分解因式:9﹙x+y+z﹚²-﹙x-y-z﹚²-搜答案-智慧作业帮

(x+2y)^2-6(x+2y)+9=(x+2y-3)²(x+y)^2-6(x+y)z+9z²=(x+y-3z)²

x^3-xyz+x^2y-x^2z=x^2(x+y)-xz(y+x)=x(x-z)(x+y)

：（1):x(x-y)+y(y-X)=(X-Y)(x-y)=(x-y)^2多项式(x+y-z)(x-y+z)-(y+z-X)(Z-x-y)的公因式是(x+y-z)

=(x-3y)²-3(x-3y)=(x-3y)(x-3y-3)

（1）(x-3(y-z))^2(2)((a+b)-2c)^2(3)8x^2(2-x)(2+x)(4)5yx^2(2xy-3y+1)（5）n(m+7)^2(6)(x-y-5)(x+y+5)(7)(5-m

：(x^2-xy)+z(x-y)=x（x-y）+z（x-y）=（x-y）（x+z）

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(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)

xyz-yz-zx-xy+x+y+z-1=yz(x-1)-z(x-1)-y(x-1)+x-1=(x-1)(yz-y-z+1)=(x-1)(y-1)(z-1)

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)

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

16(x-y)²-9(x+y)²=【4（x-y）+3(x+y)】【4(x-y)-3(x+y)】=(7x-y)(x-7y)在我回答的右上角点击【评价】,然后就可以选择【满意,问题已经

=-(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

:a(3x+9y)-x-3y=3a(x+3y)-(x+3y)=(3a-1))(x+3y)

16(X+Y)平方-9(Y-X)平方=[4(x+y)]²-[3(y-x)]²=[4x+4y-3y+3x][4x+4y+3y-3x]=(7x+y)(x+7y)