Cho a,b >0 thỏa mãn [tex]\frac{1}{a}+\frac{1}{b}=\frac{1}{2020}[/tex]. chứng minh rằng [tex]\sqrt{a+b}=\sqrt{a-2020}+\sqrt{b-2020}[/tex]
[tex]\sqrt{a+b}=\frac{a+b}{\sqrt{a+b}}=\frac{a}{\sqrt{a+b}}+\frac{b}{\sqrt{a+b}} = \sqrt{\frac{a^{2}}{a+b}}+\sqrt{\frac{b^{2}}{a+b}}[/tex]
= [tex]\sqrt{a-\frac{ab}{a+b}}+\sqrt{b-\frac{ab}{a+b}}=\sqrt{a-2020}+\sqrt{b-2020}[/tex] (đpcm)