[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_数学解答

[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

人气：290 ℃　时间：2020-05-13 20:51:03

解答

[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...少了一步