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