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