题目

如果分别以x,y的半径画同心圆(x>y),所得圆环的面积是100π,那么代数式[8(x+y)^2(x-y)^3]÷[4(x+y)(x-y)^2]^2的值为

提问时间：2020-11-23

答案

(x²-y²)π=100π
(x²-y²)=100
[8(x+y)²(x-y)²(x-y)]÷[4(x+y)(x-y)²]²
=[8(x²-y²)²(x-y)]÷[4(x²-y²)(x-y)]²
=1/[2(x-y)]