求数学高手解答！！！！

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

n(n+5)-(n-3)(n+2)
=n²+5n-(n²-n-6)
=n²+5n-n²+n+6
=6n+6
=6(n+1)
∴对于任何自然数大则n，代数式颂模n(n+5)-(n-3)(n+2)的值都滚樱棚能被6整除

n(n+5)-(n-3)(n+2)
=n^2+5n-n^2+n+6
=6n+6
=6(n+1)
6（n+1)是6的倍数培拍告。所配明以式子贺稿能被6整除

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，对于族晌拿任意兆搭自然数n，都能谨斗被6整除

n(n+5)-(n-3)(n+2)
=n^2+5n-(n^2-n-6)
=6n+6
=6*(n+1)
所以，对局毁于任何自然数胡世n，代数桐做备式n(n+5)-(n-3)(n+2)的值都能被6整除

本题属于简单的因式分解问题，具体卖伏旦做法如下～n(n+5)-(n-3)(n+2) =n²+5n-(n²-n-6) =n²+5n-n²+n+6 =6n+6 =6(n+1) ∴对于任何自中扰然数n，代数式n(n+5)-(n-3)(n+2)的值都能被6整厅散除