Punctul a)
[tex] E(x) = \dfrac{x-1}{2} - \dfrac{x-1}{x+1} \left[ x-1 : \left( 1-\dfrac{x-1}{x+1} \right) \right] \\ E(x)= \dfrac{x-1}{2} -\dfrac{x-1}{x+1} \left[ x-1 : \left( 1-\dfrac{x-1}{x+1} \right) \right] \\ E(x)= \dfrac{x-1}{2} - \dfrac{x-1}{x+1} \left( x-1: \dfrac{2}{x+1} \right) \\ E(x)= \dfrac{x-1}{2} - \dfrac{x-1}{x+1} \left( x- \dfrac{x+1}{2} \right) \\ E(x)= \dfrac{x-1}{2} - \dfrac{x-1}{x+1} \cdot \dfrac{x-1}{2} \\ E(x)= \dfrac{x-1}{2} - \dfrac{(x-1)^2}{2(x+1)} \\ E(x)= \dfrac{(x-1)(x+1) -(x-1)^2 }{2(x+1)} \\ E(x)= \dfrac{(x-1)(x+1-x+1)}{2(x+1)} \\ E(x)= \dfrac{(x-1)2}{2(x+1)} \\ \Rightarrow E(x)= \dfrac{x-1}{x+1} \ , \forall x \in \mathbb{R} \backslash \{-1,1 \} [/tex]
Punctul b)
Trebuie să arătăm ca numărul este natural, adică trebuie să calculăm.
[tex] E(\sqrt{3} )+\sqrt{3} = \dfrac{\sqrt{3} -1}{\sqrt{3} +1}+\sqrt{3} \\ = \dfrac{(\sqrt{3} -1)(\sqrt{3}-1)}{(\sqrt{3} +1)(\sqrt{3} -1)}+\sqrt{3} \\ = \dfrac{3-2\sqrt{3} +1}{3-1} +\sqrt{3} \\ = \dfrac{2(2-\sqrt{3}}{2} +\sqrt{3} \\ = 2-\sqrt{3} + \sqrt{3} \\ = 2 \in \mathbb{N} \Rightarrow \tt E(\sqrt{3} )+ \sqrt{3} \in \mathbb{N} [/tex]
