tim tat ca cac uoc -10 ;12
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tim uoc chung lon nhat cua tat ca cac so tu nhien co 9 chu so gom cac chu so tu 1 den 9
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tim uoc chung lon nhat cua tat ca cac so tu nhien co 6 chu so gom cac chu so tu 1 den 6
tim tat ca cac uoc cua 5,9,8,-13,1,-8
\(Ư\left(5\right)=\left\{1;-1;5;-5\right\}\)
\(Ư\left(9\right)=\left\{1;-1;3;-3;9;-9\right\}\)
\(Ư\left(8\right)=\left\{1;-1;2;-2;4;-4;8;-8\right\}\)
\(Ư\left(-13\right)=\left\{1;-1;13;-13\right\}\)
\(Ư\left(1\right)=\left\{1;-1\right\}\)
\(Ư\left(-8\right)=\left\{1;-1;2;-2;4;-4;8;-8\right\}\)
5 = { 5 , -5 , 1 , -1}
9={9,-9 , 3 , -3 , 1 ,-1}
8 = {8 , -8 , 4 , -4 , 2 , -2 , 1 , -1 }
-13 = {13 , -13 , 1 , -1 }
1 ={1 , -1 }
-8 = { 8 , -8 , 4 , -4 , 2 , -2 , 1 , -1}
Ư(5) = {-5; -1; 1; 5}
Ư(9) = {-9; -3; -1; 1; 3; 9}
Ư(8) ={-8; -4; -2; -1; 1; 2; 4; 8}
Ư(-13) = {-13; -1; 1; 13}
Ư(1) = {-1; 1}
Ư(-8) = {-8; -4; -2; -1; 1; 2; 4; 8}
tim tat ca cac uoc lon hon am 4 cua 12
12 = 22 . 3
Ư(12) = { 12; 6; 4; 3; 2; 1; -1; -2; -3 }
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TIM TAT CA CAC SO CO HAI CHU SO LA UOC CHUNG CUA A VA B , BIET RANG UOC CHUNG LON NHAT [A,B]=80
\(80=2^4.5\)
\(USC\left(A,B\right)=\left\{10;16;20;40;80\right\}\)
tim tat ca cac uoc chung cua hai so 3n+11va n + 2 voi n thuoc N*
tim tat ca cac so nguyen to p sao cho p^2 +59 co dung 6 uoc duong
tim tat ca cac so nguyen to p sao cho p^2 +59 co dung 6 uoc duong
tim tat ca cac so nguyen to p sao cho p-1 va p+1 deu co 6 uoc
2. tat ca cac uoc cua 8;11;19;28
3.cac so sau co bao nhieu uoc A) 45 B) 103
\(2,Ư\left(8\right)=\left\{-8;-4;-2;-1;1;2;4;8\right\}\\ Ư\left(11\right)=\left\{-11;-1;1;11\right\}\\ Ư\left(19\right)=\left\{-19;-1;1;19\right\}\\ Ư\left(28\right)=\left\{-28;-14;-7;-4;-2;-1;1;2;4;7;14;28\right\}\)
3.
\(A,45=3^2\cdot5\) nên có \(\left(2+1\right)\left(1+1\right)=6\left(ước\right)\)
\(B,103=103\) nên có \(1+1=2\left(ước\right)\)
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2. Ư(8) = {1; 2; 4; 8}
Ư(11) = {1; 11}
Ư(19) = {1; 19}
Ư(28) = {1; 2; 4; 7; 14; 28}
3. Ư(45) = {1; 3; 5; 9; 15; 45} \(\Rightarrow\) 45 có 6 ước
Ư(103) = {1; 103} \(\Rightarrow\) 103 có 2 ước
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