已知y-2x=10,求[2y(x-y)-(x-y)²+(x+y)²-2xy]÷4y的值
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解：先化简原式
[2y(x-y)-(x-y)²+(x+y)²-2xy]÷4y
=[2y(x-y)-2xy+(x+y)²-(x-y)²]÷4y
=[2y(x-y-x)+(x+y+x-y)(x+y-x+y)]÷4y
=(-2y²+4xy)÷4y
=[-2y(y-2x)]÷4y
=-(y-2x)÷2
把y-2x=10代入得，
原式=-10÷2=-5
[2y(x-y)-(x-y)²+(x+y)²-2xy]÷4y
=[2y(x-y)-2xy+(x+y)²-(x-y)²]÷4y
=[2y(x-y-x)+(x+y+x-y)(x+y-x+y)]÷4y
=(-2y²+4xy)÷4y
=[-2y(y-2x)]÷4y
=-(y-2x)÷2
把y-2x=10代入得，
原式=-10÷2=-5
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解：先化简原式
[2y(x-y)-(x-y)²+(x+y)²-2xy]÷4y
=[2y(x-y)-2xy+(x+y)²-(x-y)²]÷4y
=[2y(x-y-x)+(x²+y²+2xy)-(x²+y²-2xy)]÷4y
=[2y(-y)+(x²+y²+2xy-x²-y²+2xy)]÷4y
=(-2y²+4xy)÷4y
=4xy÷4y+(-2y²)÷4y
=x-y/2
=-(y-2x)/2
把y-2x=10代入得，
原式=-10÷2=-5
[2y(x-y)-(x-y)²+(x+y)²-2xy]÷4y
=[2y(x-y)-2xy+(x+y)²-(x-y)²]÷4y
=[2y(x-y-x)+(x²+y²+2xy)-(x²+y²-2xy)]÷4y
=[2y(-y)+(x²+y²+2xy-x²-y²+2xy)]÷4y
=(-2y²+4xy)÷4y
=4xy÷4y+(-2y²)÷4y
=x-y/2
=-(y-2x)/2
把y-2x=10代入得，
原式=-10÷2=-5
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