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\(a,ƯC\left(16,24\right)=Ư\left(8\right)=\left\{\pm1;\pm2;\pm4;\pm8\right\}\\ b,BC\left(84,108\right)=B\left(756\right)=\left\{0;756;1512;2268;...\right\}\\ c,ƯC\left(36,48,120\right)=Ư\left(12\right)=\left\{\pm1;\pm2;\pm3;\pm4;\pm6;\pm12\right\}\\ d,BC\left(42,64,70\right)=B\left(6720\right)=\left\{0;6720;13440;20160;...\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,ƯCLN\left(12,15\right)=3\\ b,ƯCLN\left(18,42\right)=6\\ c,ƯCLN\left(24,36,40\right)=2\\ d,ƯCLN\left(150,84,30\right)=6\\ e,BCNN\left(20,30\right)=60\\ f,BCNN\left(84,108\right)=756\\ g,BCNN\left(45,150\right)=450\\ h,BCNN\left(12,15,10\right)=60\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Vì ƯCLN(a,b)=15+>a=15m;b=15.n và (m;n)=1
Từ đó,suy ra : BCNN(a,b)=15.m.n=300
=>m.n=20=20.1=1.20=4.5=5.4
Xét :
*m=20;n=1=>a=300;b=15
*m=1;n=20=>a=15;b=300
*m=4;n=5=>a=60;b=75
*m=5;n=4=>a=75;b=60
Mà từ giả thuyết có:a+15=b nên (a;b)E{(75;60)}
Ta có : ƯCLN ( a , b ) = 15
=> a = 15 . m b = 15 . n. ( m , n ) = 1
=> BCNN ( a , b ) = 15 . m . n = 300
=> m . n = 300 : 15 = 20
=> m . n = 1 . 20 = 4 . 5 = 2 . 10 = 10 . 2 = 5 . 4 = 20 . 1
Xét :
- m = 1 , n = 20 => a = 15 , b = 300
- m = 20 , n = 1 => a = 300 , b = 15
- m = 4 , n = 5 => a = 60 , b = 75
- m = 5 , n = 4 => a = 75 , b = 60
- m = 2 , n = 10 => a = 30 , b = 150
- m = 10 , n = 2 => a = 150 , b = 30
Vì a + 15 = b
=> a = 60 , b = 75
(
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,BC\left(6,14\right)=B\left(42\right)=\left\{0;42;84;...\right\}\\ b,BC\left(6,20,30\right)=B\left(60\right)=\left\{0;60;120;180;...\right\}\\ c,BCNN\left(1,6\right)=6\\ d,BCNN\left(10,1,12\right)=60\\ e,BCNN\left(5,14\right)=70\)
a) 16 = 2⁴
24 = 2³.3
⇒ ƯCLN(16; 24) = 2³ = 8
⇒ ƯC(16; 24) = Ư(8) = {1; 2; 4; 8}
b) 84 = 2².3.7
108 = 2².3³
⇒ BCNN(84; 108) = 2².3³.7= 756
⇒ BC(84; 108) = B(756) = {0; 756; 1512; ...}