[tex]x^{2}+\sqrt{x-2020}= y^{2}+\sqrt{y-2020}\Rightarrow x^2-y^2+\sqrt{x-2020}-\sqrt{y-2020}=0\Rightarrow (x-y)(x+y)+\frac{x-y}{\sqrt{x-2020}+\sqrt{y-2020}}=0\Rightarrow (x-y)(x+y+\frac{1}{\sqrt{x-2020}+\sqrt{y-2020}})=0[/tex]
Vì ngoặc sau luôn lớn hơn 0 nên x = y.