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=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)
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