Fie m,n numere naturale .Aratati ca,daca a este numar natural :
  a=3n+4 +5m+8 
      supra 
      2n+3 + 2m+3   . atunci a=4 



Răspuns :

a=[tex] \frac{3n+4+5m+8}{2n+3+2m+3}[/tex]

a=4 =>4=[tex] \frac{3n+4+5m+8}{2n+3+2m+3} [/tex] => [tex] \frac{4}{1}= \frac{3n+12+5m}{2n+6+2m} [/tex]

amplifici pe [tex] \frac{4}{1} [/tex] cu 2n+6+2m

=> [tex] \frac{8n+24+8m}{2n+6+2m}= \frac{3n+12+5m}{2n+6+2m} =>

(8n+24+8m)*(2n+6+2m)=(2n+6+2m)*(3n+12+5m)

=>

 [tex]16n^{2} [/tex]+48n+16mn+48n+144+48m+16mn+48m+[tex] 16m^{2} [/tex]= [tex] 6n^{2} [/tex]+24n+10m+9n+72+30m+6mn+24M+[tex] 10m^{2} [/tex]

Asta e prima parte... o pun imediat pe a 2-a, ai putina rabdare :)