Question

1.对于任意自然数n，代数式n（n+3）-（n-4）（n-5）的值都能被4整除吗？请说明理由.

Answer 1

1.能,

∵n(n+3)-(n-4)(n-5)

=n2++3n-(n2-9n+20)

=n2+3n-n2+9n+20

=12n+20

=4(3n+5)

∴无论n,原式都能被4整除

2.∵2（a-4)(a+3)-a(2a-3)-a

=2(a2-a-12)-2a2+3a-a

=2-2a-24-2a2+3a-a

=24

∴无论a取何值,2（a-4)(a+3)-a(2a-3)-a=24

平方实在打不出来,只能用a2,n2代替