Question

Va rog ajutati-ma cu acest ex va rog

Answer 1

Răspuns:

Explicație pas cu pas:

E(x)=[1/(x+2)-1/(x-2)+(x+1)/(x²-4)]:(x²-9)/(x²+x-6)=

[1/(x+2)-1/(x-2)+(x+1)/(x²-4)]·(x²+x-6)/(x²-9)

a) E(x) are sens cand numitorii sunt diferiti de 0.  

⇒(x²-4)≠0  ⇒x≠2 si x≠-2

  (x²-9)≠0  ⇒x≠3  si x≠-3   ⇒ E(x) are sens pt. x∈R\{-3, -2, 2, 3}

b) E(x)=[1/(x+2)-1/(x-2)+(x+1)/(x²-4)]:(x²-9)/(x²+x-6)=

[1/(x+2)-1/(x-2)+(x+1)/(x²-4)]·(x²+x-6)/(x²-9)=

{[(x-2)-(x+2)+(x+1)]/(x²-4)}·(x²+x-6)/(x²-9)={(x-2-x-2+x+1)/(x²-4)}(x²+x-6)/(x²-9)=[(x-3)/(x²-4)](x²-4+x-2)/(x²-9)=[(x-3)/(x²-4)][(x-2)(x+2)+x-2]/(x²-9)=[(x-3)/(x²-4)][(x-2)(x+2+1)/(x²-9)]= [(x-3)/(x²-4)][(x-2)(x+3)/(x²-9)]=

[(x-3)(x-2)(x+3)]/[(x²-4)(x²-9)]=[(x²-9)(x-2)]/[(x²-4)(x²-9)]= (x-2)/(x²-4)=

(x-2)/(x-2)(x+2)=1/(x+2)

c) E(7)=1/(7+2)=1/9

   E(1/4)=1/(1/4+2)=1/(9/4)=4/9  

Media geometrica = √[E(7)·E(1/4)]=√(1/9)·(4/9)=√4/81=√(2²/9²)=2/9