x`2(x-y)+y`2(y-z)+z`2(z-x)因式分解如题啊

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x`2(x-y)+y`2(y-z)+z`2(z-x)因式分解如题啊x`2(x-y)+y`2(y-z)+z`2(z-x)因式分解如题啊x`2(x-y)+y`2(y-z)+z`2(z-x)因式分解如题啊当

x`2(x-y)+y`2(y-z)+z`2(z-x)因式分解如题啊
x`2(x-y)+y`2(y-z)+z`2(z-x)因式分解
如题啊

x`2(x-y)+y`2(y-z)+z`2(z-x)因式分解如题啊
当x=y时啊,原式为y`2(y-z)+y`2(z-y)+z`2(y-x) =0 根据因式定理得其中有一个因式啊就是x-y,然后啊,因为这是一个轮换对等式,而且是个3次多项式,所以还有因式（y-z）和（z-x）然后就设分解因式后他的真面目为k(x-y)(y-z)(z-x) 这是传说中的待定系数法于是令x=0,y=z,z=-1 （只要移项后左右不为0就行了,可以设任何数）带入原式,得到k=-1的于是原式分解因式为-（x-y）（y-z）（z-x）

化简(y-x)(z-x)/(x-2y+z)(x+y-2z)+（z-y)(x-y)/(x-2z+y)(y+z-2x)+(x-z)(y-z)/(y+z-2x)(x-2y+z) 计算:(y-x)(z-x)/(x-2y+z)(x+y-2z)+(z-y)(x-y)/(x+y-2z)(y+z-2x)+(x-z)(y-z)/(y+z-2x)(x-2y+z) 计算:x^2/(x-y)(x-z)+y^2/(y-x)(y-z)+z^2/(z-x)(z-y) 化简(x+Y)^2(y+z-x)(z+x-y)+(x-y)^2(x+y+z)(x+y-z) 计算题(x+y)^2(y+z-x)(z+x-y)+(x-y)^2(x+y+z)(x+y-z). 1、y-x/x²-y²2、(x-y)(y-z)(z-x)/(z-y)(y-x)(x-z) 化简(y-x)(z-x)/(x-2y+z)(x+y-2z)+（z-y)(x-y)/(xy-2z)(y+z-2x)+(x-z)(y-z)/(y+z-2x)(x-2y+z)速速回答用行列式的性质证明：y+z z+x x+y x y z x+y y+z z+x =2 z x y z+x x+y y+z y z x 这个怎么证? 化简:(2x-y-z)/(x-y)(x-z)+(2y-z-x)/(y-z)(y-x)+(2z-x-y)/(z-x)(z-y)化简:(2x-y-z)/(x-y)(x-z)+(2y-z-x)/(y-z)(y-x)+(2z-x-y)/(z-x)(z-y)我知道把分母化成一样的、但是之后怎么算? x,y,z正整数 x>y>z证明 x^2x +y^2y+z^2z>x^(y+z)*y^(x+z)*z^(x+y)x,y,z正整数 x>y>z证明 x^2x * y^2y * z^2z>x^(y+z)*y^(x+z)*z^(x+y)不是+是 * (x-2y+z)(x+y-2z)分之（y-x)(z-x) + (x+y-2z)(y+z-2x)分之(z-y)(x-y) + (y+z-2z)(x-2y+z)分之（x-z)(y-z)=?第三部分那个是（y+z-2x)(x-2y+z)分之（x-z)(y-z) 试证明(x+y-2z)+(y+z-2x)+(z+x-2y)=3(x+y-2z)(y+z-2x)(z+x-2y) 化简(y-z)^2/(x-y)(x-z)+(z-x)^2/(y-x)(y-z)+(x-y)^2/(z-x)(z-y) 已知：x^2/(z+y)+y^2/(x+z)+z^2/(x+y)=0,求x/(z+y)+y/(x+z)+z/(x+y)的值. x^2/(z+y)+y^2/(x+z)+z^2/(x+y)=0,求x/(z+y)+y/(x+z)+z/(x+y)的值已知(x+y)(x+z)=x,(y+z)(y+x)=2y,(z+x)(z+y)=3z,求x,y,z 2(x+y)(x+y-z)+3(y-z)(x+y-z) x+y+z在x^2+y^2+z^2