Question

Tính : \(\sqrt{1+2020^2+\frac{2020^2}{2021^2}}+\frac{2020}{2021}\)

Answer 1

Ta có: \(2021^2=\left(2020+1\right)^2=2020^2+2.2020.1+1^2\)

\(\Rightarrow1+2020^2=2021^2-2.2020\)

\(\Rightarrow\sqrt{1+2020^2+\frac{2020^2}{2021}}+\frac{2020}{2021}\)

\(=\sqrt{2021^2-2.2020+\frac{2020^2}{2021}}+\frac{2020}{2021}\)

\(=\sqrt{2021^2-2.2021.\frac{2020}{2021}+\left(\frac{2020}{2021}\right)^2}+\frac{2020}{2021}\)

\(=\sqrt{\left(2021-\frac{2020}{2021}\right)^2}+\frac{2020}{2021}\)

\(=2021-\frac{2020}{2021}+\frac{2020}{2021}=2021\)