Question

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

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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

Answer 2

[(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
=4x/[(2x+y)(2x-y)-1/(2x+y)
=4x/2(2x+y)-1/(2x+y)
=(2x-1)/(2x+y)
=(y+2-1)/(y+2+y)
=(y+1)/2(y+1)
=1/2