已知 x+y+z=0x+y+z=0x+y+z=0,且 x2a−b+y2b−c+z2c−a=0\dfrac{x^2}{a-b}+\dfrac{y^2}{b-c}+\dfrac{z^2}{c-a}=0a−bx2+b−cy2+c−az2=0,求 cx2+ay2+bz2abx+bcy+caz\dfrac{cx^2+ay^2+bz^2}{abx+bcy+caz}abx+bcy+cazcx2+ay2+bz2 的值。
以下是参考答案:
解:∵x+y+z=0x+y+z=0x+y+z=0 ∴z=−(x+y)z=-(x+y)z=−(x+y) ∴ x2a−b+y2b−c+(x+y)2c−a=0\dfrac{x^2}{a-b}+\dfrac{y^2}{b-c}+\dfrac{(x+y)^2}{c-a}=0a−bx2+b−cy2+c−a(x+y)2=0 即 xa−b=yb−c\dfrac{x}{a-b}=\dfrac{y}{b-c}a−bx=b−cy (后略)
解:∵x+y+z=0x+y+z=0x+y+z=0
∴z=−(x+y)z=-(x+y)z=−(x+y)
∴ x2a−b+y2b−c+(x+y)2c−a=0\dfrac{x^2}{a-b}+\dfrac{y^2}{b-c}+\dfrac{(x+y)^2}{c-a}=0a−bx2+b−cy2+c−a(x+y)2=0
即 xa−b=yb−c\dfrac{x}{a-b}=\dfrac{y}{b-c}a−bx=b−cy
(后略)
所以这个“即”后面的结论是怎么得出来的?