Răspuns :
[tex]\it x=\sqrt{(1-2\sqrt2)^2} =|1-2\sqrt2| \\\\1-2\sqrt2<0 \Rightarrow |1-2\sqrt2| = -(1-2\sqrt2) =2\sqrt2-1 =x\\\\y=\sqrt8=\sqrt{4\cdot2}=2\sqrt2\\\\2\sqrt2>2\sqrt2-1 \Rightarrow y>x[/tex]
[tex]\it x=\sqrt{(1-2\sqrt2)^2} =|1-2\sqrt2| \\\\1-2\sqrt2<0 \Rightarrow |1-2\sqrt2| = -(1-2\sqrt2) =2\sqrt2-1 =x\\\\y=\sqrt8=\sqrt{4\cdot2}=2\sqrt2\\\\2\sqrt2>2\sqrt2-1 \Rightarrow y>x[/tex]