Question

求代数式[(x+y^2-(x-y)^2+4xy]÷4xy的值，其中x=(√2)^0y=2

最好详细一点

Answer 1

[(x+y^2-(x-y)^2+4xy]÷4xy
=[x+(y-x+y)(y+x-y)+4xy]/(4xy)
=[x+x(2y-x)+4xy]/(4xy)
=(x+2xy-x^2+4xy)/(4xy)
=(x-x^2+6xy)/(4xy)……1式
因为
x=(√2)^0=1
y=2 代入1式，则
原式=3/2

Answer 2

[(x+y^2-(x-y)^2+4xy]/(4xy)
=[(x+y^2-x^2-y^2+2x+4xy]/(4xy)
=[(3x-x^2+4xy]/(4xy)
=(3-x)/(4y) +1
x=(√2)^0=1
y=2
代入 ，有 原式=3/2