[tex]b) \\ \frac{4n+9}{2n+1}= \frac{4n+2+7}{2n+1}=\frac{4n+2}{2n+1}+\frac{7}{2n+1} = \frac{2(2n+1)}{2n+1}+\frac{7}{2n+1} =2+\frac{7}{2n+1} \\ \text{2 este nr. natural} \\ 7 = 1 * 7 \\ =>\text{ trebuie sa rezolvam ecuatiile:} \\ 2n+1 = 1 => n_1 = 0 \\ 2n+1 = 7 => n_2 = 3 \\ Verificare: \\ \\ \frac{4*0+9}{2*0+1}= \frac{9}{1} = 9 \;\;(\text{numar natural}) \\ \\ \frac{4*3+9}{2*3+1}= \frac{21}{7} = 3 \;\;(\text{numar natural})[/tex]
[tex]c) \\ \frac{6n+20}{n+3}= \frac{6n+18+2}{n+3}=\frac{6n+18}{n+3}+\frac{2}{n+3} = \frac{6(n+3)}{n+3}+\frac{2}{n+3} =6+\frac{2}{n+3} \\ \text{6 este nr. natural} \\ 2 = 1 * 2 \\ =>\text{ trebuie sa rezolvam ecuatiile:} \\ n+3 = 1 => n_1 = -2 \text{ (solutie eliminata deoarece -2 nu e natural)} \\ n+3 = 2 => n_2 = -1 \text{ (solutie eliminata deoarece -1 nu e natural)} \\ \text{Concluzie: Ecuatia nu are solunie in N}[/tex]
