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\(ƯC\left(8,12\right)=\left\{\pm1;\pm2;\pm4\right\}\)
\(ƯC\left(12;15;30\right)=\left\{\pm1;\pm3\right\}\)
\(ƯC\left(60;72\right)=\left\{\pm1;\pm2;\pm3;\pm4;\pm6;\pm12\right\}\)
\(ƯC\left(24;42\right)=\left\{\pm1;\pm2;\pm3;\pm6\right\}\)
a) { 1; 2; 4 }
b) { 1; 3 }
c) { 1; 2; 3; 4; 6; 12 }
d) { 1; 2; 3; 6 }
Ý bn là tìm phần tử à:
a, ƯC(8;12)= ƯCLN (8;12)
Ta có: 8= 23 và 12 = 22.3
\(\Rightarrow\)ƯCLN(8;12)= 22= 4
\(\Rightarrow\)ƯC (8;12)= Ư(4)= {1;2;4}
b, ƯC (12;15;30)= ƯCLN (12;15;30)
Ta có: 12= 22.3
15= 3.5
30= 3.2.5
\(\Rightarrow\)ƯCLN (12;15;30)= 2.3= 6
\(\Rightarrow\)ƯC (12;15;30)= Ư(6)= {1;2;3;6}
c, ƯC (60;72)= ƯCLN (60;72)
Ta có: 60= 22.3.5 và 72= 23.32
\(\Rightarrow\)ƯCLN (60;72)= 22= 4
\(\Rightarrow\)ƯC(60;72)= Ư(4)= {1;2;4}
d, ƯC (24;42)= ƯCLN (24;42)
Ta có: 24= 23.3 và 42= 2.3.7
\(\Rightarrow\)ƯCLN (24;42)= 3
\(\Rightarrow\)ƯC (24;42)= Ư(3)= {1;3}
Chúc bn học tốt
\(a,ƯC\left(16,24\right)=Ư\left(8\right)=\left\{\pm1;\pm2;\pm4;\pm8\right\}\\ b,BC\left(84,108\right)=B\left(756\right)=\left\{0;756;1512;2268;...\right\}\\ c,ƯC\left(36,48,120\right)=Ư\left(12\right)=\left\{\pm1;\pm2;\pm3;\pm4;\pm6;\pm12\right\}\\ d,BC\left(42,64,70\right)=B\left(6720\right)=\left\{0;6720;13440;20160;...\right\}\)
a, Ta có:
Ư(48)={-48;-16;-12;-8;-4;-3;-2;-1;1;2;3;4;8;12;16;48}
Ư(64)={-64;-32;-16;-8;-4;-2;-1;1;2;4;8;16;32;64}
Ư(36)={-36;-12;-18;-9;-4;-3;-2;-1;1;2;3;4;12;18;36}
\(\Rightarrow\)ƯC(48;64;36)={-4;-2;-1;1;2;4}
Vậy ƯC(48;64;36)={-4;-2;-1;1;2;4}
a, ta có : 48 =2^4×3
64=2^6
36=2^2×3^2
Suy ra ƯCLN(48,64,36) =2^2=4
Nên ƯC(48,64,36)=Ư(4)=(1;2;4)
b, ta có:28=2^2×7
42=2×3×7
38=2×19
Suy ra ƯCLN(28;42;38)=2
Nên ƯC(28;42;38)=Ư(2)=(1;2)