合肥联发科技:关于 初中整式题

1 a^2-b^2-c^2-2bc
=a^2-(b^2+c^2+2bc)
=a^2-(b+c)^2
=(a+(b+c))(a-(b+c))
三角形，任何两边之和大于第三边，故b+c>a, 从而得知 a-(b+c)<0
而 (a+(b+c))>0
a^2-b^2-c^2-2bc<0

2 (x+2y-3/2)(x-2y+3/2)
=(x+(2y-3/2))(x-(2y-3/2))
=x^2-(2y-3/2)^2
=x^2-(4y^2-6y+9/4)
=x^2-4y^2+6y-9/4

3 (x+3)(x^2+9)(x-3)
=(x+3)(x-3)(x^2+9)
=(x^2-9)(x^2+9)
=x^4-81

1]a^2-b^2-c^2-2bc
=a^2-(b+c)^2
=[a-(b+c)][a+(b+c)]
显然a+b+c>0
由于三角形两边之和大于第三边，所以a<b+c
所以a^2-b^2-c^2-2bc>0
2](x+2y-3/2)(x-2y+3/2)
=[x+(2y-3/2)][x-(2y-3/2)]
=x^2-(2y-3/2)^2
=x^2-4y^2+6y-9/4
3](x+3)(x^2+9)(x-3)
=(x^2-9)(x^2+9)
=x^4-81

1.a^2-b^2-c^2-2bc=a^2-(b+c)^2=(a+b+c)(a-b-c)
因为a<b+c ，所以a+b+c>0,a-b-c<0
所以（a^2-b^2-c^2-2bc）为负数

2.(x+2y-3/2)(x-2y+3/2) ＝x^2-(2y-3/2)^2=x^2-4y^2-9/4+6y

3.(x+3)(x^2+9)(x-3)=(x^2+9)(x^2-9)=x^4-81

1、a^2-b^2-c^2-2bc=a^2-(b+c)^2=(a-b-c)(a+b+c)根据三角形，两边之和大于第三边。即b+c>a,所以a-b-c<0.所以a^2-b^2-c^2-2bc<0
2、(x+2y-3/2)(x-2y+3/2)=[x+(2y-3/2)][x-(2y-3/2)]=x^2-(2y-3/2)^2=x^2-(4y^2+9/4-6y)
3、(x+3)(x^2+9)(x-3)=(x^2+9)(x^2-9)=x^4-81

1、a2-b2-c2-2bc
=a2-(b2+c2+2bc)
=a2-(b+c)2
=(a+b+c)(a-b-c)
因为a+b+c>0
a-b-c<0
所以(a+b+c)(a-b-c)<0 即原式<0
2、(x+2y-3/2)(x-2y+3/2)
=(x+(2y-3/2))(x-(2y-3/2))
=x^2-(2y-3/2)^2
=x^2-(4y^2-6y+9/4)
=x^2-4y^2+6y-9/4
3、(x+3)(x2+9)(x-3)
=(x+3)(x-3)(x2+9)
=(x2-9)(x2+9)
=x4-81
其中字母后方的数为未知数的幂，如：a2就是a的二次方；x4就是x的四次方。以此类推。

:(x+3)(x^2+9)(x-3)=(x^2-9)(x^2+9)=x^4-9^2