[(x²+y²)-(x-y)²+2y(x-y)]÷4y=1,求4x/4x²-y²-1/2x+y 的值
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[(x²+y²)-(x-y)²+2y(x-y)]÷4y=1,求4x/4x²-y²-1/2x+y的值[(x²+y²)-(x-y)
[(x²+y²)-(x-y)²+2y(x-y)]÷4y=1,求4x/4x²-y²-1/2x+y 的值
[(x²+y²)-(x-y)²+2y(x-y)]÷4y=1,求4x/4x²-y²-1/2x+y 的值
[(x²+y²)-(x-y)²+2y(x-y)]÷4y=1,求4x/4x²-y²-1/2x+y 的值
解析:
由题意可知y≠0
已知[(x²+y²)-(x-y)²+2y(x-y)]÷4y=1,那么:
(x²+y²)-(x²-2xy+y²)+2xy-2y²=4y
即4xy-2y²=4y
两边同除以2y可得:2x-y=2
所以:4x/(4x²-y²)-1/(2x+y)
=[4x-(2x-y)]/(4x²-y²)
=(2x+y)/(4x²-y²)
=1/(2x-y)
=1/2