$y=1-x\\ =>\sqrt{x^2+y^2}\\=\sqrt{x^2+(1-x)^2}\\=\sqrt{2x^2-2x+1}\\=\sqrt{(\sqrt{2}x)^2-2.\sqrt{2}.\dfrac{1}{\sqrt{2}}x+(\dfrac{1}{\sqrt{2}})^2+\dfrac{1}{2}}\\=\sqrt{(\sqrt{2}x-\dfrac{1}{\sqrt{2}})^2+\dfrac{1}{2}}\\ \ge \dfrac{1}{\sqrt{2}}\\ =>min_y=\dfrac{1}{\sqrt{2}};$ không tồn tại max