\(\sqrt{\left(n+1\right)^2}+\sqrt{n^2}=\left(n+1\right)^2-n^2\)
\(VT=\left|n+1\right|+\left|n\right|=n+1+n=2n+1\)
\(VP=\left(n+1+n\right)\left(n+1-n\right)=\left(2n+1\right)\cdot1=2n+1\)
\(\Rightarrow VT=VP\)
Vậy \(\sqrt{\left(n+1\right)^2}+\sqrt{n^2}=\left(n+1\right)^2-n^2\) đúng với \(n\in N\)