Răspuns:
D = {0,1}
Explicație pas cu pas:
[tex]x = \dfrac{2k^2-k-1}{2k-1} = \dfrac{k(2k-1)-1}{2k-1} = k - \dfrac{1}{2k-1}\\ \\ \dfrac{1}{2k-1}\in \mathbb{Z}\Rightarrow 2k-1 \in \{-1,1\} \Rightarrow 2k \in \{0,2\} \Rightarrow \\ \\ \Rightarrow k \in \{0,1\}[/tex]
[tex]k = 0 \Rightarrow x = \dfrac{0-0-1}{0-1} = 1 \\\\ k = 1 \Rightarrow x = \dfrac{2-1-1}{2-1} = 0[/tex]
=> x ∈ {0,1}
