Ta có:
\(n^4+6n^3+11n^2+6n\)
\(=n\left(n^3+6n^2+11n+6\right)\)
\(=n\left(n^3+n^2+5n^2+5n+6n+6\right)\)
\(=n\left[n^2\left(n+1\right)+5n\left(n+1\right)+6\left(n+1\right)\right]\)
\(=n\left(n+1\right)\left(n^2+5n+6\right)\)
\(=n\left(n+1\right)\left(n^2+3n+2n+6\right)\)
\(=n\left(n+1\right)\left[n\left(n+3\right)+2\left(n+3\right)\right]\)
\(=n\left(n+1\right)\left(n+2\right)\left(n+3\right)\)
Vì tích 4 số nguyên liên tiếp luôn chia hết cho 24
\(\Rightarrow n\left(n+1\right)\left(n+2\right)\left(n+3\right)\) chia hết cho 24
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