2个回答
展开全部
|x-y-1|+(xy+2)²=0求(-2xy+2x+3y)-(3xy+2y-2x)-(x+4y+xy)
|x-y-1|+(xy+2)²=0
绝对值和平方的值恒为非负值,只有两个都为0的时候,和才为0
所以x-y-1=0,xy+2=0
=>x-y=1.xy=-2
(-2xy+2x+3y)-(3xy+2y-2x)-(x+4y+xy)
=-2xy+2x+3y-3xy-2y+2x-x-4y-xy
=-6xy+3x-3y
=6xy+3(x-y)
=6*(-2)+3*1
=-12+3
=-9
|x-y-1|+(xy+2)²=0
绝对值和平方的值恒为非负值,只有两个都为0的时候,和才为0
所以x-y-1=0,xy+2=0
=>x-y=1.xy=-2
(-2xy+2x+3y)-(3xy+2y-2x)-(x+4y+xy)
=-2xy+2x+3y-3xy-2y+2x-x-4y-xy
=-6xy+3x-3y
=6xy+3(x-y)
=6*(-2)+3*1
=-12+3
=-9
推荐律师服务:
若未解决您的问题,请您详细描述您的问题,通过百度律临进行免费专业咨询