[tex]= \frac{8x-16}{ x^{2} +x+1} * \frac{ x^{3} -1}{ x^{2} -4x+4} = \frac{8(x-2)}{ x^{2} +x+1} * \frac{(x-1)( x^{2} +x+1)}{ (x-2)^{2} } = \frac{8}{1} * \frac{(x-1)}{(x-2)} = \frac{8(x-1)}{x-2} [/tex]
iar daca x-2=0 =>x=2, deci ec nu are solutii => DVA=R\{2}
cum vreti voi, eu l-am stabilit dupa ce am simplificat, lucru care se face la limite,initial se simplifica sau se da factor comun fortat apoi calculele, bine in functie de cazuri, in deoseb la 0/0 , in caz ca ati uitat, dar in fine, [tex] x^{3} -1 \neq 0 => x \neq 1[/tex]
