Răspuns:
[tex]\red{\boxed{x \in \Big\{5,4,1 \Big\} }}[/tex]
Explicație pas cu pas:
[tex]G = \Big\{ x \;\Big| \; x \in \mathbb{N}\; , \; x = 5-a^2\; , \; a \in \mathbb{N} \Big\}[/tex]
[tex]x \in \mathbb{N} \Rightarrow x \geq 0 \Rightarrow 5-a^2 \geq 0 \Rightarrow -a^2 \geq -5 \;\Big| \cdot (-1)[/tex]
[tex]\left.\begin{aligned} a^2 \leq 5 \Rightarrow a \leq \sqrt{5} \;, \sqrt{5} \in \mathbb{I} \;\;\;\\\\\sqrt{4} \leq \sqrt{5}\; , \; \sqrt{4} \in \mathbb{R} \;,\; a \in \mathbb{N} \; \end{aligned} \right\} \Rightarrow a \in \Big\{ 0,1,2\Big\}[/tex]
[tex]\left.\begin{aligned} a = 0\;\; \\ \\ a = 1\;\; \\ \\ a=2 \;\;\end{aligned} \right\} \Rightarrow \begin{cases} x =5-0^2=5-0=5 \in \mathbb{N} \\\\ x = 5-1^2=5-1=4 \in \mathbb{N} \\\\ x=5-2^2 = 5-4=1\in \mathbb{N} \end{cases}[/tex]
Schema grafica a raspunsurilor (nu trebuie scrisa :) )
[tex]\red{\boxed{\Bigg \Uparrow \;\;\Bigg| \bigoplus \Bigg \rangle \Bigg \lfloor \quad \begin{bmatrix} \mathbf { 1} & \Downarrow & \mathbf{5} \\ \Rightarrow & \mathbf{x} & \Leftarrow \\ \mathbf{4} & \Uparrow & \mathbf{1} \end{bmatrix} \quad \Bigg \rceil \Bigg \rangle \bigoplus \Bigg | \;\;\Bigg \Uparrow }}[/tex]
