已知(2x+1)*2+y*2+2y+1=0 求{(x*2+y*2)-(x-y)*2+2y(x-y)}/(2y)

题目

已知(2x+1)*2+y*2+2y+1=0 求{(x*2+y*2)-(x-y)*2+2y(x-y)}/(2y)

答案

(2x+1)^2+y^2+2y+1=0
(2x+1)^2+(y+1)^2=0
(2x+1)^2=0,(y+1)^2=0
x=-1/2,y=-1
{(x^2+y^2)-(x-y)^2+2y(x-y)}/(2y)
=[x^2+y^2-(x^2+y^2-2xy)+2xy-2y^2]/(2y)
=[x^2+y^2-x^2-y^2+2xy+2xy-2y^2]/(2y)
=[4xy-2y^2]/(2y)
=2y(2x-y)/(2y)
=2x-y
=2*(-1/2)-(-1)
=-1+1
=0