分解因式2x²y²-4y³z

A=y(x-y)(x-z)+z(x-y)(z-x)=y(x-y)(x-z)-z(x-y)(x-z)=(x-y)(x-z)(y-z)x/2=y/3=z/4=tx=2t,y=3t,z=4t代入A/（xyz

x²-4y²+x-2y=(x²-4y²)+(x-2y)=(x-2y)(x+2y)+(x-2y)=(x-2y)(x+2y+1)很高兴为您解答,祝你学习进步!【学习

原式=x^4-(2y^2+2z^2)x^2+y^4+z^4-2y^2z^2=x^4-(2y^2+2z^2)x^2+(y^2-z^2)^2=x^4-(2y^2+2z^2)x^2+(y+z)^2(y-z)

A=y(x-y)(x-z)+z(x-y)(z-x)=y(x-y)(x-z)-z(x-y)(x-z)=(y-z)(x-y)(x-z)令x/2=y/3=z/4=k则x=2k,y=3k,z=4kA/（xyz

(x+y+z)³+(3x-2y-3z)³-(4x-y-2z)³=(x+y+z)³+[(3x-2y-3z)-(4x-y-2z)][(3x-2y-3z)²

x^2-x-4y^2+2y=(x^2-4y^2)-(x-2y)=(x+2y)(x-2y)-(x-2y)=(x-2y)(x+2y-1)

最后一项应该是(4x-y-2z)³,且(x+y+z)+(3x-2y-3z)=4x-y-2z(x+y+z)³+(3x-2y-3z)³-(4x-y-2x)³=(x+

原式=2[(x²+2xy+y²)-4z²]=2[(x+y)²-(2z)²]=2(x+y+2z)(x+y-2z)

题目有问题吧,我感觉应该是(x²+y²-z²)²-4x²y²(x²+y²-z²)²-4x²

(x^2+y^2-z^2)^2-4x^2y^2利用平方差公式(x^2+y^2-z^2)^2-4x^2y^2=[(x^2+y^2-z^2)+2xy][(x^2+y^2-z^2)-2xy]=[x^2+y^

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

x(x-y-z)-y(y-x+z)-z(z-x+y),=x(x-y-z)+y(x-y-z)+z(x-y-z)=(x-y-z)(x+y+z)再问：为啥x(x-y-z)+y(x-y-z)+z(x-y-z)

x^2(y-z)+y^2(z-x)+z^2(x-y)=x^2y-x^2z+y^2z-y^2x+z^2x-z^2y=y(x^2-z^2)-xz(x-z)-y^2(x-z)=y(x+z)(x-z)-xz(

4x²-4xyz+y²z²-4z²=(4x²-4xyz+y²z²)-4z²=(2x-yz)²-(2z)

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

(x+y)^4+(x+y)^2-20＝[﹙x＋y﹚²＋5﹚][﹙x＋y﹚²－4]＝﹙x²＋2xy＋y²＋5﹚﹙x＋y－2﹚﹙x＋y＋2﹚

(x+y+z)3+(3x-2y-3z)3-(4x-y-2z)3=(4x-y-2z)[(x+y+z)^2-2(x+y+z)(3x-2y-3z)+(3x-2y-3z)^2]-(4x-y-2z)3=(4x-

=-(3x-2y-3z)[(x+y+z)^2-(x+y+z)(4x-y-2z)+(4x-y-2z)^2]+(3x-2y-3z)^3然后提取公因式就好了

=x²(y-z)+y²(z-x)+z²(x-z+z-y)=(y-z)(x²-z²)+(z-x)(y²-z²)=(y-z)(x-z)

9(x+y+z)²-(x-y-z)²=[3(x+y+z)]²-(x-y-z)²=[3(x+y+z)+(x-y-z)]×[3(x+y+z)-(x-y-z)]=(4x+2y