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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 }
a: \(Ư\left(16\right)=\left\{1;2;4;8;16\right\}\)
\(Ư\left(24\right)=\left\{1;2;3;4;6;8;12;24\right\}\)
\(ƯC\left(16;24\right)=\left\{1;2;4;8\right\}\)
b: \(Ư\left(20\right)=\left\{1;2;4;5;10;20\right\}\)
\(Ư\left(32\right)=\left\{1;2;4;8;16;32\right\}\)
\(ƯC\left(20;32\right)=\left\{1;2;4\right\}\)
a) Ư (12) = {1;2;3;4;6;12}
Ư (16) = {1;2;4;8;16}
Ư (24) = {1;2;3;4;6;8;12;24}
ƯC (12; 16; 24) = {1;2;4}
b) ƯC (5; 15; 35) = {1;5}
c) BC (8; 12; 24) = {0;24;48}
d) BC (5; 15; 35) = {0;105;210;…}
a) Ư(8) = {1;2;4;8}; Ư(12) = {1;2;3;4;6;12} => ƯC(8;12) = {1;2;4;}
b) Ư(24) = {1;2;3;4;6;8;12;24}; Ư(32) = {1;2;4;8;16;32} => ƯC(24; 32) = {1;2;4;8;}
c) Ư(7) = {1;7} ; Ư(10) = {1;2;5;10} => ƯC(7;10) = {1}
d) 8 = 23; 10 = 2.5 => BCNN (8;10) = 23.5 = 40 => BC(8;10) = B(40) = {0;40;80;...}
e) 25 = 52 => BCNN(2;3;25) = 2.3.52 = 150 => BC (2;3;25) = B(150) = {0;150; 300; ...}
2) N = {0;1;2;3;...}; N* = {1;2;3;....} => N giao N* = {1;2;3;...} = N*
a) Ư(8) = {1;2;4;8}; Ư(12) = {1;2;3;4;6;12} => ƯC(8;12) = {1;2;4;}
Vâu b,c,d,e tương tự nha bn
2) N = {0;1;2;3;...}; N* = {1;2;3;....} => N giao N* = {1;2;3;...} = N*
hok tốt
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