问题 解答题
计算及解方程:
(
1
4
x3y2)2•(4x2y)3-3(-x2y)5x2y2

②(x+y)2+(x-y)2-(x+y)(x-y);
x3-2x[
1
2
x2-3(
1
3
x-1)]=2(x-3)2

④(2x-3)2-(-2x+3)2
答案

①原式=

1
16
x6y4•64x6y3+3x12y7=4x12y7+3x12y7=7x12y7

②原式=x2+2xy+y2+x2-2xy+y2-x2+y2=x2+3y2

③x3-2x(

1
2
x2-x+3)=2(x-3)2

2x(x-3)-2(x-3)2=0,

即(x-3)[2x-2(x-3)]=0,

解得:x=3;

④(2x-3)2-(-2x+3)2=[(2x-3)+(-2x+3)][(2x-3)-(-2x+3)]=0.

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