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a, Ta có: 10 = 2.5; 12 = 2 2 .3
BCNN(10,12) = 2 2 .3.5 = 60
b, Ta có: 24 = 2 3 .3; 56 = 2 3 .7
BCNN(24;56) = 2 3 .3.7 = 168
c, Ta có: 84 = 2 2 .3.7; 108 = 2 2 . 3 3
BCNN(84;108) = 2 2 . 3 3 .7
d, Ta có: 33 = 3.11; 44 = 4.11; 55 = 5.11
BCNN(33;44;55) = 3.4.5.11 = 660
e, Ta có: 8 = 2 3 ; 18 = 2. 3 2 ; 30 = 2.3.5
BCNN(8,18,30) = 2 3 . 3 2 . 5 = 360
f, Ta có: 40 = 2 3 . 5 ; 28 = 2 2 .7; 140 = 2 2 .5.7
BCNN(40;28;140) = 2 3 . 5 . 7 = 280
a: \(42=2\cdot3\cdot7;70=2\cdot5\cdot7\)
=>\(BCNN\left(42;70\right)=2\cdot3\cdot5\cdot7=210\)
=>\(BC\left(42;70\right)=B\left(210\right)=\left\{0;210;420;...\right\}\)
b: \(70=2\cdot5\cdot7;180=3^2\cdot5\cdot2^2\)
=>\(BCNN\left(70;180\right)=2^2\cdot3^2\cdot5\cdot7=1260\)
=>\(BC\left(70;180\right)=\left\{1260;2520;...\right\}\)
c: \(5=5;7=7;8=2^3\)
=>\(BCNN\left(5;7;8\right)=5\cdot7\cdot8=280\)
=>\(BC\left(5;7;8\right)=\left\{280;560;...\right\}\)
d: \(12=2^2\cdot3;18=3^2\cdot2\)
=>\(BCNN\left(12;18\right)=2^2\cdot3^2=36\)
=>\(BC\left(12;18\right)=\left\{36;72;...\right\}\)
e: \(15=3\cdot5;18=3^2\cdot2\)
=>\(BCNN\left(15;18\right)=3^2\cdot2\cdot5=90\)
=>\(BC\left(15;18\right)=\left\{90;180;...\right\}\)
f: \(84=2^2\cdot3\cdot7;108=3^3\cdot2^2\)
=>\(BCNN\left(84;108\right)=2^2\cdot3^3\cdot7=756\)
=>\(BC\left(84;108\right)=\left\{756;1512;...\right\}\)
j: \(33=3\cdot11;44=2^2\cdot11;55=5\cdot11\)
=>\(BCNN\left(33;44;55\right)=3\cdot2^2\cdot5\cdot11=660\)
=>\(BC\left(33;44;55\right)=\left\{660;1320;...\right\}\)
g: \(1=1;12=2^2\cdot3;27=3^3\)
=>\(BCNN\left(1;12;27\right)=1\cdot2^2\cdot3^3=108\)
=>\(BC\left(1;12;27\right)=\left\{108;216;...\right\}\)
n: \(5=5;9=3^2;11=11\)
=>\(BCNN\left(5;9;11\right)=5\cdot3^2\cdot11=495\)
=>\(BC\left(5;9;11\right)=\left\{495;990;...\right\}\)
a)15va 18
15=3.5
18=3^2.2
BCNN(15;18)=2.3^2.5=90
d)33;44va55
33=3.11
44=4.11
55=5.11
BCNN(33;44;55)=3.4.5.11=660
c)4;14 va 26
4=2^2
14=2.7
26=2.13
BCNN(4;14;26)=2^2.7.13=364
\(a.15=3.5;18=2.3^2\)
\(BCNN\left(15;18\right)=2.3^2.5=90\)
\(b.33=3.11;44=2^2.11;55=5.11\)
\(BCNN\left(33;44;55\right)=2^2.3.5.11=660\)
a) Ta có: 6 = 2.3; 14 = 2.7
=> BCNN(6, 14) = 2.3.7 = 42
=> BC(6, 14) = B(42) = {0; 42; 84; 126;... }
b) Ta có: 6 = 2.3; 20 = 22.5; 30 = 2.3.5
=> BCNN(6, 20, 30) = 22.3.5 = 60
=> BC(6, 20, 30) = B(60) = {0; 60; 120; 180; 240;...}.
c) Vì hai số 1 và 6 là hai số nguyên tố cùng nhau => BCNN(1, 6) = 1.6 = 6.
d) Ta có: 10 = 2.5
12 = 22.3
=> \(BCNN(10, 1, 12) = 2^2.3.5 = 60.\)
e) Vì hai số 5 và 14 là hai số nguyên tố cùng nhau => BCNN(5, 14) = 5 . 14 = 70.
a: BCNN(60;280)=840
b: BCNN(84;108)=756
c: BCNN(5;8;15)=120
d: BCNN(12;16;48)=48
a) Ta có 15 = 3 . 5 ; 18 = 2 . 3 2 → B C N N ( 15 ; 18 ) = 2 . 3 2 . 5 = 90 .
b) Ta có 84 = 2 2 . 3 . 7 ; 108 = 2 2 . 3 3 ;
→ B C N N ( 84 ; 104 ) = 2 2 . 3 3 . 7 = 756 .
c) 660.
d) 360.