Question

arătați ca numărul x este natural
$x = \sqrt{12 - 6 \sqrt{3} } + \sqrt{12 + 6 \sqrt{3} } + \sqrt{6 + 4 \sqrt{2} } - \sqrt{6 - 4 \sqrt{2} } - 2 \sqrt{2}$
Urgent!!!!!!
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Answer 1

$\it 12-6\sqrt3=9+3-6\sqrt3=3^2-6\sqrt3+(\sqrt3)^2=(3-\sqrt3)^2\\ \\ Analog:\\ \\ 12+6\sqrt3=(3+\sqrt3)^2\\ \\ 6+4\sqrt2=(2+\sqrt2)^2\\ \\ 6-4\sqrt2=(2-\sqrt2)^2$

$\it Folosind\ formula\ \sqrt{a^2}=a,\ \forall a>0,\ vom\ avea:\\ \\ x=3-\sqrt3+3+\sqrt3+2+\sqrt2-2+\sqrt2-2\sqrt2=6\in\mathbb{N}$