Răspuns :
Răspuns:
Explicație pas cu pas:
vom aplica foemula a²-b²=(a-b)(a+b) la numaratorul (x+2)²-3² si numitorul x²-5²
[tex]E(x)=\frac{(x+2)^{2}-3^{2}}{(x^{2}-5^{2}}:\frac{x-1}{x-5}=\frac{(x+2-3)(x+2+3)}{(x-5)(x+5)}*\frac{x-5}{x-1}=\frac{(x-1)(x+5)}{(x-5)(x+5)}*\frac{x-5}{x-1}=1[/tex]
[tex]\it E(x)=\dfrac{(x+2)^2-3^2}{x^2-5^2}:\dfrac{x-1}{x-5}=\dfrac{(x+2-3)(x+2+3)}{(x-5)(x+5)}\cdot\dfrac{x-5}{x-1}=\dfrac{x+5}{x+5}=1[/tex]