原式=[(x^2-y^2)/2]/[x-(2xy-y^2)/2]是这个吧?
如果是这个的话,原式=[(x^2-y^2)/x]/[(x^2-2xy+y^2)/x]
=(x^2-y^2)/(x^2-2xy+y^2)
=[(x+y)(x-y)]/[(x-y)^2]
=(x+y)/(x-y)
将x=2010,y=2011带入原式,原式=-4021
原式=[(x^2-y^2)/2]/[x-(2xy-y^2)/2]是这个吧?
如果是这个的话,原式=[(x^2-y^2)/x]/[(x^2-2xy+y^2)/x]
=(x^2-y^2)/(x^2-2xy+y^2)
=[(x+y)(x-y)]/[(x-y)^2]
=(x+y)/(x-y)
将x=2010,y=2011带入原式,原式=-4021