Răspuns :
[tex]\it x*y=xy-\sqrt2(x+y)+2+\sqrt2=yx-\sqrt2(y+x)+2+\sqrt2=y*x\\ \\ Pentru\ e- element\ neutru, \ are\ loc\ rela\c{\it t}ia:\ e*x=x*e=x,\ \ \forall x\in\mathbb{R}\\ \\ Folosim\ rezultatul\ de\ la\ b):\\ \\ x*e=x \Rightarrow (x-\sqrt2)(e-\sqrt2)+\sqrt2=x \Rightarrow (x-\sqrt2)(e-\sqrt2)=x-\sqrt2 \Rightarrow \\ \\ \Rightarrow (x-\sqrt2)(e-\sqrt2)-(x-\sqrt2)=0 \Rightarrow (x-\sqrt2)(e-\sqrt2-1)=0 \Rightarrow \\ \\ \Rightarrow e-\sqrt2-1=0 \Rightarrow e=1+\sqrt2[/tex]