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1244690mery85312
a fost răspuns

Demonstrați că :
a.(2n+3, 5n+7) = 1
b.(n+3, 2n+7) = 1
c.(4n+5, 5n+6) = 1
d.(2n+1, 3n+2) = 1
e.(5n+6, 6n+7) = 1
f.(3n+4, 5n+7) = 1​

Răspuns :

A) 2n+3=1
2n=1-3
2n=-2
n=-1

5n+7=1
5n=1-7
5n=-6
n=-6/5

B) n+3=1
n=-2

2n+7=1
2n=1-7
2n=-6
n=-3

C)4n+5=1
4n=-4
n=-1

5n+6=1
5n=-5
n=-1

D)2n+1=1
2n=0

3n+2=1
3n=1-2
3n=-1
n=-1/3

E)5n+6=1
5n=1-6
5n=-5
n=-1

6n+7=1
6n=1-7
6n=-6
n=-1

F) 3n+4=1
3n=1-4
3n=-3
n=-1

5n+7=1
5n=1-7
5n=-6
n=-6/5

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