Răspuns:
Explicație pas cu pas:
a) E(x)= [(2x-1)²-(x-5)²]/(x³+4x²-4x-16)= [(2x-1-x+5)(2x-1+x-5)]/[x(x²-4) +4(x²-4)]= [(x+4)(3x-6)]/[(x²-4)(x+4)= 3(x-2)/(x²-4)= 3(x-2)/(x+2)(x-2)= 3/(x+2)
E(x)=3/(x+2)
b) F(x)= (x-3)²/(x³-x-2x²-2x)= (x-3)²/[x(x²-1)-2x(x+1)]= (x-3)²/[x(x-1)(x+1)-2x(x+1)]=(x-3)²/[x(x+1)(x-1-2)] = (x-3)²/[x(x+1)(x-3)]= (x-3)/[x(x+1)]
F(x)= (x-3)/[x(x+1)] E(x)=3/(x+2)
Pentru a aduce cele 2 expresii la numitor comun, amplificam fiecare fractie dupa cum urmeaza:
E(x)=3/(x+2)= 3x(x+1)/[x(x+1)(x+2)]
F(x)= (x-3)/[x(x+1)] = (x-3)(x+2)/[x(x+1)(x+2)]
E(x)=3x(x+1)/[x(x+1)(x+2)] F(x)= (x-3)(x+2)/[x(x+1)(x+2)]
Acum cele 2 expresii au acelasi numitor.
