Ta có:
\(\left(n+1\right)\left(n-1\right)=\left(n+1\right)n-\left(n+1\right)=n^2+n-n-1=n^2-1\)
\(\Rightarrow n^2\left(n+1\right)\left(n-1\right)=n^2\left(n^2-1\right)=n^4-n^2\)
Vậy \(n^4-n^2=n^2\left(n+1\right)\left(n-1\right)\)
Bài mk cũng gần giống Dung:
VP=(n+1)(n-1)=n.(n-1)+1(n-1)=n^2-n+n-1=n^2+0-1=n^2-1
Khi đó n^2.(n+1(n-1)=n^2.(n^2-1)=n^4-n^2=VT
=>VT=VP (đpcm)