Question

[•๖ۣۜ长υɀ༄]

Cho \(n\in N\).CMR:\(\sqrt{\left(2n+1\right)^2}+\sqrt{4n^2}=\left(2n+1\right)^2-4n^2\).

Viết đẳng thức trên với n=1,2,3,4,5,6,7

Answer 1

[Nobi Nobita]

Ta có: \(VT=\sqrt{\left(2n+1\right)^2}+\sqrt{4n^2}=\sqrt{\left(2n+1\right)^2}+\sqrt{\left(2n\right)^2}\)

\(=\left|2n+1\right|+\left|2n\right|\)

Vì \(n\inℕ\)\(\Rightarrow2n+1>0\); \(2n\ge0\)

\(\Rightarrow\left|2n+1\right|=2n+1\)và \(\left|2n\right|=2n\)

\(\Rightarrow VT=2n+1+2n=4n+1\)

Ta có: \(VP=\left(2n+1\right)^2-4n^2=\left(2n+1\right)^2-\left(2n\right)^2\)

\(=\left(2n+1-2n\right)\left(2n+1+2n\right)=4n+1\)

\(\Rightarrow VT=VP\)\(\Rightarrowđpcm\)