maximo divisor comum de: (24,36) (24 ,32,36) (12,28)
Kaiane0506:
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Soluções para a tarefa
Respondido por
60
24l2
12l2
6 l2
3 l3
1
36l2
18l2
9 l3
3 l3
1
mdc 24= 2(sobre 3)x3
mdc 36= 2(sobre 2)x 3(sobre 2)
mdc(24,36)= 3
24l2
12l2
6 l2
3 l3
1
32l2
16l2
6 l2
3 l3
1
36l2
18l2
9 l3
3 l3
1
mdc 24= 2(sobre 3)x3
mdc 32= 2(sobre 5)
mdc 36= 2(sobre 2)x 3(sobre 2)
mdc(24,32,36)= 2x2
mdc(24,32,36)=4
12l2
6 l2
3 l3
1
28l2
14l2
7 l7
1
mdc 12= 2(sobre 2) x3
mdc 28= 2(sobre 2) x7
mdc(12,28)= 2x2
mdc(12,28)=4
Ufa! Aqui está :]
12l2
6 l2
3 l3
1
36l2
18l2
9 l3
3 l3
1
mdc 24= 2(sobre 3)x3
mdc 36= 2(sobre 2)x 3(sobre 2)
mdc(24,36)= 3
24l2
12l2
6 l2
3 l3
1
32l2
16l2
6 l2
3 l3
1
36l2
18l2
9 l3
3 l3
1
mdc 24= 2(sobre 3)x3
mdc 32= 2(sobre 5)
mdc 36= 2(sobre 2)x 3(sobre 2)
mdc(24,32,36)= 2x2
mdc(24,32,36)=4
12l2
6 l2
3 l3
1
28l2
14l2
7 l7
1
mdc 12= 2(sobre 2) x3
mdc 28= 2(sobre 2) x7
mdc(12,28)= 2x2
mdc(12,28)=4
Ufa! Aqui está :]
Respondido por
49
24/2 36/2
12/2 18/2
6/2 9/3
3/3 3/3
1 1
2x2x3= 12
24/2 32/2 36/2
12/2 16/2 18/2
6/2 8/2 9/3
3/3 4/2 3/3
1 2/2 1
1
2x2= 4
12/2 28/2
6/2 14/2
3/3 7/7
1 1
2x2= 4
12/2 18/2
6/2 9/3
3/3 3/3
1 1
2x2x3= 12
24/2 32/2 36/2
12/2 16/2 18/2
6/2 8/2 9/3
3/3 4/2 3/3
1 2/2 1
1
2x2= 4
12/2 28/2
6/2 14/2
3/3 7/7
1 1
2x2= 4
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