(1)原式=(x+2)(5x-3)
(2)原式=(a²+b²-1+2ab)(a²+b²-1-2ab)
=[(a+b)²-1][(a-b)²-1]
=(a+b+1)(a+b-1)(a-b+1)(a-b-1)
(3)①1×4+1+4=9
4×9+4+9=49
9×49+9+49=499
②设能,则ab+a+b=1999
ab+a+b+1=2000
(a+1)(b+1)=2000
令a=1,则b=999
令mn+m+n=999
(m+1)(n+1)=1000
令m=1,则n=499
∴能
(1)原式=(x+2)(5x-3)
(2)原式=(a²+b²-1+2ab)(a²+b²-1-2ab)
=[(a+b)²-1][(a-b)²-1]
=(a+b+1)(a+b-1)(a-b+1)(a-b-1)
(3)①1×4+1+4=9
4×9+4+9=49
9×49+9+49=499
②设能,则ab+a+b=1999
ab+a+b+1=2000
(a+1)(b+1)=2000
令a=1,则b=999
令mn+m+n=999
(m+1)(n+1)=1000
令m=1,则n=499
∴能