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UCLN(16;42;86)=2
UC(16;42;86)={1;2;-1;-2}
BCNN(16;42;86)=14448
BC(16;42;86)={14448;28896;...}
16= 24 ; 42= 2.3.7 ; 86= 2.43
UCLN(16;42;86)=2
UC(16;42;86)={1;2;-1;-2}
BCNN(16;42;86)=14448
BC(16;42;86)={14448;28896;...}
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Ta có:100 = 22 .52 ; 150 = 2.3.52 ;125 = 53
ƯCLN(100;150;125) = 52
BCNN(100;150;125) = 22.3.53 = 1500
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46=2.23
54=2.33
112=24.7
UCLN(46,54.112)=2
UC(46,54,112)=U(2)=(1,2)
BCNN=24.33.23.7=68552
BC(46,54,112)=B(68552)=(0,68552,137104,....)
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a: \(24=2^3\cdot3;40=2^3\cdot5\)
=>\(ƯCLN\left(24;40\right)=2^3=8\)
=>\(ƯC\left(24;40\right)=\left\{1;-1;2;-2;4;-4;8;-8\right\}\)
b: \(12=2^2\cdot3;52=2^2\cdot13\)
=>\(ƯCLN\left(12;52\right)=2^2=4\)
=>\(ƯC\left(12;52\right)=\left\{1;-1;2;-2;4;-4\right\}\)
c: \(45=3^2\cdot5;180=2^2\cdot3^2\cdot5\)
=>\(ƯCLN\left(45;180\right)=3^2\cdot5=45\)
=>\(ƯC\left(45;180\right)=\left\{1;-1;3;-3;5;-5;9;-9;15;-15;45;-45\right\}\)
d: \(27=3^3;45=3^2\cdot5;75=3\cdot5^2\)
=>\(ƯCLN\left(27;45;75\right)=3\)
=>\(ƯC\left(27;45;75\right)=\left\{1;-1;3;-3\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bg :
-Tìm ƯCLN + ƯC :
36 = 22 . 32
18 = 2 . 32
24 = 23 . 3
\(\RightarrowƯCLN\left(36,18,24\right)=2.3=6\)
\(\RightarrowƯC\left(36,18,24\right)=Ư\left(6\right)=\left\{1,2,3,6\right\}\)
-Tìm BCNN và BC :
-Phân tích ra thừa số nguyên số như ở trên ((:
\(\Rightarrow BCNN\left(36,18,24\right)=2^3.3^2=72\)
\(\Rightarrow BC\left(36,18,24\right)=B\left(72\right)=\left\{0,72,...\right\}\)
P/s : Hok tốt a~
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1
BC(15;25)>400= (0;75;150)
UC(108;180)<15=(18;36)
A thuộc BCNN(126;198)=1386
1 thuộc UCLN(90:126)=18
2
UCLN(40;52;70)=2; BCNN(40;52;70)=3640
UCLN(12;18)=6
UCLN(60;72)=12
3
A=(15;16;17;18;19;20;21;22;23;24;25;26;27;28;29;30;31;32;33;34;35;36;37;38;39)
số phần tử:(39-15):1+1=25
tổng:(39+15)x25:2=675
tick mình nha bạn HAPPY NEW YEAR!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
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Câu10: m là số tự nhiên nhỏ nhất khác 0 mà m đều chia hết cho cả a và b thì:
A. m ϵ BC(a;b).
B. m ϵ UC(a;b).
C. m = UCLN(a;b).
D. m = BCNN(a;b)
ta có :
a, 40=23*5
52=22*13
ƯCLN ( 40;52)=4
BCNN(40.52)=520
a) 40 = 23 . 5
52 = 22 . 13
=> ƯCLN(40, 52) = 22 = 4
=> BCNN(40, 52) = 23 . 5 . 13 = 520
ƯC(40, 52) = Ư(4) = { 1 ; 2 ; 4 }
BC(40, 52) = B(520) = { 0 ; 520 ; 1040 ; 1560 ; 2060 ; ... }
b) 40 = 23 . 5
72 = 23 . 32
196 = 22 . 72
=> ƯCLN(40, 72, 196) = 22 = 4
=> BCNN(40, 72, 196) = 23 . 32 . 5 . 72 = 17 640
ƯC(40, 72, 196) = Ư(4) = { 1 ; 2 ; 4 }
BC(40, 72, 196) = B(17 640) = { 0 ; 17 640 ; 35 820 ; ... }