{[xˆ(-2)+yˆ(-2)]/[xˆ(-2/3)+yˆ(

发布时间：2021-03-09 21:56:16

{[xˆ(-2)+yˆ(-2)]/[xˆ(-2/3)+yˆ(-2/3)]}-{[xˆ(-2)-yˆ(-2)]/xˆ(-2/3)-yˆ(-2/3)}

网友回答

原式=[x^﹙-2/3﹚+y^﹙-2/3﹚][(x^﹙﹣4/3﹚-x^﹙﹣2/3﹚y^﹙-2/3﹚+y^﹙-4/3)] /
(x^﹙﹣2/3﹚+y^﹙-2/3)-[x^﹙﹣2/3﹚-y^﹙-2/3)][(x^﹙﹣4/3﹚+x^﹙﹣2/3﹚y^﹙-2/3﹚+y^﹙-4/3) ]/[x^﹙﹣2/3﹚-y^﹙-2/3)]
=x^﹙﹣4/3﹚-x^﹙﹣2/3﹚y^﹙-2/3﹚+y^﹙-4/3﹚-x^﹙﹣4/3﹚-x^﹙﹣2/3﹚y^﹙-2/3﹚-y^﹙-4/3﹚
=﹣2x^﹙﹣2/3﹚y^﹙-2/3﹚
======以下答案可供参考======
供参考答案1:
我看不清呀供参考答案2:
分子是立方和和立方差
原式=(x^2/3+y^-2/3)(x^4/3-x^2/3y^-2/3+y^-4/3) /(x^2/3+y^-2/3)-(x^2/3-y^-2/3)(x^4/3+x^2/3y^-2/3+y^-4/3) /(x^2/3-y^-2/3)
=x^4/3-x^2/3y^-2/3+y^-4/3-x^4/3-x^2/3y^-2/3-y^-4/3
=-2x^2/3y^-2/3
以上问题属网友观点，不代表本站立场，仅供参考！