Question

Fie m,n numere naturale. Arătați că, dacă a este număr natural: 
a= $\frac{3n+4}{2n+3} + \frac{5m+8}{2m+3}$, atunci a=4 . :)

Answer 1

[tex]a= \frac{2n+3}{2n+3} + \frac{n+1}{2n+3} + \frac{2(2m+3)}{2m+3} + \frac{m+2}{2m+3} \\ a=1 + \frac{n+1}{2n+3} + 2 + \frac{m+2}{2m+3}\\ Notam\ t=\frac{n+1}{2n+3} + \frac{m+2}{2m+3}\in N\\ 2t=\frac{2n+2}{2n+3} + \frac{2m+4}{2m+3}\\ 2t=\frac{2n+3-1}{2n+3} + \frac{2m+3+1}{2m+3}\\ 2t=1-\frac{1}{2n+3} +1+ \frac{1}{2m+3}\\ 2t-2=-\frac{1}{2n+3} + \frac{1}{2m+3}\\ 2(t-1)= \frac{1}{2m+3}-\frac{1}{2n+3} \\ 2(t-1)= \frac{2n+3-2m-3}{(2m+3)(2n+3)} \\ 2(t-1)=\frac{2(n-m)}{(2m+3)(2n+3)} \\ t-1=\frac{n-m}{(2m+3)(2n+3)}\in N[/tex]
[tex]n-m=0=>t=1\\ a=1+2+1=4[/tex]