已知,x^2+y^2-6x+2y+10=0,求(x+2y)^2(x-2y)^2-(x-2y)(x^2+4y^2)(x+2y)的值._数学解答

已知,x^2+y^2-6x+2y+10=0,求(x+2y)^2(x-2y)^2-(x-2y)(x^2+4y^2)(x+2y)的值.

人气：245 ℃　时间：2020-03-19 20:51:19

解答

x^2+y^2-6x+2y+10=(x-3)^2+(y+1)^2=0在实数范围内,显然,x=3y=-1(x+2y)^2(x-2y)^2-(x-2y)(x^2+4y^2)(x+2y)=[(x+2y)(x-2y)]^2 - [(x^2+4y^2)(x^2-4y^2)]=(x^2-4y^2)(-8y^2)=(3*3-4)*(-8)=-40