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如果分别以x y为半径画同心圆,(x>y)所得圆环的面积为14派(就是那个圆周率),求代数式[-6(x+y)的平方(x-y)的三次方]的平方/[3(x+y)(x-y)的平方]的平方的值
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人气:412 ℃ 时间:2020-04-29 23:14:18
解答
^是几次方的意思,如5^3=125
圆环面积=πx^2-πy^2=π(x+y)(x-y)=14π
所以(x+y)(x-y)=14
原式=[-6(x+y)^2*(x-y)^3]^2/[3(x+y)(x-y)^2]^2
=36(x+y)^4*(x-y)^6/[9(x+y)^2*(x-y)^4]
=4(x+y)^2*(x-y)^2
=4[(x+y)(x-y)]^2
=196*4
=784
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