已知2x=y,求代数式【(x²+y²)-(x-y)²+2y(x-y)】÷4y的值.

已知2x=y,求代数式【(x²+y²)-(x-y)²+2y(x-y)】÷4y的值.

因为y=2x，
所以【(x²+y²)-(x-y)²+2y(x-y)】÷4y
=[5x^2-x^2+4x(x-2x)])÷8x
=0÷8x=0

【(x²+y²)-(x-y)²+2y(x-y)】÷4y
=【x²+y²-x²+2xy-y²+2xy-2y²】÷4y
=[4xy-2y²]÷4y
=x-y/2
因为2x=y,所以x=y/2
所以，原式的值等于0

答：
2x=y
[(x²+y²)-(x-y)²+2y(x-y)] ÷(4y)
=(x²+y²-x²+2xy-y²+2xy-2y²)÷(4y)
=(4xy-2y²)÷(4y)
=x-y/2
=y/2-y/2
=0