已知2x+y=4,求代数式[(x+y)²-(x-y)²-2y(x-1/2y)]÷4y的值

已知2x+y=4,求代数式[(x+y)²-(x-y)²-2y(x-1/2y)]÷4y的值
已知2x+y=4,求代数式[(x+y)²-(x-y)²-2y(x-1/2y)]÷4y的值

已知2x+y=4,求代数式[(x+y)²-(x-y)²-2y(x-1/2y)]÷4y的值
已知2x+y=4
y=4-2x
代数式[(x+y)²-(x-y)²-2y(x-1/2y)]÷4y
=[(x+y+x-y)(x+y-x+y)-2y(x-1/2y)]÷4y
=[(2x)(2y)-2y(x-1/2y)]÷4y
=2y*[2x-x+1/2y]÷4y
=[x+1/2y]÷2
=[x+1/2*(4-2x)]÷2
=[x+2-x]÷2
=2÷2
=1

[(x+y)²-(x-y)²-2y(x-1/2y)] ÷4y
=[x²+2xy+y²-(x²-2xy+y²)-(2xy-y²)] ÷4y
=(x²+2xy+y²-x²+2xy-y²-2xy+y²) ÷4y
=(2xy+y²) ÷4y
=y(2x+y) ÷4y
已知2x+y=4,
∴4y ÷4y=1