Question

Calculati c.m.m.d.c. si c.m.m.m.c. al numerelor:

a) a si a+1 , a∈N
b) 2a+3 si 3a+4 , a∈Z

Answer 1

a) Deoarece a si a+1 sunt numere naturale consecutive, sunt prime intre ele : 
d/a si d/a+1 ⇒ d/(a+1)-a ⇒ d/a+1-a ⇒ d/1 ⇒ d=1 ⇒ (a;a+1)=1
deoarece (a;a+1)·[a;a+1]=a·(a+1) ⇒ [a;a+1]=a·(a+1)
b) Fie d/2a+3⇒d/3·(2a+3)⇒d/6a+9  (1)
     si  d/3a+4⇒d/2·(3a+4)⇒d/6a+8  (2) .Din (1) si (2) ⇒
d/(6a+9)-(6a+8) ⇒ d/6a+9-6a-8 ⇒d/1 ⇒d=1 ⇒ (2a+3;3a+4)=1 deoarece
(2a+3;3a+4)·[2a+3;3a+4]=(2a+3)·(3a+4) ⇒[2a+3;3a+4]=(2a+3)·(3a+4)