已知2x+y=4,求代数式[(x+y)²-(x-y)²-2y(x-1/2y)]÷4y的值 (还没学因式分解)
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已知2x+y=4,求代数式[(x+y)²-(x-y)²-2y(x-1/2y)]÷4y的值 (还没学因式分解)
已知2x+y=4,求代数式[(x+y)²-(x-y)²-2y(x-1/2y)]÷4y的值 (还没学因式分解)
已知2x+y=4,求代数式[(x+y)²-(x-y)²-2y(x-1/2y)]÷4y的值 (还没学因式分解)
[(x+y)²-(x-y)²-2y(x-1/2y)]÷4y=[(x²+2xy+y²)-(x²-2xy+y²)-2xy+y²]÷4y=(2xy+y²)÷4y=y(2x+y)÷4y
=4/4=1
[(x+y)²-(x-y)²-2y(x-1/2y)]÷4y
=[(x²+2xy+y²)-(x²-2xy+y²)-2xy+y²]÷4y
=(2xy+y²)÷4y
=y(2x+y)÷4y
=4/4
=1