Question

计算(x-y)^2/(z-x)(z-y)+(y-z)^2/(x-y)(x-z)+(z-x)^2/(y-x)(y-z) 怎么做?

好加分

Answer 1

(x-y)^2/(z-x)(z-y)+(y-z)^2/(x-y)(x-z)+(z-x)^2/(y-x)(y-z)
=(x-y)^3/(x-y)(z-x)(z-y)+(y-z)^3/(x-y)(y-z)(x-z)+(z-x)^3/(y-x)(y-z)(z-x)
=3(y^2x-x^2y+z^2y-y^2z+x^2z-z^2x)/(y-x)(y-z)(z-x)
=3{(z^2y-z^2x)+(y^2x-x^2y)+(x^2z-y^2z)}/(y-x)(y-z)(z-x)
=3{z^2(y-x)+xy(y-x)-z(y^2-x^2)}/(y-x)(y-z)(z-x)
=3{z^2(y-x)+xy(y-x)-z(y-x)(y+x)}/(y-x)(y-z)(z-x)
=3{z^2+xy-z(y+x)}/(y-z)(z-x)
=3(z^2+xy-zy-zx)/(y-z)(z-x)
=3{(z^2-zx)+(xy-zy)}/(y-z)(z-x)
=3{z(z-x)-y(z-x)}/(y-z)(z-x)
=3(z-y)/(y-z)
=-3(y-z)/(y-z)
=-3

Answer 2

x-y=a
y-z=b
z-x=c
那么a+b+c = 0
原式 = -(a^2/bc+b^2/ac+c^2/ab) = -(a^3+b^3+c^3)/abc
由于a^3+b^3+c^3 = 3abc + (a+b+c)(a^2+b^2+c^2-ab-ac-bc) = 3abc
所以原式 = -3abc / abc = 3

Answer 3

(x-y)^2/(z-x)(z-y)+(y-z)^2/(x-y)(x-z)+(z-x)^2/(y-x)(y-z)

=[-(x-y)^3-(y-z)^3-(z-x)^3]/[(x-y)(y-z)(z-x)]
=[-x^3+3x^2y-3xy^2+y^3-y^3+3y^2z-3yz^2+z^3-z^3+3z^2x-3zx^2+x^3]/[(x-y)(y-z)(z-x)]
=[3x^2y-3xy^2+3y^2z-3yz^2+3z^2x-3zx^2]/[(x-y)(y-z)(z-x)]

=3[x^2y-xy^2+y^2z-yz^2+z^2x-zx^2]/[(x-y)(y-z)(z-x)]
=3[xy(x-y)+yz(y-z)+zx(z-x)]/[(x-y)(y-z)(z-x)]

只能这样了，去掉立方项