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Bài toán 1 : Viết các tập hợp sau.
a)Ư:(6,9,12) d) B(23) ; B(10) ; B(8)
b)Ư(7) ; Ư(18) ; Ư(10) e) B(3) ; B(12) ; B(9)
c)Ư(15) ; Ư(16) ; Ư(250 g) B(18) ; B(20) ; B(14)
\(a)\)
\(Ư (6) = \) \(\left\{1;2;3;6\right\}\)
\(Ư\left(9\right)=\left\{1;3;9\right\}\)
\(Ư\left(12\right)=\left\{1;2;3;4;6;12\right\}\)
\(b)\)
\(Ư\left(7\right)=\left\{1;7\right\}\)
\(Ư\left(18\right)=\left\{1;2;3;6;9;18\right\}\)
\(Ư\left(10\right)=\left\{1;2;5;10\right\}\)
\(c)\)
\(Ư\left(15\right)=\left\{1;3;5;15\right\}\)
\(Ư\left(16\right)=\left\{1;2;4;8;16\right\}\)
\(Ư\left(250\right)=\left\{1;2;5;10;25;50;125;250\right\}\)
\(d)\)
\(B\left(23\right)=\left\{0;23;46;69;...\right\}\)
\(B\left(10\right)=\left\{0;10;20;30;...\right\}\)
\(B\left(8\right)=\left\{0;8;16;24;...\right\}\)
\(e)\)
\(B\left(3\right)=\left\{0;3;6;9;...\right\}\)
\(B\left(12\right)=\left\{0;12;24;36;...\right\}\)
\(B\left(9\right)=\left\{0;9;18;27;...\right\}\)
\(g)\)
\(B\left(18\right)=\left\{0;18;36;54;...\right\}\)
\(B\left(20\right)=\left\{0;20;40;60;...\right\}\)
\(B\left(14\right)=\left\{0;14;28;42;...\right\}\)
a)
\(Ư\left(12\right)=\left\{1;2;3;4;6;12\right\}\)
\(Ư\left(18\right)=\left\{1;2;3;6;9;18\right\}\)
=> \(ƯC\left(12;18\right)=\left\{1;2;3;6;\right\}\)
b) \(B\left(15\right)=\left\{0;15;30;45;60;90;105...\right\}\)
\(B\left(21\right)=\left\{0;21;42;63;84;105;...\right\}\)
=> \(BC\left(15;21\right)=\left\{0;105;...\right\}\)
c)
\(20=2^2.5\)
\(24=2^3.3\)
\(28=2^2.7\)
=> \(BCNN\left(20;24;28\right)=2^3.3.5.7=840\)
=> \(BC\left(20;24;28\right)=B\left(840\right)=\left\{0;840;1680;...\right\}\)
Câu hỏi của Công Chúa Họ Kim - Toán lớp 6 - Học toán với OnlineMath
a, Ư(12)=\(\left\{\pm1;\pm2;\pm3;\pm4\pm6\pm12\right\}\)
Ư(18)=\(\left\{\pm1;\pm2;\pm3;\pm6;\pm9;\pm18\right\}\)
ƯC(12;18)=\(\left\{\pm1;\pm2;\pm3;\pm6\right\}\)
c,ƯC(45;20;72)=\(\left\{\pm1\right\}\)
Câu hỏi của Công Chúa Họ Kim - Toán lớp 6 - Học toán với OnlineMath
1)
\(Ư\left(6\right)=\left\{\pm1;\pm2;\pm3;\pm6\right\}\)
\(Ư\left(14\right)=\left\{\pm1;\pm2;\pm7;\pm14\right\}\)
\(Ư\left(17\right)=\left\{\pm1;\pm17\right\}\)
\(Ư\left(1\right)=\left\{\pm1\right\}\)
2)
a)
\(Ư\left(12\right)=\left\{\pm1;\pm2;\pm3;\pm4;\pm6;\pm12\right\}\)
b)
\(Ư\left(18\right)=\left\{\pm1;\pm2;\pm3;\pm6;\pm9;\pm18\right\}\)
c)
\(Ư\left(24\right)=\left\{\pm1;\pm2;\pm3;\pm4;\pm6;\pm8;\pm12;\pm24\right\}\)
\(\text{Ta có:}\)\(x>8\)\(\Rightarrow\)\(x\in\left\{12;24\right\}\)
a)Ư(8)={1;-1;2;-2;4;-4;8;-8}
Ư(12)={1;-1;2;-2;3;-3;4;-4;6;-6;12;-12}
ƯC(8;12)={1;-1;2;-2;4;-4}
b)B(8)={8;-8;16;-16;24;-24;32;-32;40;-40.....}
B(12)={12;-12;24;-24;36;-36.............}
BC(8;12)={24;-24;...................}
a, Ư(8)= {1;2;4;8 }
Ư(12)= { 1;2;3;4;6;12 }
ƯC(8,12)= { 1;2;4 }
b, B(8)= {8;16;24;32;40;48;...}
B(12)= { 12;24;36;48;60;... }
BC(8,12)= { 24;48 }