Question

#Capacitate #EvaluareNationala

Scrieti si calculele

Answer 1

$E(x) = \dfrac{(x+2)^2-9}{x^2-25} : \dfrac{x-1}{x-5} = \dfrac{(x+2)^2-3^2}{x^2-5^2} : \dfrac{x-1}{x-5} =$

$\rightarrow \boxed{ la prima fractie aplicam formula: a^2-b^2=(a-b)(a+b)}$

$\\ \\ =\dfrac{\Big[(x+2)-3\Big]\Big[(x+2)+3\Big]}{(x-5)(x+5)}: \dfrac{x-1}{x-5} = \\ \\ = \dfrac{(x+2-3)(x+2+3)}{(x-5)(x+5)} : \dfrac{x-1}{x-5} = \dfrac{(x-1)(x+5)}{(x-5)(x+5)} : \dfrac{x-1}{x-5} = \\ \\ = \dfrac{x-1}{x-5} : \dfrac{x-1}{x-5} = \dfrac{x-1}{x-5}\cdot \dfrac{x-5}{x-1} = 1$

Answer 2

(x+2)²-9=(x+2-3)(x+2+3)=(x-1)(x+5)
x²-25=(x+5)(x-5)
E(x)=[(x-1)(x+5)]/[(x+5)(x-5)]*x-5/x-1
E(x)=(x-1)/(x-5)*(x-5)/(x-1)(se simplifica totul
E(x)=1