Răspuns:
[tex]\boldsymbol{ \red{N = 1 \in \Bbb{N}}}[/tex]
Explicație pas cu pas:
[tex]b=\bigg(\dfrac{^{2)} 2}{5} + \dfrac{1}{10}\bigg) \cdot \dfrac{2}{3} = \dfrac{4+1}{10} \cdot \dfrac{2}{3} = \dfrac{5^{(5} }{10} \cdot \dfrac{2}{3} = \dfrac{1}{\not2} \cdot \dfrac{\not2}{3} = \dfrac{1}{3}[/tex]
[tex]N = \dfrac{\dfrac{1}{3} + \dfrac{1}{3}}{2 \cdot \sqrt{\dfrac{1}{3} \cdot \dfrac{1}{3}} } = \dfrac{\dfrac{1+1}{3}}{2 \cdot \sqrt{\bigg(\dfrac{1}{3}\bigg)^2} } = \dfrac{\dfrac{2}{3}}{2 \cdot \dfrac{1}{3}} =\\[/tex]
[tex]= \dfrac{\dfrac{2}{3}}{\dfrac{2}{3}} = \dfrac{2}{3} : \dfrac{2}{3} = \dfrac{2}{3} \cdot \dfrac{3}{2} = \bf 1 \in \Bbb{N}[/tex]
