Question

如果2x-y=0,求代数式[(x2+y2)-(x-y)2+2y(x-y)]/4y的值

Answer 1

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

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

=(4xy-2y^2)/4y

=x-1/2y

=1/2(2x-y)

=1/2*0

=0