Va rog dau coroana!
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[tex]\sqrt{3^{19}-2\times 3^{16}}\\\sqrt{3^{16}(3^2-2)}\\\sqrt{3^{16}(27-2)}\\\sqrt{3^{16}\times 25}=3^8\times 5 {\displaystyle \in }{\displaystyle \mathbb {Q} }[/tex]
[tex]\it \sqrt{3^{19}-3^{16}-3^{16}}=\sqrt{3^{16}(3^3-1-1)}=\sqrt{3^{16}(27-1-1)}=\\ \\ \\ =\sqrt{3^{16}\cdot25}=\sqrt{(3^8)^2\cdot5^2}=3^8\cdot5\ \in\mathbb{Q}[/tex]