因式分解:9x^y^-16z^(要过程)

(X+Y)(X+Z)(Y+Z)+XYZ=(X2+XY+XZ+YZ)(Y+Z)+XYZ=(X2Y+XY2+XYZ+Y2Z+X2Z+XYZ+XZ2+YZ2)+XYZ=(X2Y+X2Z+XYZ)+(Y2Z+

(x+y+z)^3-(y+z-x)^3-(z+x-y)^3-(x+y-z)^3=((x+y+z)^3-(y+z-x)^3)-((z+x-y)^3+(x+y-z)^3)=(x+y+z-y-z+x)((x

16x的16次方-y的四次方z的四次方=(4x^8-y^2z^2)(4x^8+y^2z^2)=(2x^4-yz)(2x^4+yz)(4x^8+2y^2z^2)

错了吧,这个不能分解16yz+25x²-16y²-4z²=25x²-(16y²-16yz+4z²)=(5x)²-(4y-2z)&s

(X+Y)(X+Z)(Y+Z)+XYZ=(X2+XY+XZ+YZ)(Y+Z)+XYZ=(X2Y+XY2+XYZ+Y2Z+X2Z+XYZ+XZ2+YZ2)+XYZ=(X2Y+X2Z+XYZ)+(Y2Z+

x"'(x+1)(y-z)+y"'(y+1)(z-x)+z"'(z+1)(x-y)=x"'(x+1)(y-z)-y"'(y+1)(x-z)+(z^4+z"')(x-y)这样就看到,前两个有x"'-y"

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

9x²y²-16z²=(3xy+4z)(3xy-4z)平方差公式

=(x+y+z)^2+yz(y+z+x)=(x+y+z)(x+y+z+yz)

因式分解分为以下四种情况：提取公因式法,乘法公式法,分组分解法,十字相乘法.此题不符合任一形式,所以不能再分解.

题目是否错了应该改为(x+y+z)^2-4(x-y-z)^2(x+y+z)^2-4(x-y-z)^2=[(x+y+z)-2(x-y-z)][(x+y+z)+2(x-y-z)]=(3Y+3Z-X)(3X

x+2y-7z=a3x-4y+6z=b4x-2y-z=c则a+b=ca^3+b^-c^3=(a+b)^-3a^2b-3ab^2-c^3=-3ab(a+b)=-3abc=-3(x+2y-7z)(3x-4

(4x-3y)²-16y²=(4x-3y+4y)(4x-3y-4y)=(4x+y)(4x-7y)x²+2x(y-z)+(z-y)²=x²+2x(y-z

你令x=y+z代入原式得到原式=0,可知原式有因式y+z-x,类似的还有剩下两个对称的因式.除以这三个因式后剩下个x+y+z,具体除法用长除就可以,或者剩下的是个一次对称因式,对比下系数就知道肯定是x

y-x吧25(x-y)²－10(y-x)+1=25(x-y)²+10(x-y)+1=[5(x-y)+1]²=（5x-5y+1）²

1题设y-z=a,z-x=b,x-y=c,则a+b+c=0,y+z-2x=b-cx+z-2y=c-ax+y-2z=a-b于是原式=（y+z-2x)^2+(z+x-2y)^2+(x+y-2z)^2=(b

25xy^2z^2(x+y-z)-30xyz(z-x-y)^2+5xyz^3(z-x-y)=25xy^2z^2(x+y-z)+30xyz(x+y-z)^2-5xyz^3(x+y-z)=5xyz(x+y

原式=(x-2y)^-z^=(x-2y-z)(x-2y+z)

看就不几个小时你的问题就over了,我一个初中生就班门弄斧一下吧.该式为轮换式,当x+y=-z时原式=0,故有因式(x+y+z),再用多项式除法易知另一项,所以原式=(x+y+z)(xy+yz+zx)