初二一道数学题不会~~~~

（x²-4）×（x²-10x+21）+100

= (x+2)(x-2)(x-3)(x-7)+100

= (x^2 - 4)((x-5)^2 - 4) + 100

= x^2(x-5)^2 - 4(x^2+(x-5)^2) + 16 + 100

= (x(x-5))^2 - 4(2x^2-10x)-100+16+100

= (x(x-5))^2 - 8x(x-5)+16

= (x(x-5) - 4)^2

>= 0

原式=(x-2)(x+2)(x-3)(x-7)+100
=[(x-2)(x-3)][(x+2)(x-7)]+100
=(x^2-5x+6)(x^2-5x-14)+100

设x^2-5x=y

则原式=(y+6)(y-14)+100
=y^2-8y-84+100
=y^2-8y+16
=(y-4)^2

所以，原式为一个完全平方数，非负数。

我在纳闷一楼二楼都答对了为什么不给分？？？？
（x²-4）×（x²-10x+21）+100
= (x^2 - 4)((x-5)^2 - 4) + 100

= x^2×(x-5)^2 - 4(x^2+(x-5)^2) + 16 + 100

= (x×(x-5))^2 - 4(2x^2-10x)-100+16+100

= (x×(x-5))^2 - 8x(x-5)+16

= (x×(x-5) - 4)^2

所以原式的值为非负数