[tex]\text{Prin rationalizare fractiei obtinem:}\\\dfrac{a\sqrt{2}+b}{\sqrt 2-1}=\dfrac{(a\sqrt 2+b)(\sqrt 2+1)}{2-1}= (a\sqrt 2+b)(\sqrt 2+1)= \\=2a+b\sqrt 2+b+a\sqrt 2= \sqrt{2}(a+b) +2a+b\\\text{Evident, }2a+b \in \mathbb{Q}, \forall a,b\in \mathbb{Q}. \text{Singura conditie impusa este}\\\sqrt{2}(a+b) \in \mathbb{Q}. \text{ Aceasta are loc daca }a+b=0,\text{ deci }a=-b[/tex]
