已知2x-y=10,求代数式[(x的2次方+y的2次方)-(x-y)的2次方+2y(x-y)]÷4y的值.

已知2x-y=10,求代数式[(x^2+y^2)-(x-y)^2+2y(x-y)]/(4y)的值.

解 把y=2x-10 代入，得
[-(10 - x)^2 + x^2 + 2*(10 - x)*(-10 + 2*x) + (-10 + 2*x)^2] / (4y)
=[40x - 200] / (8x - 40) = 5

2x - y = 10

[(x^2 + y^2) - (x - y)^2 + 2y(x - y)]/(4y)
= [(x^2 + y^2) - (x^2 - 2xy + y^2) + 2y(x - y)]/(4y)
= [2xy + 2y(x - y)]/(4y)
= [2y(2x - y)]/(4y)
= (2x - y)/2
= 5本回答被提问者采纳

原式=(4xy-2y^2)/4y
=x-y/2
因为 2x-y=10
所以 x-y/2=5
所以 原式=5

分子展开消去,为4xy-2y^2
约去分母的2y,结果为10/2=5

[(x^2 + y^2) - (x - y)^2 + 2y(x - y)]/(4y)
= [(x^2 + y^2) - (x^2 - 2xy + y^2) + 2y(x - y)]/(4y)
= [2xy + 2y(x - y)]/(4y)
= [2y(2x - y)]/(4y)
= (2x - y)/2
= 5