已知2x-y=10求[x2+y2-(x-y)2+2y(x-y)]÷4y

已知2x-y=10求[x2+y2-(x-y)2+2y(x-y)]÷4y
已知2x-y=10求[x2+y2-(x-y)2+2y(x-y)]÷4y

已知2x-y=10求[x2+y2-(x-y)2+2y(x-y)]÷4y
原式=[x^2+y^2-(x^2+y^2-2xy)+2xy-2y^2]÷4y
=(4xy-2y^2)÷4y
=2y(2x-y)÷4y
=(2x-y)÷2
代入2x-y=10,
原式=5