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a)16=24
24=23.3
=>UCLN(16,24)=23=8
b)180=22.32.5
234=2.32.13
=>UCLN(180,234)=2.32=18
c)60=22.3.5
90=2.32.5
135=33.5
=>UCLN(60,90,135)=3.5=15
a) ƯCLN (16, 24) = 8, ƯC (16, 24) = {1; 2; 4; 8};
b) Ta có 180 = 22 . 32 . 5; 234 = 2 . 32 . 13;
ƯCLN (180, 234) = 2 . 32 = 18, ƯC (180, 234) = {1; 2; 3; 6; 9; 18};
c) Ta có 60 = 22 . 3 . 5; 90 = 2 . 32 . 5; 135 = 33 . 5. Do đó
ƯCLN (60, 90, 135) = 3 . 5 = 15; ƯC (60, 90, 135) = {1; 3; 5; 15}.
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a)16=24
24=23.3
=>UCLN(16,24)=23=8
b)180=22.32.5
234=2.32.13
=>UCLN(180,234)=2.32=18
c)60=22.3.5
90=2.32.5
135=33.5
=>UCLN(60,90,135)=3.5=15
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a, Ta có : \(Ư\left(16\right)=1;2;4;8;16\) và \(Ư\left(24\right)=1;2;3;4;6;8;12;24\)
\(\RightarrowƯC\left(16;24\right)=1;2;4;8\)
b, Ta có : \(\hept{\begin{cases}Ư\left(60\right)=1;2;3;4;5;6;10;12;15;20;20;60\\Ư\left(90\right)=1;2;3;5;6;9;10;15;18;30;45;90\end{cases}}\)
\(\RightarrowƯC\left(60;90\right)=1;2;3;5;6;15;30\)
c, Ta có : \(\hept{\begin{cases}Ư\left(18\right)=1;2;3;6;9;18\\Ư\left(30\right)=1;2;3;5;6;10;15;30\\Ư\left(42\right)=1;2;3;6;7;14;21;42\end{cases}}\)\(\RightarrowƯC\left(18;30;42\right)=1;2;3;6\)
d,Ta có : \(\hept{\begin{cases}Ư\left(26\right)=1;2;13;26\\Ư\left(39\right)=1;3;13;39\\Ư\left(48\right)=1;2;4;6;8;12;16;24;48\end{cases}}\)\(\RightarrowƯC\left(26;39;48\right)=1\)
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\(ƯC\left(8,12\right)=\left\{\pm1;\pm2;\pm4\right\}\)
\(ƯC\left(12;15;30\right)=\left\{\pm1;\pm3\right\}\)
\(ƯC\left(60;72\right)=\left\{\pm1;\pm2;\pm3;\pm4;\pm6;\pm12\right\}\)
\(ƯC\left(24;42\right)=\left\{\pm1;\pm2;\pm3;\pm6\right\}\)
Ta có :
60 = 22.3.5
90 = 2.32.5
135=33.5
=> ƯCLN(60,90,135) = 3.5=15
=> ƯC(60,90,135)=Ư(15)={1;3;5;15}