Răspuns:
[tex]a) \ x^2+3x-10 = x^2 + 5x - 2x - 10 =\\[/tex]
[tex]= x(x + 5) - 2(x + 5) = \bf (x + 5)(x - 2)\\[/tex]
[tex]b) \ E(x) = \dfrac{x + 5}{(x + 5)(x - 2)} : \bigg(\dfrac{x + 4}{3(x - 2)} - \dfrac{^{3)} 1}{x - 2}\bigg) \cdot \bigg(\dfrac{x + 2}{x + 2} - \dfrac{1}{x + 2}\bigg) =\\[/tex]
[tex]= \dfrac{1}{x - 2} : \dfrac{x + 4 - 3}{3(x - 2)} \cdot \dfrac{x + 2 - 1}{x + 2} = \dfrac{1}{x - 2} : \dfrac{x + 1}{3(x - 2)} \cdot \dfrac{x + 1}{x + 2} \\[/tex]
[tex]= \dfrac{1}{x - 2} \cdot \dfrac{3(x - 2)}{x + 1} \cdot \dfrac{x + 1}{x + 2} = \dfrac{3}{x + 2}\\[/tex]
Deci:
[tex]\boldsymbol{E(x) = \dfrac{3}{x + 2}} \ \forall x \in \Bbb {R} - \{-5,-2,-1,2\}[/tex]
