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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}.
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}.
a)
Ta có:
\(24=2^3.3\)
\(30=2.3.5\)
\(\RightarrowƯCLN\left(24;30\right)=2.3=6\)
\(\RightarrowƯC\left(24;30\right)=Ư\left(6\right)=\left\{1;2;3;6\right\}\)
b)
Ta có:
\(42=2.3.7\)
\(98=2.7^2\)
\(\RightarrowƯCLN\left(42;98\right)=2.7=14\)
\(\RightarrowƯC\left(42;98\right)=Ư\left(14\right)=\left\{1;2;7;14\right\}\)
c)
Ta có:
\(180=2^2.3^2.5\)
\(234=2.3^2.13\)
\(\RightarrowƯCLN\left(180;234\right)=2.3^2=18\)
\(\RightarrowƯC\left(180;234\right)=Ư\left(18\right)=\left\{1;2;3;6;9;18\right\}\)
a. ƯCLN(24;30) = 6
ƯC(24, 30) = {1; 2; 3; 6}
b. UCLN( 42 ; 98)= 14
ƯC(42;98) = \(\left\{1;2;7;14\right\}\)
c.UCLN( 180 ; 234 ) = 18
ƯC(180;234) = \(\left\{1;2;;6;9;18\right\}\)
a) Ta có: 16=4^2=2^4
24=3.8=3.2^3
ƯCLN(16,24)=2^3=8
ƯC(16,24)=Ư(8)={1,2,4,8}
b) Ta có: 180=2^2.3^2.5
243=2.3^2.13
60=2^2.3.5
90=2.3^2.5
135=3^3.5
ƯCLN(234,180,60,90,135)=3
ƯC(234,180,60,90,135)=Ư(3)={1,3}
a, 16= 24
24= 23.3
\(\Rightarrow\)ƯCLN (16; 24)= 23= 8
\(\Rightarrow\)ƯC (16; 24)= Ư (8)= {1;2;4;8}
b, 180= 22.32.55
234= 2.33.13
\(\Rightarrow\)ƯCLN (180; 234)= 2.32= 18
\(\Rightarrow\)ƯC (180; 234)= Ư (18)= {1;2;3;6;9;18}
c, 60= 22.3.5
90= 32.2.5
135= 33.5
\(\Rightarrow\)ƯCLN (60;90;135)= 3.5= 15
\(\Rightarrow\)ƯC (60;90;135)= Ư (15)= {1;3;5;15}
a)
Ư(5) = {1; -1; 5; -5}
Ư(10) = {1; -1; 2; -2; 5; -5; 10; -10}
Ư(15) = {1; -1; 3; -3; 5; -5; 15; -15}
ƯC(5; 10; 15) = {1; -1; 5; -5}
B(5) = {0; 5; -5; 10; -10...}
B(10) = {0; 10; -10; 20; -20...}
B(15) = {0; 15; -15; 30; -30...}
BC(5; 10) = {0; 10; -10; 20; -20...}
b)
120; 180
120 = \(2^3\). 3 . 5
180 = \(2^2\). \(3^2\). 5
\(\Rightarrow\)ƯCLN(120; 180) = \(2^2\). 3 . 5 = 4 . 3 . 5 = 60
\(\Rightarrow\)ƯC(120; 180) = Ư(60) = {1; -1; 2; -2; 3; -3; 4; -4; 5; -5; 6; -6; 10; -10; 20; -20; 30; -30; 60; -60}
c)
20; 50
20 = \(2^2\). 5
50 = 2 . \(5^2\)
\(\Rightarrow\)BCNN(20; 50) = \(2^2\). \(5^2\)= 4 . 25 = 100
\(\Rightarrow\)BC(20; 50) = B(100) = {0; 100; -100; 200; -200...}
ok nhé!