Question

Intr-o jumatate de ora am pregatire, am nevoie urgentaa!

Fie expresia E(x) = (x+2/x-5 - 14/x^2 -8x +15) :1/x^2 - 9, unde x e R\{-3,3,5}
a) Aratati ca x^2 - 8x +15 = (x-3) (x-5) , oricare ar fi x e R.
b) Demonstrati ca E(x) = (x+3) (x+4) ,oricare ar fi x e R\{-3,3,5}
c)Aratati ca E(a) este numar par, pentru orice a e N - {3,5}

Answer 1

a)
 x^2 - 8x +15 =x^2 -3x-5x+15=

=x(x-3)-5(x-3)=(x-3)(x-5)

b)
 E(x) = [(x+2)/(x-5) - 14/(x^2 -8x +15)] :[1/(x^2 - 9)]=

= [(x+2)/(x-5) - 14/(x-3)(x-5)]*(x^2 - 9)=

= [(x+2)/(x-5) - 14/(x-3)(x-5)]*(x-3)(x+3)=

=(x+2)(x-3)(x+3)/(x-5) -14(x-3)(x+3)/(x-3)(x-5)=

=(x+2)(x^2-9)/(x-5) -14(x+3)/(x-5)=

=(x^3 -9x+2x^2-18-14x-42)/(x-5)=

=(x^3 +2x^2 -23x-60)/(x-5)=

=(x^3 -5x^2 +7x^2 -35x +12x-60)/(x-5)=

=[x^2 (x-5) +7x(x-5) +12(x-5)]/(x-5)=

=(x-5)(x^2 +7x +12)/(x-5)=

=x^2 +7x +12=(x^2 +3x)+(4x+12)=

=x(x+3)+4(x+3)=(x+3)(x+4)

c)
E(a)=(a+3)(a+4)

E(a)=un produs de numere consecutive, care intotdeauna este un nr. par.

Exemple: 2x3=6,  9x10=90, 101x102=10302, etc.