Răspuns :

Elyelyzid

Răspuns:

Pfiu... Urât exercițiu. Am folosit formula radicalilor compuși

Vezi imaginea Elyelyz
Vezi imaginea Elyelyz
Targoviste44id

[tex]\it m=\Big(\dfrac{1}{\sqrt2}+\dfrac{1}{\sqrt5}\Big)\cdot\sqrt{10}= \dfrac{\sqrt5+\sqrt2}{\sqrt{10}}\cdot\sqrt{10}=\sqrt5+\sqrt2\\ \\ \\ n=\sqrt{7-2\sqrt{10}}=\sqrt{5+2-2\sqrt{10}}=\sqrt{(\sqrt5)^2-2\sqrt{10}+(\sqrt2)^2}=\\ \\ \\ =\sqrt{(\sqrt5-\sqrt2)^2}=\sqrt5-\sqrt2\\ \\ \\ m\cdot n=(\sqrt5+\sqrt2)(\sqrt5-\sqrt2)=(5-2)=3 \in \mathbb{N}[/tex]

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