Răspuns:
[tex]\boldsymbol {\red{E(x) = 1}}[/tex]
Explicație pas cu pas:
a) Înlocuim x = 1
[tex]E(x) = \bigg( \dfrac{4}{2} - \dfrac{2}{4} \bigg) : \dfrac{6}{2} - \dfrac{ - 2}{4} = \dfrac{8 - 2}{4} \cdot \dfrac{2}{6} + \dfrac{1}{2} = \dfrac{6}{4} \cdot \dfrac{2}{6} + \dfrac{1}{2} = \dfrac{1}{2} + \dfrac{1}{2} = \bf 1[/tex]
[tex]b) \ E(x) = \bigg( \dfrac{^{x + 3)} x+3}{x+1} - \dfrac{^{x + 1)} x+1}{x+3} \bigg) : \dfrac{2(x+2)}{x(x + 1)} - \dfrac{x-3}{x+3} = \\ [/tex]
[tex]= \dfrac{(x + 3)^{2} - (x + 1)^{2}}{(x + 1)(x + 3)} \cdot \dfrac{x(x + 1)}{2(x+2)} - \dfrac{x-3}{x+3} \\ [/tex]
[tex]= \dfrac{x^{2} + 6x + 9 - (x^{2} + 2x + 1)}{(x + 1)(x + 3)} \cdot \dfrac{x(x + 1)}{2(x+2)} - \dfrac{x-3}{x+3} \\ [/tex]
[tex]= \dfrac{4(x + 2)}{(x + 1)(x + 3)} \cdot \dfrac{x(x + 1)}{2(x+2)} - \dfrac{x-3}{x+3} \\ [/tex]
[tex]= \dfrac{2x}{x + 3} - \dfrac{x-3}{x+3} = \dfrac{2x - (x - 3)}{x + 3} = \dfrac{2x - x + 3}{x+3} \\ [/tex]
[tex]= \dfrac{x + 3}{x+3} = \bf 1[/tex]
⇒ E(x) are o valoare constantă oricare ar fi x ∈ R \{-3,-2,-1,0}
