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+y²-2xy)+2xy-2y²)]÷4y=1
(x²+y²-x²-y²+2xy+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)
=1/2,4,[(x²+y²)-(x-y)²+2y(x-y)]÷4y=1
(x²+y²-x²-y²+2xy+2xy-2y²)÷4y=1
(4xy-2y²﹚÷4y=1
x-½y=1
2x-y=2
2x=y+2
4x²-y²分之4x减2x+y分之1,0,已知[(x²+y²)-(x-y)²+2y(x-y)]÷4y=1,求4x²-y²分之4x减2x+y分之1的值
急!马上要,答的快且对的加分!