Bài 9: Tìm BC thông qua tìm BCNN
a) 10 và 24
c) 20, 35, 60
b) 48, 120
d) 18, 24, 32
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a 10=2.5
24=23.3
BCNN(10,24)=23.5.3=120
→BC(10,24)=B(120)={0;120;240;360;480;....}
b 20=22.5
35=7.5
60=22.3.5
BCNN(20,35,60)=22.3.5.7=420
→BC(20,35,60)=B(420)={0;420;840;1260;...}
c 48=24.3
120=23.3.5
BCNN(48,120)=24.3.5=240
→BC(48,120)=B(240)={0;240;480;720;...}
d
18=2.32
24=23.3
32=25
BCNN(18,24,32)=25.32=288
→BC(18,24,32)=B(288)={0;288;576;864;...}
\(a,ƯC\left(40,24\right)=Ư\left(8\right)=\left\{...\right\}\\ b,ƯC\left(12,52\right)=Ư\left(4\right)=\left\{...\right\}\\ c,ƯC\left(36,990\right)=Ư\left(18\right)=\left\{...\right\}\\ d,ƯC\left(54,36\right)=Ư\left(9\right)=\left\{...\right\}\\ e,ƯC\left(10,20,70\right)=Ư\left(10\right)=\left\{...\right\}\\ f,ƯC\left(25,55,75\right)=Ư\left(5\right)=\left\{...\right\}\\ g,ƯC\left(80,144\right)=Ư\left(16\right)=\left\{...\right\}\\ h,ƯC\left(63,2970\right)=Ư\left(9\right)=\left\{...\right\}\\ i,ƯC\left(65,125\right)=Ư\left(5\right)=\left\{...\right\}\\ j,ƯC\left(9,18,72\right)=Ư\left(9\right)=\left\{...\right\}\\ k,ƯC\left(24,36,60\right)=Ư\left(12\right)=\left\{...\right\}\\ l,ƯC\left(16,42,86\right)=Ư\left(2\right)=\left\{..\right\}\)
\(BCNN\left(9,24,35\right)=2520\\ BCNN\left(18,36,108\right)=108\)
g) Ta có: 54 = 2 . 33
36 = 22.32
=>ƯCLN(54; 36) = 2.32 = 18
=> ƯC(54;36) = Ư(18) = {1; 2; 3; 6; 9; 18}
j) Ta có: 9 = 32
18 = 2.32
72 = 23 . 32
=> ƯCLN(9; 18; 72) = 32 = 9
=> ƯC(9; 18; 72) = Ư(9) = {1; 3; 9}
h) Ta có: 10 = 2.5
20 = 22 . 5
60 = 22 . 3.5
=> ƯCLN(10; 20; 60) = 2.5 = 10
=> ƯC(10; 20; 60) = Ư(10) = {1; 2; 5; 10}
g) Ta có : 54 = 2.33
36 = 22.32
=> ƯCLN ( 54; 36 ) = 2.32 = 18
=> ƯC ( 54;36 ) = Ư(18) = ( 1; 2; 3; 6; 9; 18 )
j) Ta có : 9 = 32
18 = 2.32
72 = 23.32
=> ƯCLN ( 9; 18; 72 ) = 32 = 9
h) Ta có : 10 = 2.5
20 = 22.5
60 = 22.3.5
=> ƯCLN ( 10; 20; 60 ) = 2.5 = 10
=> ƯC ( 10; 20; 60 ) = Ư ( 10 ) - { 1; 2; 5; 10 }
1+2+3+4+5+6+7+8+9+10+11+12+13+14+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+40+41+42+43+44+45+46+47+48+49+50=1275
a: \(12=2^2\cdot3;18=3^2\cdot2\)
=>\(ƯCLN\left(12;18\right)=2\cdot3=6\)
b: \(24=2^3\cdot3;48=2^4\cdot3\)
=>\(ƯCLN\left(24;48\right)=2^3\cdot3=24\)
c: \(44=2^2\cdot11;121=11^2\)
=>\(ƯCLN\left(44;121\right)=11\)
d: \(36=3^2\cdot2^2;108=3^3\cdot2^2;224=2^5\cdot7\)
=>\(ƯCLN\left(36;108;224\right)=2^2=4\)
a, Ta có :
\(12=2\cdot2\cdot3=2^2\cdot3\)
\(18=2\cdot3\cdot3=3^2\cdot2\)
Thừa số nguyên tố chung là : \(2;3\)
\(ƯCLN\) \(\left(12;18\right)=2\cdot3=6\)
⇒ \(ƯCLN\) \(\left(12;18\right)\) là \(6\)
b, Ta có:
\(24=2\cdot2\cdot2\cdot3=2^3\cdot3\)
\(48=2\cdot2\cdot2\cdot2\cdot3=2^4\cdot3\)
Thừa số nguyên tố chung là : \(2;3\)
\(ƯCLN\) \(\left(24;48\right)\) \(=2^3\cdot3=8\cdot3=24\)
⇒ \(ƯCLN\) \(\left(24;48\right)\) là \(24\)
c, Ta có
\(44=2\cdot2\cdot11=2^2\cdot11\)
\(121=11\cdot11=11^2\)
Thừa số nguyên tố chung là : \(11\)
\(ƯCLN\) \(\left(44;121\right)\) \(=11\)
⇒ \(ƯCLN\) là \(11\)
d, Ta có :
\(36=2\cdot2\cdot3\cdot3=2^2\cdot3^3\)
\(108=2\cdot2\cdot3\cdot3\cdot3=2^2\cdot3^3\)
\(224=2\cdot2\cdot2\cdot2\cdot2\cdot7=2^5\cdot7\)
Thừa số nguyên tố chung là : \(2\)
\(ƯCLN\) \(\left(36;108;224\right)\) \(=2\cdot2=2^2=4\)
⇒ \(ƯCLN\) \(\left(36;108;224\right)\) là \(4\)
\(#thaolinh\)