Question

已知{(x²+y²)-(x-y)²+2y(x-y)}÷4y=1,求4x²-y&#？

Answer 1

{(x²+y²)-(x-y)²+2y(x-y)}÷4y=1
{x²+y²-(x²-2xy+y²)+2xy-2y²}÷4y=1
{x²+y²-x²+2xy-y²+2xy-2y²}÷4y=1
{4xy-2y²}÷4y=1
2y{2x-y}÷4y=1
{2x-y}÷2=1
2x-y=2
4x/(4x²-y²)-1/(2x+y)
=4x/[(2x-y)(2x+y)]-(2x-y)/[(2x-y)(2x+y)]
=[4x-(2x-y)]/[(2x-y)(2x+y)]
=(4x-2x+y)/[(2x-y)(2x+y)]
=(2x+y)/[(2x-y)(2x+y)]
=1/(2x-y)
=1/2,5,已知{(x²+y²)-(x-y)²+2y(x-y)}÷4y=1,求4x²-y²分之4x -- 2x+y分之1 的值!