Răspuns
Explicație pas cu pas:
Se rationalizeaza numitorii.
[tex]x=\dfrac{1}{\sqrt{2}+1}=\dfrac{\sqrt 2-1}{2-1}=\sqrt{2}-1\\y=\dfrac{1}{2\sqrt{2}-2}=\dfrac{2\sqrt 2+2}{8-4}=\dfrac{2(\sqrt 2+1)}{4}=\dfrac{\sqrt 2+1}{2}\\\dfrac{x}{y}+\dfrac{4y}{x}=\dfrac{\sqrt{2}-1}{\frac{\sqrt 2+1}{2}}}+4\dfrac{\frac{\sqrt 2+1}{2}}{\sqrt 2 -1 } = 2 \dfrac{\sqrt 2-1}{\sqrt 2+1}+2\dfrac{\sqrt 2+1}{\sqrt 2-1}= \\=2\left(\dfrac{\sqrt 2-1}{\sqrt 2+1}+\dfrac{\sqrt 2+1}{\sqrt 2-1}\right)=2\left(\dfrac{(\sqrt 2-1)^2+(\sqrt 2+1)^2}{(\sqrt 2-1)(\sqrt 2+1)}\right) =[/tex]
[tex]=2\left(\dfrac{2-2\sqrt 2+1+2+2\sqrt 2+1}{2-1} \right)=2\cdot 6= 12\in \mathbb{Z}[/tex]
