Question

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 

Answer 1

a=$\frac{3n+4+5m+8}{2n+3+2m+3}$

a=4 =>4=$\frac{3n+4+5m+8}{2n+3+2m+3}$ => $\frac{4}{1}= \frac{3n+12+5m}{2n+6+2m}$

amplifici pe $\frac{4}{1}$ cu 2n+6+2m

=> $\frac{8n+24+8m}{2n+6+2m}= \frac{3n+12+5m}{2n+6+2m} => (8n+24+8m)*(2n+6+2m)=(2n+6+2m)*(3n+12+5m) =>  [tex]16n^{2}$+48n+16mn+48n+144+48m+16mn+48m+$16m^{2}$= $6n^{2}$+24n+10m+9n+72+30m+6mn+24M+$10m^{2}$

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