Question

计算:[x^2+y^2)-(x-y)^2+2y(x-y)÷4y,若,2x-y的算术平方根是4,求代数式值的立方根

Answer 1

:[x^2+y^2)-(x-y)^2+2y(x-y)]÷4y
=(x²+y²-x²+2xy-y²+2xy-2y²)÷4y
=(4xy-2y²)÷4y
=(2x-y)÷2
=4²÷2
=8
∴代数式值的立方根=³√8=2

追问

我很笨。。。- -~为什么要除2

追答

:[x^2+y^2)-(x-y)^2+2y(x-y)]÷4y
=(x²+y²-x²+2xy-y²+2xy-2y²)÷4y
=(4xy-2y²)÷4y
=2y(2x-y)÷4y
=2(2x-y)÷4
=(2x-y)÷2
=4²÷2
=8

这样清楚了吧？
呵呵，我很聪明。