Question

整式乘法（2x－y)^2(-2x-y)^2-(4x^2+y^2)(4x^2-y^2)

Answer 1

（2x－y)^2(-2x-y)^2-(4x^2+y^2)(4x^2-y^2)

=(2x-y)^2(2x+y)^2-(16x^4-y^4)

=[(2x-y)(2x+y)]^2-16x^4+y^4

=(4x^2-y^2)^2-16x^4+y^4

=16x^4-8x^2y^2+y^4-16x^4+y^4

=-8x^2y^2+2y^4

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

其中x＝1/29,y=2/29

=-2*4/29^2*[4/29^2-4/29^2]

=0

=