已知xy是方程组①x+2y=4②x－y=－5的解,求1－x－y/x+2y÷x²－y²/x²+4xy+4y的值

①x+2y=4,②x-y=-5,
——》x=-2,y=3,
1-[(x-y)/(x+2y)]÷[(x^2-y^2)/(x^2+4xy+4y^2)]
=1-[(x-y)/(x+2y)]*[(x+2y)^2/(x-y)(x+y)]
=1-(x+2y)/(x+y)
=-y/(x+y)
=-3/(-2+3)
=-3.