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1,Tìm ƯC(-12,18)
Ta có :Ư(-12)={1;-1;2;-2;3;-3;4;-4;6;-6;12;-12}
Ư(18)={1;-1;2;-2;3;-3;6;-6;8;-8;9;-9;18;-18}
=>ƯC(-12,18)={1;-1;2;-2;3;-3;6;-6}
Vậy ƯC(-12;18)={1;-1;2;-2;3;-3;6;-6}
2,Tìm BC(-15,20)
Ta có:B(-15)={...;-30;-15;0;15;30;...}
B(20)={...;-40;-20;0;20;40;...}
=>BC(-15;20)={...;0;...}
Vậy BC(-15;20)={...;0;...}
\(ƯC\left(-12;18\right)\)
\(-12=-\left(2^2.3\right)\)
\(18=2.3^2\)
\(ƯCLN\left(-12;18\right)=-3.2=-6\)
\(\RightarrowƯC\left(-12;18\right)=Ư\left(-6\right)=\left\{-1;1;-2;2;-3;3;-6;6\right\}\)
\(BC\left(-15;20\right)\)
\(-15=-\left(3.5\right)\)
\(20=4.5\)
\(BCNN\left(-15;20\right)=3.4.5=60\)
\(\Rightarrow BC\left(-15;20\right)=B\left(60\right)=\left\{0;60;120;180;240;...\right\}\)
Tìm ƯC(90;252) như sau
Ta có:
90=2x3^2x5 ; 252=2^2x3^2x7
ƯCLN( 90;252 )= 2x3^2 = 18
ƯC( 90;252 )= Ư(18)
= { 1;2;3;6;9;18 }
Tìm BC( 90;252 )
Ta có:
90= 2x3^2x5 ; 252= 2^2x3^2x7
BCNN ( 90;252 )= 2^2x3^2x5x7=1260
BC ( 90;252 )= B( 1260)
= { 0; 1260; 2520;....}
( NHỚ TICK CHO MK NHÉ CÁC BN!!! )
a) \(ƯC_{a;a+1}=\left\{1\right\}\)
b) \(BC_{2;5}=\left\{10;20;30;40;50....\right\}\)
a,ƯCLN(24; 36)
Ta có:
24= 23.3
36= 2.33
Vậy: ƯCLN(24; 36)= 2.3= 6
ƯC( 24; 56)
Ta có:
Ư(24)= {1;2;3;4;6;8;12;24}
Ư(56)= {1;2;4;7;8;14;28;56}
ƯC(6; 8)={1;2;4;8}
b,BCNN(10; 12)
Ta có:
10=2.5
12=22.3
Vậy: BCNN(10; 12)=22.3.5=60
BC(10; 12)
Ta có:
B(10)={0;10;20;30;40;50;60;...}
B(12)={0;12;24;36;48;60;...}
BC(10; 12)={0;60;...}
9=3^2. 18=2.3^2. 36=2^2.3^2
UCLN(9,18,36)= 3^2=9
=> UC(9,18,36)={1,3,9}
BCNN( 9,18,36)=3^2..2^2=36
=> BC(9,18,36)={0,36,72,108,144,...}
Phân tích mỗi số ra TSNT:
68 = 22 x 17
264 = 23 x 3 x 11
15 = 3 x 5
=> BCNN (68;264;15) = 23 x 5 x 3 x11 x17 = 22440
=> BCNN (68;264;15) = B(22440) = { 0 ; 22440 ; 44880 ; 67321 ; 89760 ; .......}
=> UCLN (68;264;15) = 1
=> UC (68;264;15) = 1
Ta có: 68 = 22 . 17
264 = 23. 3 . 11
15 = 3 . 5
=> BCNN(68, 264, 15) = 23 . 3 . 5 . 11 . 17 = 22440
=> ƯCLN(68, 264, 15) = 1
Khi đó:
BC (68, 264, 15) = B(22440) = {0 ; 22440 ; 44880 ; 67320 ; ...}
ƯC(68, 264, 15) = Ư(1) = {1}
`x ∈ ƯC (150,35)`
Ta có:
`150 = 2 . 3 . 5^2`
`35 = 5 . 7 `
`=> UCLN(150;35) = 5`
Hay `x ∈ U(5) = {-5;-1;1;5}`
Vậy ....
`x ∈ BC (25,30,40)`
Ta có:
`25 = 5^2`
`30 = 2.3.5`
`40 = 2^3 . 5`
`=> BCNN = 5^2 . 2^3 . 3 = 600`
`=> x ∈ B(600) = {0;600;1200;....}`
Vậy ....