Question

Fie E(x)=$\frac{ x^{2} -25}{ x^{2} +5x} + \frac{10}{ x^{2} -5x} : \frac{2}{x-5}$.
Aratatica E(X)=1,pentru orice x∈R/(-5;05).

Answer 1

(x-5)(x+5)/x(x+5)+10/x(x-5):2/(x-5)=
(x-5)/x+10/x(x-5)(x-5)/2=
(x-5)/x+5/x=
x-5+5/x=
x/x=1

Answer 2

$\displaystyle E(x)= \frac{x^2-25}{x^2+5x} + \frac{10}{x^2-5x} : \frac{2}{x-5} \\ \\E(x)= \frac{(x+5)(x-5)}{x(x+5)} + \frac{10}{x(x-5)} \cdot \frac{x-5}{2} \\ \\ E(x)= \frac{x-5}{x} + \frac{10}{x(x-5)} \cdot \frac{x-5}{2} \\ \\ E(x)= \frac{x-5}{x} + \frac{5}{x}\\ E(x)= \frac{x-5+5}{x} \\ \\ E(x)= \frac{x}{x} \\ \\ E(x)=1$