Question

[Lê Bùi]

Cm A là số tự nhiên

A=\(\sqrt{8000^2+8000^2.8001^2+8001^2}\)

Answer 1

[Son Goku]

\(A=\sqrt{8000^2+8000^28001^2+8001^2}\)

Đặt \(8000=a\Rightarrow8001=a+1\\ \)

\(\Rightarrow A=\sqrt{a^2+a^2\left(a+1\right)^2+\left(a+1\right)^2}=\sqrt{\left(2a+1\right)^2}=2a+1\)

Tự lm nốt nha.