Aiuto plss

(√x - √(2·y))·(√x + √(2·y)) - (√x + √(2·y))^2 + √(8·x·y)=

=(x - 2·y) - (x + 2·√(2·x·y) + 2·y) + 2·√(2·x·y)=

=- 4·y

(a + b) × (a - b) = a^2 - b^2.

(a + b)^2 = a^2 + 2ab + b^2.

l'elevamento a potenza 2 elimina la radice quadrata;

[radicequadrata(x)]^2 = x; [radicequadrata(2y)]^2 = 2y.

x - 2y - x - 2y - 2 radice(x * 2y) + radice(8xy) =

= - 4y - radice(4 * 2xy) + radice(8xy) =

= - 4y - radice(8xy) + radice(8xy)  = - 4y.

Ciao @green

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$\small \left(\sqrt{x}-\sqrt{2y}\right)·\left(\sqrt{x}+\sqrt{2y}\right)-\left(\sqrt{x}+\sqrt{2y}\right)^2+\sqrt{8xy}=$

$\small =x+\sqrt{2xy}-\sqrt{2xy}-2y-\left(x+2\sqrt{2xy}+2y\right)+\sqrt{8xy}=$

$\small =x\cancel{+\sqrt{2xy}}\cancel{-\sqrt{2xy}}-2y-\left(x+\sqrt{4·2xy}+2y\right)+\sqrt{8xy}=$

$\small =x-2y-\left(x+\sqrt{8xy}+2y\right)+\sqrt{8xy}=$

$\small =x-2y-x-\sqrt{8xy}-2y+\sqrt{8xy}=$

$\small =\cancel{x}-2y\cancel{-x}\cancel{-\sqrt{8xy}}-2y\cancel{+\sqrt{8xy}}=$

$\small =-4y$