\(=\dfrac{x+\sqrt{xy}+y-\sqrt{xy}}{\sqrt{x}+\sqrt{y}}:\left(\dfrac{x}{\sqrt{y}\left(\sqrt{x}+\sqrt{y}\right)}-\dfrac{y}{\sqrt{x}\left(\sqrt{x}-\sqrt{y}\right)}-\dfrac{x+y}{\sqrt{xy}}\right)\)
\(=\dfrac{x+y}{\sqrt{x}+\sqrt{y}}:\dfrac{x\sqrt{x}\left(\sqrt{x}-\sqrt{y}\right)-y\sqrt{y}\left(\sqrt{x}+\sqrt{y}\right)-\left(x^2-y^2\right)}{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)}\)
\(=\dfrac{x+y}{\sqrt{x}+\sqrt{y}}\cdot\dfrac{\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}{x^2-x\sqrt{xy}-y\sqrt{xy}-y^2-x^2+y^2}\)
\(=\dfrac{\sqrt{xy}\left(x+y\right)\cdot\left(\sqrt{x}-\sqrt{y}\right)}{-\sqrt{xy}\left(x+y\right)}=-\sqrt{x}+\sqrt{y}\)(1)
Khi x=3 và \(y=4+2\sqrt{3}\) vào (1), ta được:
\(=-\sqrt{3}+\sqrt{4+2\sqrt{3}}=-\sqrt{3}+\sqrt{3}+1=1\)