fac
artificiu de calcul;
adun si sca d numere convenabile pt ca cel de sus sa se imparta exact la cel de
jos
imi convine ca 1=6-5, pt ca 6 e dublul lui 3 , asa cum 2x e dublul lui x
2(x=3): (x+3)=2, care e deja nr.natural
(2x+6-5)/(x+3) ∈N, 2-5/ (2x+1)∈N
2x+1∈D5
D5= {1;5}
2x+1=1...x=0 2-5/1=-3∉N deci 0 nu e bun
2x+1=5...x=2 2-5/5 =2-1=1∈N deci 2 e bun
x=2
3x+14=3x+6+8
(3x+6+8)/(x+2)=3 + 8/(x+2)
x+2∈Db= {1;2;4;8}
x∈{-1;0;2;6}∩N={0;2;6}
c)
(x+3) | (3x+25)
⇒(x+3) | (3x+9+16)
[3(x+3) +16]/(x+3)∈N
3+16/(x+3)∈N
dar 3∈N
ramane ca 16/(x+3)∈N
x+3 ∈{ 1;2;4;8.16}
x∈{-2;-1;1;5;13} ∩ N= {1;5;13}
