[3x^2-(x+y)(y-x)][2(-x-y)(y-x)+2x^2+3y^2],其中x=-2分之1,y=1.5

来源：学生作业帮助网编辑：作业帮时间：2024/04/27 14:04:53

[3x^2-(x+y)(y-x)][2(-x-y)(y-x)+2x^2+3y^2],其中x=-2分之1,y=1.5
[3x^2-(x+y)(y-x)][2(-x-y)(y-x)+2x^2+3y^2],其中x=-2分之1,y=1.5

[3x^2-(x+y)(y-x)][2(-x-y)(y-x)+2x^2+3y^2],其中x=-2分之1,y=1.5
[3x^2-(x+y)(y-x)][2(-x-y)(y-x)+2x^2+3y^2]
=[3x^2+(x+y)(x-y)][-2(y+x)(y-x)+2x^2+3y^2]
=(3x^2+x^2-y^2)[-2(y^2-x^2)+2x^2+3y^2]
=(4x^2-y^2)(-2y^2+2x^2+2x^2+3y^2)
=(4x^2-y^2)(4x^2+y^2)
=(4x^2)^2-(y^2)^2
=16x^4-y^4
=16×(-1/16)-(3/2)^4
=-1-81/16
=-97/16

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