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
问题描述：
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
答
由圆面积得x^2π－y^2π＝100π
x^2－y^2＝100，即(x+y)(x-y)＝100
根据直角三角形，三边的关系得x=12.5 y=7.5
[8(x+y)^2(x-y)^3]÷[4(x+y)(x-y)^2]^2
=[8(x+y)^2(x-y)^2(x-y)]÷[4^2(x+y)^2(x-y)^2(x-y)^2]
=[8×100^2(x-y)]÷[16×100^2(x-y)^2]
=[8×100^2(x-y)]÷[8×100^2(x-y)]×2(x-y)
=1÷[2(x-y)]
=(x+y)÷2(x+y)(x-y)
=(x+y)/200
=20/200
=0.1
答
(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)]