Question

求数学高手解答！！！！

求证：对于任何自然数n，代数式n(n+5)-(n-3)(n+2)的值都能被6整除。

Answer 1

n(n+5)-(n-3)(n+2)
=n²+5n-(n²-n-6)
=6n+6
=6(n+1)
∴[n(n+5)-(n-3)(n+2)]÷6＝6(n+1)÷6＝n+1
∵n为自然数，
∴n+1为自然数
∴代数式n(n+5)-(n-3)(n+2)的值都能被6整除

Answer 2

n(n+5)-(n-3)(n+2)
=n²+5n-(n²-n-6)
=n²+5n-n²+n+6
=6n+6
=6(n+1)
∴对于任何自然数n，代数式n(n+5)-(n-3)(n+2)的值都能被6整除

Answer 3

n(n+5)-(n-3)(n+2)
=n^2+5n-n^2+n+6
=6n+6
=6(n+1)
6（n+1)是6的倍数。所以式子能被6整除

Answer 4

n(n+5)-(n-3)(n+2)
=n^2+5n-(n^2+2n-3n-6)
=n^2+5n-n^2+n+6
=6n+6
=n*(n+1)
所以，对于任意自然数n，都能被6整除

Answer 5

n(n+5)-(n-3)(n+2)
=n^2+5n-(n^2-n-6)
=6n+6
=6*(n+1)
所以，对于任何自然数n，代数式n(n+5)-(n-3)(n+2)的值都能被6整除

Answer 6

本题属于简单的因式分解问题，具体做法如下～n(n+5)-(n-3)(n+2) =n²+5n-(n²-n-6) =n²+5n-n²+n+6 =6n+6 =6(n+1) ∴对于任何自然数n，代数式n(n+5)-(n-3)(n+2)的值都能被6整除