Răspuns:
Explicație pas cu pas:
a. amplificam pe prima xu x+2 si a doua pe x+3
obtinem sus la numarator
x+2-x-3=-1
si la numitor (x+2)(x+3)=x²+5x+6
b. [tex](\frac{x}{(x+3)^{2} } -\frac{1}{x+3} +\frac{1}{x+2}[/tex]
lucram pe rand prima paranteza
aducem la acelasi numitor comun
[tex]\frac{x^{2} +2x-x^{2} -5x-6+x^{2} +6x+9}{(x+3)^{2} (x+2)}[/tex]
[tex]\frac{x^{2} +3x+3}{(x+3)^{2} (x+2)}[/tex]
3x²+15x+18==3(x²+5x+6)=3(x+3)(x+2)
E(x)=[tex]\frac{x^{2} +3x+3}{(x+3)^{2} (x+2)} \cdot \frac{3(x+2)(x+3)}{x^{2} +3x+3 }[/tex]
observam ca in E(x) x²+3x+3 se simplifica si (x+3)(x+2) se simplifica
ne ramane
E(x)=[tex]\frac{3}{x+3}[/tex]
c. [tex]\frac{3}{x+3}[/tex]∈Z daca x+3={1,-1,3,-3}
adica x={-2,-4,0,-6}
