLN\left(40,24\right)=2^3=8\)
\(ƯC\left(40,24\right)=Ư\left(8\right)=\left\{1;2;4;8\right\}\)
b, \(11=11\)
\(15=3\cdot5\)
\(ƯCLN\left(11,15\right)=1\)
\(ƯC\left(11,15\right)=Ư\left(1\right)=\left\{1\right\}\)
c, \(12=2^2\cdot3\)
\(15=3\cdot5\)
\(10=2\cdot5\)
\(ƯCLN\left(12,15,10\right)=1\)
\(ƯC\left(12,15,10\right)=Ư\left(1\right)=\left\{1\right\}\)
d, \(9=3^2\)
\(18=2\cdot3^2\)
\(72=2^3\cdot3^2\)
\(ƯCLN\left(9,18,72\right)=3^2=9\)
\(ƯC\left(9,18,72\right)=Ư\left(9\right)=\left\{1;3;9\right\}\)